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Example 1.4.5. Suppose\n\n\[ f\left( x\right) = \left\{ \begin{array}{ll} {3x} + 5, & \text{ if }x \leq 1 \\ {10} - {2x}, & \text{ if }x > 1 \end{array}\right. \]\n\nIf \( \epsilon \) is a positive infinitesimal, then
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\[ f\left( {1 + \epsilon }\right) = 3\left( {1 + \epsilon }\right) + 5 = 8 + {3\epsilon } \simeq 8 = f\left( 1\right) ,\]\n\nso \( f \) is continuous from the right at \( x = 1 \), and\n\n\[ f\left( {1 - \epsilon }\right) = 3\left( {1 - \epsilon }\right) + 5 = 8 - {3\epsilon } \simeq 8 = f\left( 1\right) ,\]\n\nso \( f \) is continuous from the left at \( x = 1 \) as well. Hence \( f \) is continuous at \( x = 1 \) .
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Yes
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Suppose that both \( f \) and \( g \) are continuous at the real number \( c \) and we let \( s\left( x\right) = f\left( x\right) + g\left( x\right) \) . If \( \epsilon \) is any infinitesimal, then
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\[ s\left( {c + \epsilon }\right) = f\left( {c + \epsilon }\right) + g\left( {c + \epsilon }\right) \simeq f\left( c\right) + g\left( c\right) = s\left( c\right) ,\] and so \( s \) is also continuous at \( c \) .
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Yes
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Since\n\n\[ \n{\left( x + \epsilon \right) }^{3} = {x}^{3} + 3{x}^{2}\epsilon + {3x}{\epsilon }^{2} + {\epsilon }^{3} \simeq {x}^{3} \n\]\n\nfor any real number \( x \) and any infinitesimal \( \epsilon \), it follows that \( g\left( x\right) = {x}^{3} \) is continuous on \( \left( {-\infty ,\infty }\right) \) .
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From the previous theorems, it then follows that\n\n\[ \nh\left( x\right) = 5{x}^{2} + 3{x}^{3} \n\]\n\nis continuous on \( \left( {-\infty ,\infty }\right) \) .
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No
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The function \( f\left( x\right) = {x}^{2} \) attains neither a maximum nor a minimum value on the interval \( \left( {0,1}\right) \) .
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Indeed, given any point \( a \) in \( \left( {0,1}\right), f\left( x\right) > f\left( a\right) \) whenever \( a < x < 1 \) and \( f\left( x\right) < f\left( a\right) \) whenever \( 0 < x < a \) .
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Yes
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Example 1.5.9. Let\n\n\[ f\\left( x\\right) = \\left\\{ \\begin{array}{ll} \\frac{1}{x}, & \\text{ if } - 1 \\leq x < 0\\text{ or }0 < x \\leq 1, \\\\ 0, & \\text{ if }x = 0. \\end{array}\\right.\n\]\n\nSee Figure 1.5.5. Then \( f \) does not have a maximum value: if \( a \\leq 0 \), then \( f\\left( x\\right) > f\\left( a\\right) \) for any \( x > 0 \), and if \( a > 0 \), then \( f\\left( x\\right) > f\\left( a\\right) \) whenever \( 0 < x < a \) . Similarly, \( f \) has no minimum value: if \( a \\geq 0 \), then \( f\\left( x\\right) < f\\left( a\\right) \) for any \( x < 0 \) , and if \( a < 0 \), then \( f\\left( x\\right) < f\\left( a\\right) \) whenever \( a < x < 0 \) .
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The problem this time is that \( f \) is not continuous at \( x = 0 \). Indeed, if \( \\epsilon \) is an infinitesimal, then \( f\\left( \\epsilon \\right) \) is infinite, and, hence, not infinitesimally close to \( f\\left( 0\\right) = 0 \) .
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Yes
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If \( y = {x}^{2} \), then, for any nonzero infinitesimal \( {dx} \)
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\[ {dy} = {\left( x + dx\right) }^{2} - {x}^{2} = \left( {{x}^{2} + {2xdx} + {\left( dx\right) }^{2}}\right) - {x}^{2} = \left( {{2x} + {dx}}\right) {dx}. \] Hence \[ \frac{dy}{dx} = {2x} + {dx} \simeq {2x} \] and so the derivative of \( y \) with respect to \( x \) is \[ \frac{dy}{dx} = {2x} \]
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Yes
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If \( f\left( x\right) = {4x} \), then, for any nonzero infinitesimal \( {dx} \)
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\n\[
\frac{f\left( {x + {dx}}\right) - f\left( x\right) }{dx} = \frac{4\left( {x + {dx}}\right) - {4x}}{dx} = \frac{4dx}{dx} = 4.
\]
\nHence \( {f}^{\prime }\left( x\right) = 4 \) . Note that this implies that \( f\left( x\right) \) has a constant rate of change: every change of one unit in \( x \) results in a change of 4 units in \( f\left( x\right) \) .
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Yes
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If \( y = {x}^{2} \), then we saw above that \( \frac{dy}{dx} = {2x} \) . Hence the rate of change of \( y \) with respect to \( x \) when \( x = 3 \) is
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\[ {\left. \frac{dy}{dx}\right| }_{x = 3} = \left( 2\right) \left( 3\right) = 6 \]
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Yes
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If \( f\left( x\right) = {x}^{\frac{2}{3}} \), then for any infinitesimal \( {dx} \) ,
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\[ f\left( {0 + {dx}}\right) - f\left( 0\right) = f\left( {dx}\right) = {\left( dx\right) }^{\frac{2}{3}}, \] which is infinitesimal. Hence \( f \) is continuous at \( x = 0 \). Now if \( {dx} \neq 0 \), then \[ \frac{f\left( {0 + {dx}}\right) - f\left( 0\right) }{dx} = \frac{{\left( dx\right) }^{\frac{2}{3}}}{dx} = \frac{1}{{\left( dx\right) }^{\frac{1}{3}}}. \] Since this is infinite, \( f \) does not have a derivative at \( x = 0 \). In particular, this shows that a function may be continuous at a point, but not differentiable at that point.
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Yes
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If \( f\left( x\right) = \sqrt{x} \), then, as we have seen above, for any \( x > 0 \) and any nonzero infinitesimal \( {dx} \),
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\[ f\left( {x + {dx}}\right) - f\left( x\right) = \sqrt{x + {dx}} - \sqrt{x} \] \[ = \left( {\sqrt{x + {dx}} - \sqrt{x}}\right) \frac{\sqrt{x + {dx}} + \sqrt{x}}{\sqrt{x + {dx}} + \sqrt{x}} \] \[ = \frac{\left( {x + {dx}}\right) - x}{\sqrt{x + {dx}} + \sqrt{x}} \] \[ = \frac{dx}{\sqrt{x + {dx}} + \sqrt{x}}. \] It now follows that \[ \frac{f\left( {x + {dx}}\right) - f\left( x\right) }{dx} = \frac{1}{\sqrt{x + {dx}} + \sqrt{x}} \simeq \frac{1}{\sqrt{x} + \sqrt{x}} = \frac{1}{2\sqrt{x}}. \] Thus \[ {f}^{\prime }\left( x\right) = \frac{1}{2\sqrt{x}}. \]
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Yes
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The function \( f\left( x\right) = \sqrt{x} \) is differentiable on \( \left( {0,\infty }\right) \) .
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Note that \( f \) is not differentiable at \( x = 0 \) since \( f\left( {0 + {dx}}\right) = f\left( {dx}\right) \) is not defined for all infinitesimals \( {dx} \).
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Yes
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Theorem 1.7.2. If \( f \) and \( g \) are both differentiable and \( s\left( x\right) = f\left( x\right) + g\left( x\right) \) , then
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\[ {s}^{\prime }\left( x\right) = {f}^{\prime }\left( x\right) + {g}^{\prime }\left( x\right) \]
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Yes
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If \( y = {x}^{2} + \sqrt{x} \), then, using our results from the previous section,
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\[ \frac{dy}{dx} = \frac{d}{dx}\left( {x}^{2}\right) + \frac{d}{dx}\left( \sqrt{x}\right) = {2x} + \frac{1}{2\sqrt{x}}. \]
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Yes
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Theorem 1.7.3. If \( c \) is a real constant, \( f \) is differentiable, and \( g\left( x\right) = {cf}\left( x\right) \) , then
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\[ {g}^{\prime }\left( x\right) = c{f}^{\prime }\left( x\right) \]
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Yes
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Example 1.7.3. If \( y = 5{x}^{2} \), then
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\n\[
\frac{dy}{dx} = 5\frac{d}{dx}\left( {x}^{2}\right) = 5\left( {2x}\right) = {10x}.
\]
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Yes
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Theorem 1.7.4. If \( f \) and \( g \) are both differentiable and \( p\left( x\right) = f\left( x\right) g\left( x\right) \), then
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\[ {p}^{\prime }\left( x\right) = f\left( x\right) {g}^{\prime }\left( x\right) + g\left( x\right) {f}^{\prime }\left( x\right) . \]
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Yes
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We may use the product rule to find a formula for the derivative of a positive integer power of \( x \) .
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We first note that if \( y = x \), then, for any infinitesimal \( {dx} \), \[ {dy} = \left( {x + {dx}}\right) - x = {dx} \] and so, if \( {dx} \neq 0 \), \[ \frac{dy}{dx} = \frac{dx}{dx} = 1 \] Thus we have \[ \frac{d}{dx}x = 1 \] (1.7.18) as we should expect, since \( y = x \) implies that \( y \) changes at exactly the same rate as \( x \). Using the product rule, it now follows that \[ \frac{d}{dx}{x}^{2} = \frac{d}{dx}\left( {x \cdot x}\right) = x\frac{d}{dx}x + x\frac{d}{dx}x = x + x = {2x}, \] (1.7.19) in agreement with a previous example. Next, we have \[ \frac{d}{dx}{x}^{3} = x\frac{d}{dx}{x}^{2} + {x}^{2}\frac{d}{dx}x = 2{x}^{2} + {x}^{2} = 3{x}^{2} \] (1.7.20) and \[ \frac{d}{dx}{x}^{4} = x\frac{d}{dx}{x}^{3} + {x}^{3}\frac{d}{dx}x = 3{x}^{3} + {x}^{3} = 4{x}^{3}. \] (1.7.21) At this point we might suspect that for any integer \( n \geq 1 \), \[ \frac{d}{dx}{x}^{n} = n{x}^{n - 1} \] (1.7.22) This is in fact true, and follows easily from an inductive argument: Suppose we have shown that for any \( k < n \), \[ \frac{d}{dx}{x}^{k} = k{x}^{k - 1}. \] (1.7.23) Then \[ \frac{d}{dx}{x}^{n} = x\frac{d}{dx}{x}^{n - 1} + {x}^{n - 1}\frac{d}{dx}x \] \[ = x\left( {\left( {n - 1}\right) {x}^{n - 2}}\right) + {x}^{n - 1} \] \[ = n{x}^{n - 1}\text{.} \] (1.7.24) We call this result the power rule.
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Yes
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When \( n = {34} \), the power rule shows that
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\[ \frac{d}{dx}{x}^{34} = {34}{x}^{33} \]
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Yes
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Example 1.7.6. If \( f\left( x\right) = {14}{x}^{5} \), then, combining the power rule with our result for constant multiples,
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\[ {f}^{\prime }\left( x\right) = {14}\left( {5{x}^{4}}\right) = {70}{x}^{4}. \]
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Yes
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If \( n < 0 \) is an integer, then
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\( \frac{d}{dx}{x}^{n} = \frac{d}{dx}\left( \frac{1}{{x}^{-n}}\right) = - \frac{1}{{x}^{-{2n}}} \cdot \left( {-n{x}^{-n - 1}}\right) = n{x}^{n - 1}. \)
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Yes
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If \( f\left( x\right) = 3{x}^{2} - \frac{5}{{x}^{7}} \) then find \( f'\left( x\right) \).
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then \( f\left( x\right) = 3{x}^{2} - 5{x}^{-7} \), and so \[ {f}^{\prime }\left( x\right) = {6x} + {35}{x}^{-8} = {6x} + \frac{35}{{x}^{8}}. \]
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Yes
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Theorem 1.7.8. If \( f \) and \( g \) are differentiable, \( g\left( x\right) \neq 0 \), and\n\n\[ q\left( x\right) = \frac{f\left( x\right) }{g\left( x\right) } \]\n\n(1.7.36)\n\nthen\n\n\[ {q}^{\prime }\left( x\right) = \frac{g\left( x\right) {f}^{\prime }\left( x\right) - f\left( x\right) {g}^{\prime }\left( x\right) }{{\left( g\left( x\right) \right) }^{2}}. \]\n\n\( \left( {1.7.37}\right) \)
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\[ {q}^{\prime }\left( x\right) = \frac{g\left( x\right) {f}^{\prime }\left( x\right) - f\left( x\right) {g}^{\prime }\left( x\right) }{{\left( g\left( x\right) \right) }^{2}}. \]
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Yes
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If\n\n\[ f\left( x\right) = \frac{3{x}^{2} - {6x} + 4}{{x}^{2} + 1} \]\n\nthen\n\n\[ {f}^{\prime }\left( x\right) = \frac{\left( {{x}^{2} + 1}\right) \left( {{6x} - 6}\right) - \left( {3{x}^{2} - {6x} + 4}\right) \left( {2x}\right) }{{\left( {x}^{2} + 1\right) }^{2}} \]
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\n\n\[ = \frac{6{x}^{3} - 6{x}^{2} + {6x} - 6 - 6{x}^{3} + {12}{x}^{2} - {8x}}{{\left( {x}^{2} + 1\right) }^{2}} \]\n\n\[ = \frac{6{x}^{2} - {2x} - 6}{{\left( {x}^{2} + 1\right) }^{2}}. \]
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Yes
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Example 1.7.12. We may use either 1.7.30 or 1.7.37 to differentiate\n\n\[ y = \frac{5}{{x}^{2} + 1}. \]
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In either case, we obtain\n\n\[ \frac{dy}{dx} = - \frac{5}{{\left( {x}^{2} + 1\right) }^{2}}\frac{d}{dx}\left( {{x}^{2} + 1}\right) = - \frac{10x}{{\left( {x}^{2} + 1\right) }^{2}}. \]
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Yes
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Theorem 1.7.9. If \( y \) is a differentiable function of \( u \) and \( u \) is a differentiable function of \( x \), then\n\n\[ \frac{dy}{dx} = \frac{dy}{du}\frac{du}{dx} \]
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Not that if we let \( y = f\left( u\right), u = g\left( x\right) \), and\n\n\[ h\left( x\right) = f \circ g\left( x\right) = f\left( {g\left( x\right) }\right) ,\]\n\nthen\n\n\[ \frac{dy}{dx} = {h}^{\prime }\left( x\right) ,\frac{dy}{du} = {f}^{\prime }\left( {g\left( x\right) }\right) ,\text{ and }\frac{du}{dx} = {g}^{\prime }\left( x\right) .\]\n\nHence we may also express the chain rule in the form\n\n\[ {h}^{\prime }\left( x\right) = {f}^{\prime }\left( {g\left( x\right) }\right) {g}^{\prime }\left( x\right) .\]
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Yes
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If \( y = 3{u}^{2} \) and \( u = {2x} + 1 \), then
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\n\[
\frac{dy}{dx} = \frac{dy}{du}\frac{du}{dx} = \left( {6u}\right) \left( 2\right) = {12u} = {24x} + {12}.
\]\n\nWe may verify this result by first finding \( y \) directly in terms of \( x \), namely,\n\n\[
y = 3{u}^{2} = 3{\left( 2x + 1\right) }^{2} = 3\left( {4{x}^{2} + {4x} + 1}\right) = {12}{x}^{2} + {12x} + 3,
\]\n\nand then differentiating directly:\n\n\[
\frac{dy}{dx} = \frac{d}{dx}\left( {{12}{x}^{2} + {12x} + 3}\right) = {24x} + {12}.
\]
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Yes
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If \( h\left( x\right) = \sqrt{{x}^{2} + 1} \), then \( h\left( x\right) = f\left( {g\left( x\right) }\right) \) where \( f\left( x\right) = \sqrt{x} \) and \( g\left( x\right) = {x}^{2} + 1 \).
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Since\n\n\[ \n{f}^{\prime }\left( x\right) = \frac{1}{2\sqrt{x}}\text{ and }{g}^{\prime }\left( x\right) = {2x}, \n\]\n\nit follows that\n\n\[ \n{h}^{\prime }\left( x\right) = {f}^{\prime }\left( {g\left( x\right) }\right) {g}^{\prime }\left( x\right) = \frac{1}{2\sqrt{{x}^{2} + 1}} \cdot {2x} = \frac{x}{\sqrt{{x}^{2} + 1}}. \n\]
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Yes
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Example 1.7.16. With \( n = {10} \) and \( g\left( x\right) = {x}^{2} + 3 \), we have
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\[ \frac{d}{dx}{\left( {x}^{2} + 3\right) }^{10} = {10}{\left( {x}^{2} + 3\right) }^{9}\left( {2x}\right) = {20x}{\left( {x}^{2} + 3\right) }^{9}. \]
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Yes
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If \( f\left( x\right) = \frac{15}{{\left( {x}^{4} + 5\right) }^{2}} \), then what is \( f'(x) \)?
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then we may apply the previous result with \( n = - 2 \) and \( g\left( x\right) = {x}^{4} + 5 \) to obtain\n\n\[ {f}^{\prime }\left( x\right) = - {30}{\left( {x}^{4} + 5\right) }^{-3}\left( {4{x}^{3}}\right) = - \frac{{120}{x}^{3}}{{\left( {x}^{4} + 5\right) }^{3}}. \]
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Yes
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Example 1.7.18. With \( r = \frac{1}{2} \) in the previous theorem, we have
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\[ \frac{d}{dx}\sqrt{x} = \frac{1}{2}{x}^{-\frac{1}{2}} = \frac{1}{2\sqrt{x}} \] in agreement with our earlier direct computation.
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Yes
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If \( y = {x}^{\frac{2}{3}} \), then
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\n\[
\frac{dy}{dx} = \frac{2}{3}{x}^{-\frac{1}{3}} = \frac{2}{3{x}^{\frac{1}{3}}}.
\]\n\nNote that \( \frac{dy}{dx} \) is not defined at \( x = 0 \), in agreement with our earlier result showing that \( y \) is not differentiable at 0 .
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Yes
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If \( f\left( x\right) = \sqrt{{x}^{2} + 1} \)
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\[ {f}^{\prime }\left( x\right) = \frac{1}{2}{\left( {x}^{2} + 1\right) }^{-\frac{1}{2}}\left( {2x}\right) = \frac{x}{\sqrt{{x}^{2} + 1}}. \]
|
Yes
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Example 1.7.21. If\n\n\\[ \n g\\left( t\\right) = \\frac{1}{{t}^{4} + 5} \n\\]\n\nthen
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\\[ \n {g}^{\\prime }\\left( t\\right) = \\left( {-1}\\right) {\\left( {t}^{4} + 5\\right) }^{-2}\\left( {4{t}^{3}}\\right) = - \\frac{4{t}^{3}}{{\\left( {t}^{4} + 5\\right) }^{2}}. \n\\]
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Yes
|
Using the chain rule,
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\[ \frac{d}{dx}\cos \left( {4x}\right) = - \sin \left( {4x}\right) \frac{d}{dt}\left( {4x}\right) = - 4\sin \left( {4x}\right) . \]
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Yes
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If \( f\left( t\right) = {\sin }^{2}\left( t\right) \), then, again using the chain rule,
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\[ {f}^{\prime }\left( t\right) = 2\sin \left( t\right) \frac{d}{dt}\sin \left( t\right) = 2\sin \left( t\right) \cos \left( t\right) . \]
|
Yes
|
If \( g\left( x\right) = \cos \left( {x}^{2}\right) \), then
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\[ {g}^{\prime }\left( x\right) = - \sin \left( {x}^{2}\right) \left( {2x}\right) = - {2x}\cos \left( {x}^{2}\right) . \]
|
Yes
|
Example 1.7.25. If \( f\left( x\right) = {\sin }^{3}\left( {4x}\right) \), then, using the chain rule twice,
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\[ {f}^{\prime }\left( x\right) = 3{\sin }^{2}\left( {4x}\right) \frac{d}{dx}\sin \left( {4x}\right) = {12}{\sin }^{2}\left( {4x}\right) \cos \left( {4x}\right) . \]
|
Yes
|
If \( f\left( x\right) = {x}^{5} - 6{x}^{2} + 5 \), then
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\[ {f}^{\prime }\left( x\right) = 5{x}^{4} - {12x} \]\n\nIn particular, \( {f}^{\prime }\left( {-\frac{1}{2}}\right) = \frac{101}{16} \), and so the equation of the line tangent to the graph of \( f \) at \( x = - \frac{1}{2} \) is\n\n\[ y = \frac{101}{6}\left( {x + \frac{1}{2}}\right) + \frac{111}{32}. \]
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Yes
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Theorem 1.9.1. If \( f \) is differentiable on \( \left( {a, b}\right) \) and attains a maximum, or a minimum, value at \( c \), then \( {f}^{\prime }\left( c\right) = 0 \) .
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Now suppose \( f \) is continuous on \( \left\lbrack {a, b}\right\rbrack \), differentiable on \( \left( {a, b}\right) \), and \( f\left( a\right) = \) \( f\left( b\right) \) . If \( f \) is a constant function, then \( {f}^{\prime }\left( c\right) = 0 \) for all \( c \) in \( \left( {a, b}\right) \) . If \( f \) is not constant, then there is a point \( c \) in \( \left( {a, b}\right) \) at which \( f \) attains either a maximum or a minimum value, and so \( {f}^{\prime }\left( c\right) = 0 \) .
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No
|
Theorem 1.9.2. If \( f \) is continuous on \( \left\lbrack {a, b}\right\rbrack \), differentiable on \( \left( {a, b}\right) \), and \( f\left( a\right) = \) \( f\left( b\right) \), then there is a real number \( c \) in \( \left( {a, b}\right) \) for which \( {f}^{\prime }\left( c\right) = 0 \) .
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More generally, suppose \( f \) is continuous on \( \left\lbrack {a, b}\right\rbrack \) and differentiable on \( \left( {a, b}\right) \). Let \[ g\left( x\right) = f\left( x\right) - \frac{f\left( b\right) - f\left( a\right) }{b - a}\left( {x - a}\right) - f\left( a\right) . \] Note that \( g\left( x\right) \) is the difference between \( f\left( x\right) \) and the corresponding \( y \) value on the line passing through \( \left( {a, f\left( a\right) }\right) \) and \( \left( {b, f\left( b\right) }\right) \) . Moroever, \( g \) is continuous on \( \left\lbrack {a, b}\right\rbrack \), differentiable on \( \left( {a, b}\right) \), and \( g\left( a\right) = 0 = g\left( b\right) \) . Hence Rolle’s theorem applies to \( g \), so there must exist a point \( c \) in \( \left( {a, b}\right) \) for which \( {g}^{\prime }\left( c\right) = 0 \) . Now \[ {g}^{\prime }\left( c\right) = {f}^{\prime }\left( x\right) - \frac{f\left( b\right) - f\left( a\right) }{b - a}, \] so we must have \[ 0 = {g}^{\prime }\left( c\right) = {f}^{\prime }\left( c\right) - \frac{f\left( b\right) - f\left( a\right) }{b - a}. \] That is, \[ {f}^{\prime }\left( c\right) = \frac{f\left( b\right) - f\left( a\right) }{b - a}, \] which is our desired connection between instantaneous and average rates of change, known as the mean-value theorem.
|
Yes
|
Consider the function \( f\left( x\right) = {x}^{3} - {3x} + 1 \) on the interval \( \left\lbrack {0,2}\right\rbrack \) . By the mean-value theorem, there must exist at least one point \( c \) in \( \left\lbrack {0,2}\right\rbrack \) for which \[ {f}^{\prime }\left( c\right) = \frac{f\left( 2\right) - f\left( 0\right) }{2 - 0} = \frac{3 - 1}{2} = 1. \]
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Now \( {f}^{\prime }\left( x\right) = 3{x}^{2} - 3 \), so \( {f}^{\prime }\left( c\right) = 1 \) implies \( 3{c}^{2} - 3 = 1 \) . Hence \( c = \sqrt{\frac{4}{3}} \) . Note that this implies that the tangent line to the graph of \( f \) at \( x = \sqrt{\frac{4}{3}} \) is parallel to the line through the endpoints of the graph of \( f \), that is, the points \( \left( {0,1}\right) \) and \( \left( {2,3}\right) \) . See Figure 1.9.1.
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Yes
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Let \( f\left( x\right) = 2{x}^{3} - 3{x}^{2} - {12x} + 1 \) . Then
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\[ {f}^{\prime }\left( x\right) = 6{x}^{2} - {6x} - {12} = 6\left( {{x}^{2} - x - 2}\right) = 6\left( {x - 2}\right) \left( {x + 1}\right) . \] Hence \( {f}^{\prime }\left( x\right) = 0 \) when \( x = - 1 \) and when \( x = 2 \) . Now \( x - 2 < 0 \) for \( x < 2 \) and \( x - 2 > 0 \) for \( x > 2 \), while \( x + 1 < 0 \) for \( x < - 1 \) and \( x + 1 > 0 \) when \( x > - 1 \) . Thus \( {f}^{\prime }\left( x\right) > 0 \) when \( x < - 1 \) and when \( x > 2 \), and \( {f}^{\prime }\left( x\right) < 0 \) when \( - 1 < x < 2 \) . It follows that \( f \) is increasing on the intervals \( \left( {-\infty , - 1}\right) \) and \( \left( {2,\infty }\right) \), and decreasing on the interval \( \left( {-1,2}\right) \) .
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Yes
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Let \( f\left( x\right) = x + 2\sin \left( x\right) \). Then \( {f}^{\prime }\left( x\right) = 1 + 2\cos \left( x\right) \), and so \( {f}^{\prime }\left( x\right) < 0 \) when, and only when,
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\[ \cos \left( x\right) < - \frac{1}{2} \] For \( 0 \leq x \leq {2\pi } \), this occurs when, and only when, \[ \frac{2\pi }{3} < x < \frac{4\pi }{3} \] Since the cosine function has period \( {2\pi } \), if follows that \( {f}^{\prime }\left( x\right) < 0 \) when, and only when, \( x \) is in an interval of the form \[ \left( {\frac{2\pi }{3} + {2\pi n},\frac{4\pi }{3} + {2\pi n}}\right) \] for \( n = 0, \pm 1, \pm 2,\ldots \) Hence \( f \) is decreasing on these intervals and increasing on intervals of the form \[ \left( {-\frac{2\pi }{3} + {2\pi n},\frac{2\pi }{3} + {2\pi n}}\right) \]
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Yes
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Theorem 1.10.1. If \( f \) is a continuous function on a closed and bounded interval \( \left\lbrack {a, b}\right\rbrack \), then the maximum and minimum values of \( f \) occur at either (1) stationary points in the open interval \( \left( {a, b}\right) ,\left( 2\right) \) singular points in the open interval \( \left( {a, b}\right) \) , or (3) the endpoints of \( \left\lbrack {a, b}\right\rbrack \) .
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Hence we have the following procedure for optimizing a continuous function \( f \) on an interval \( \left\lbrack {a, b}\right\rbrack \) :\n\n(1) Find all stationary and singular points of \( f \) in the open interval \( \left( {a, b}\right) \) .\n\n(2) Evaluate \( f \) at all stationary and singular points of \( \left( {a, b}\right) \), and at the endpoints \( a \) and \( b \) .\n\n(3) The maximum value of \( f \) is the largest value found in step (2) and the minimum value of \( f \) is the smallest value found in step (2).
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Yes
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Consider the function \( g\left( t\right) = t - 2\cos \left( t\right) \) defined on the interval \( \left\lbrack {0,{2\pi }}\right\rbrack \) .
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\[ {g}^{\prime }\left( t\right) = 1 + 2\sin \left( t\right) \] and so \( {g}^{\prime }\left( t\right) = 0 \) when \[ \sin \left( t\right) = - \frac{1}{2} \] For \( t \) in the open interval \( \left( {0,{2\pi }}\right) \), this means that either \[ t = \frac{7\pi }{6} \] or \[ t = \frac{11\pi }{6} \] That is, the stationary points of \( g \) in \( \left( {0,{2\pi }}\right) \) are \( \frac{7\pi }{6} \) and \( \frac{11\pi }{6} \) . Note that \( g \) is differentiable at all points in \( \left( {0,{2\pi }}\right) \), and so there are no singular points of \( g \) in \( \left( {0,{2\pi }}\right) \) . Hence to identify the extreme values of \( g \) we need evaluate only \[ g\left( 0\right) = - 2 \] \[ g\left( \frac{7\pi }{6}\right) = \frac{7\pi }{6} + \sqrt{3} \approx {5.39724} \] \[ g\left( \frac{11\pi }{6}\right) = \frac{11\pi }{6} - \sqrt{3} \approx {4.02753} \] and \[ g\left( {2\pi }\right) = {2\pi } - 2 \approx {4.28319}. \] Thus \( g \) has a maximum value of \( {5.39724} \) at \( t = \frac{7\pi }{6} \) and a minimum value of -2 at \( t = 0 \) .
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Yes
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Suppose we inscribe a rectangle \( R \) inside the ellipse \( E \) with equation\n\n\[ 4{x}^{2} + {y}^{2} = {16} \]\n\nas shown in Figure 1.10.2. If we let \( \left( {x, y}\right) \) be the coordinates of the upper right-hand corner of \( R \), then the area of \( R \) is\n\n\[ A = \left( {2x}\right) \left( {2y}\right) = {4xy}. \]\n\nSince \( \left( {x, y}\right) \) is a point on the upper half of the ellipse, we have\n\n\[ y = \sqrt{{16} - 4{x}^{2}} = 2\sqrt{4 - {x}^{2}} \]\n\nand so\n\n\[ A = {8x}\sqrt{4 - {x}^{2}}. \]\n\nNow suppose we wish to find the dimensions of \( R \) which maximize its area. That is, we want to find the maximum value of \( A \) on the interval \( \left\lbrack {0,2}\right\rbrack \) .
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Now\n\n\[ \frac{dA}{dx} = {8x} \cdot \frac{-{2x}}{2\sqrt{4 - {x}^{2}}} + 8\sqrt{4 - {x}^{2}} = \frac{-8{x}^{2} + 8\left( {4 - {x}^{2}}\right) }{\sqrt{4 - {x}^{2}}} = \frac{{32} - {16}{x}^{2}}{\sqrt{4 - {x}^{2}}}. \]\n\nHence \( \frac{dA}{dx} = 0 \), for \( x \) in \( \left( {0,2}\right) \), when \( {32} - {16}{x}^{2} = 0 \), that is, when \( x = \sqrt{2} \). Thus the maximum value of \( A \) must occur at \( x = 0, x = \sqrt{2} \), or \( x = 2 \). Evaluating, we have\n\n\[ {\left. A\right| }_{x = 0} = 0 \]\n\n\[ {\left. A\right| }_{x = \sqrt{2}} = 8\sqrt{2}\sqrt{2} = {16} \]\n\nand\n\n\[ {\left. A\right| }_{x = 2} = 0 \]\n\nHence the rectangle \( R \) inscribed in \( E \) with the largest area has area 16 when \( x = \sqrt{2} \) and \( y = 2\sqrt{2} \). That is, \( R \) is \( 2\sqrt{2} \) by \( 4\sqrt{2} \).
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Yes
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Consider the problem of finding the extreme values of\n\n\[ y = 4{x}^{2} + \frac{1}{x} \]\n\non the interval \( \left( {0,\infty }\right) \) .
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Since\n\n\[ \frac{dy}{dx} = {8x} - \frac{1}{{x}^{2}} \]\n\nwe see that \( \frac{dy}{dx} < 0 \) when, and only when,\n\n\[ {8x} < \frac{1}{{x}^{2}} \]\n\nThis is equivalent to\n\n\[ {x}^{3} < \frac{1}{8} \]\n\nso \( \frac{dy}{dx} < 0 \) on \( \left( {0,\infty }\right) \) when, and only when, \( 0 < x < \frac{1}{2} \). Similarly, we see that \( \frac{dy}{dx} > 0 \) when, and only when, \( x > \frac{1}{2} \). Thus \( y \) is a decreasing function of \( x \) on the interval \( \left( {0,\frac{1}{2}}\right) \) and an increasing function of \( x \) on the interval \( \left( {\frac{1}{2},\infty }\right) \), and so must have an minimum value at \( x = \frac{1}{2} \). Note, however, that \( y \) does not have a maximum value: given any \( x = c \), if \( c < \frac{1}{2} \) we may find a larger value for \( y \) by using any \( 0 < x < c \), and if \( c > \frac{1}{2} \) we may find a larger value for \( y \) by using any \( x > c \). Thus we conclude that \( y \) has a minimum value of 3 at \( x = \frac{1}{2} \), but does not have a maximum value.
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Yes
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Consider the problem of finding the shortest distance from the point \( A = \left( {0,1}\right) \) to the parabola \( P \) with equation \( y = {x}^{2} \) . If \( \left( {x, y}\right) \) is a point on \( P \) (see Figure 1.10.4), then the distance from \( A \) to \( \left( {x, y}\right) \) is\n\n\[ D = \sqrt{{\left( x - 0\right) }^{2} + {\left( y - 1\right) }^{2}} = \sqrt{{x}^{2} + {\left( {x}^{2} - 1\right) }^{2}}.\]
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Our problem then is to find the minimum value of \( D \) on the interval \( \left( {-\infty ,\infty }\right) \) . However, to make the problem somewhat easier to work with, we note that, since \( D \) is always a positive value, finding the minimum value of \( D \) is equivalent to finding the minimum value of \( {D}^{2} \) . So letting\n\n\[ z = {D}^{2} = {x}^{2} + {\left( {x}^{2} - 1\right) }^{2} = {x}^{2} + {x}^{4} - 2{x}^{2} + 1 = {x}^{4} - {x}^{2} + 1,\]\n\nour problem becomes that of finding the minimum value of \( z \) on \( \left( {-\infty ,\infty }\right) \) . Now\n\n\[ \frac{dz}{dx} = 4{x}^{3} - {2x} \]\n\nso \( \frac{dz}{dx} = 0 \) when, and only when,\n\n\[ 0 = 4{x}^{3} - {2x} = {2x}\left( {2{x}^{2} - 1}\right) ,\]\n\nthat is, when, and only when, \( x = - \frac{1}{\sqrt{2}}, x = 0 \), or \( x = \frac{1}{\sqrt{2}} \) . Now \( {2x} < 0 \) when \( - \infty < x < 0 \) and \( {2x} > 0 \) when \( 0 < x < \infty \), whereas \( 2{x}^{2} - 1 < 0 \) when \( - \frac{1}{\sqrt{2}} < x < \frac{1}{\sqrt{2}} \) and \( 2{x}^{2} - 1 > 0 \) either when \( x < - \frac{1}{\sqrt{2}} \) or when \( x > \frac{1}{\sqrt{2}} \) . Taking the product of \( {2x} \) and \( 2{x}^{2} - 1 \), we see that \( \frac{dz}{dx} < 0 \) when \( x < - \frac{1}{\sqrt{2}} \) and when \( 0 < x < \frac{1}{\sqrt{2}} \), and \( \frac{dz}{dx} > 0 \) when \( - \frac{1}{\sqrt{2}} < x < 0 \) and when \( x > \frac{1}{\sqrt{2}} \) . It follows that \( z \) is a decreasing function of \( x \) on \( \left( {-\infty , - \frac{1}{\sqrt{2}}}\right) \) and on \( \left( {0,\frac{1}{\sqrt{2}}}\right) \), and is an increasing function of \( x \) on \( \left( {-\frac{1}{\sqrt{2}},0}\right) \) and on \( \left( {\frac{1}{\sqrt{2}},\infty }\right) \) .\n\nIt now follows that \( z \) has a local minimum of \( \frac{3}{4} \) at \( x = - \frac{1}{\sqrt{2}} \), a local maximum of 1 at \( x = 0 \), and another local minimum of \( \frac{3}{4} \) at \( x = \frac{1}{\sqrt{2}} \) . Note that \( \frac{3}{4} \) is the minimum value of \( z \) both on the the interval \( \left( {-\infty ,0}\right) \) and on the interval \( \left( {0,\infty }\right) \) ; since \( z \) has a local maximum of 1 at \( x = 0 \), it follows that \( \frac{3}{4} \) is in fact the minimum value of \( z \) on \( \left( {-\infty ,\infty }\right) \) . Hence we may conclude that the minimum distance from \( A \) to \( P \) is \( \frac{\sqrt{3}}{2} \), and the points on \( P \) closest to \( A \) are \( \left( {-\frac{1}{\sqrt{2}},\frac{1}{2}}\right) \) and \( \left( {\frac{1}{\sqrt{2}},\frac{1}{2}}\right) \) . Note, however, that \( z \) does not have a maximum value, even though it has a local maximum value at \( x = 0 \) .
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Yes
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Using \( a = 2 \) and \( b = 1 \) ,(1.11.1) becomes, after multiplying both sides of the equation by 4 ,\n\n\[ \n{x}^{2} + 4{y}^{2} = 4 \n\]
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Differentiating both sides of this equation by \( x \), and remembering to use the chain rule when differentiating \( {y}^{2} \), we obtain\n\n\[ \n{2x} + {8y}\frac{dy}{dx} = 0 \n\]\n\nSolving for \( \frac{dy}{dx} \), we have\n\n\[ \n\frac{dy}{dx} = - \frac{x}{4y} \n\]\n\nwhich is defined whenever \( y \neq 0 \) (corresponding to the points \( \left( {-2,0}\right) \) and \( \left( {2,0}\right) \) , at which, as we saw above, the slope of the tangent lines is undefined). For\n\nexample, we have\n\n\[ \n{\left. \frac{dy}{dx}\right| }_{\left( {x, y}\right) = \left( {1,\frac{\sqrt{3}}{2}}\right) } = - \frac{1}{2\sqrt{3}} \n\]\n\nand so the equation of the line tangent to the ellipse at the point \( \left( {1,\frac{\sqrt{3}}{2}}\right) \) is\n\n\[ \ny = - \frac{1}{2\sqrt{3}}\left( {x - 1}\right) + \frac{\sqrt{3}}{2}. \n\]\n\nSee Figure 1.11.2.
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Yes
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Consider the hyperbola \( H \) with equation\n\n\[ \n{x}^{2} - {4xy} + {y}^{2} = 4 \n\]
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Differentiating both sides of the equation, remembering to treat \( y \) as a function of \( x \), we have\n\n\[ \n{2x} - {4x}\frac{dy}{dx} - {4y} + {2y}\frac{dy}{dx} = 0. \n\]\n\nSolving for \( \frac{dy}{dx} \), we see that\n\n\[ \n\frac{dy}{dx} = \frac{{4y} - {2x}}{{2y} - {4x}} = \frac{{2y} - x}{y - {2x}}. \n\]\n\nFor example,\n\n\[ \n{\left. \frac{dy}{dx}\right| }_{\left( {x, y}\right) = \left( {2,0}\right) } = \frac{2}{4} = \frac{1}{2}. \n\]\n\nHence the equation of the line tangent to \( H \) at \( \left( {2,0}\right) \) is\n\n\[ \ny = \frac{1}{2}\left( {x - 2}\right) \n\]
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Yes
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Suppose oil is being poured onto the surface of a calm body of water. As the oil spreads out, it forms a right circular cylinder whose volume is\n\n\[ V = \pi {r}^{2}h \]\n\nwhere \( r \) and \( h \) are, respectively, the radius and height of the cylinder. Now suppose the oil is being poured out at a rate of 10 cubic centimeters per second and that the height remains a constant 0.25 centimeters. Then the volume of the cylinder is increasing at a rate of 10 cubic centimeters per second, so\n\n\[ \frac{dV}{dt} = {10}{\mathrm{\;{cm}}}^{3}/\mathrm{{sec}} \]\n\nat any time \( t \) .
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Now with \( h = {0.25} \),\n\n\[ V = {0.25\pi }{r}^{2} \]\n\nso\n\n\[ \frac{dV}{dt} = \frac{1}{2}{\pi r}\frac{dr}{dt} \]\n\nHence\n\n\[ \frac{dr}{dt} = \frac{2}{\pi r}\frac{dV}{dt} = \frac{20}{\pi r}\mathrm{\;{cm}}/\mathrm{{sec}}. \]\n\nFor example, if \( r = {10} \) centimeters at some time \( t = {t}_{0} \), then\n\n\[ {\left. \frac{dr}{dt}\right| }_{t = {t}_{0}} = \frac{20}{10\pi } = \frac{2}{\pi } \approx {0.6366}\mathrm{\;{cm}}/\mathrm{{sec}}. \]
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Yes
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Suppose ship \( A \), headed due north at 20 miles per hour, and ship \( B \), headed due east at 30 miles per hour, both pass through the same point \( P \) in the ocean, ship \( A \) at noon and ship \( B \) two hours later (see Figure 1.11.4). If we let \( x \) denote the distance from \( A \) to \( {Pt} \) hours after noon, \( y \) denote the distance from \( B \) to \( {Pt} \) hours after noon, and \( z \) denote the distance from \( A \) to \( {Bt} \) hours after noon, then, by the Pythagorean theorem,\n\n\[ \n{z}^{2} = {x}^{2} + {y}^{2} \n\]
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Differentiating this equation with respect to \( t \), we find\n\n\[ \n{2z}\frac{dz}{dt} = {2x}\frac{dx}{dt} + {2y}\frac{dy}{dt} \n\]\n\nor\n\n\[ \nz\frac{dz}{dt} = x\frac{dx}{dt} + y\frac{dy}{dt} \n\]\n\nFor example, at 4 in the afternoon, that is, when \( t = 4 \), we know that\n\n\[ \nx = \left( 4\right) \left( {20}\right) = {80}\text{miles,} \n\]\n\n\[ \ny = \left( 2\right) \left( {30}\right) = {60}\text{miles,} \n\]\n\nand\n\n\[ \nz = \sqrt{{80}^{2} + {60}^{2}} = {100}\text{ miles,} \n\]\n\nso\n\n\[ \n{100}\frac{dz}{dt} = {80}\frac{dx}{dt} + {60}\frac{dy}{dt}\text{ miles/hour. } \n\]\n\nSince at any time \( t \) ,\n\n\[ \n\frac{dx}{dt} = {20}\text{ miles }/\text{ hour } \n\]\n\nand\n\n\[ \n\frac{dy}{dt} = {30}\text{ miles }/\text{ hour,} \n\]\n\nwe have\n\n\[ \n{\left. \frac{dz}{dt}\right| }_{t = 4} = \frac{\left( {80}\right) \left( {20}\right) + \left( {60}\right) \left( {30}\right) }{100} = {34}\text{ miles/hour. } \n\]
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Yes
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If \( y = 4{x}^{5} - 3{x}^{2} + 4 \), then
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\[ \frac{dy}{dx} = {20}{x}^{4} - {6x} \] and so \[ \frac{{d}^{2}y}{d{x}^{2}} = {80}{x}^{3} - 6 \]
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Yes
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If\n\n\[ f\\left( x\\right) = \\frac{1}{x} \]\n\nthen\n\n\[ {f}^{\\prime }\\left( x\\right) = - \\frac{1}{{x}^{2}} \]
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\[ {f}^{\\prime \\prime }\\left( x\\right) = \\frac{2}{{x}^{3}} \]\n\n\[ {f}^{\\prime \\prime \\prime }\\left( x\\right) = - \\frac{6}{{x}^{4}} \]\n\nand\n\n\[ {f}^{\\left( 4\\right) }\\left( x\\right) = \\frac{24}{{x}^{5}}. \]
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No
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If \( f\left( x\right) = 2{x}^{3} - 3{x}^{2} - {12x} + 1 \), then
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\[ {f}^{\prime }\left( x\right) = 6{x}^{2} - {6x} - {12} \] and \[ {f}^{\prime \prime }\left( x\right) = {12x} - 6 \] Hence \( {f}^{\prime \prime }\left( x\right) < 0 \) when \( x < \frac{1}{2} \) and \( {f}^{\prime \prime }\left( x\right) > 0 \) when \( x > \frac{1}{2} \), and so the graph of \( f \) is concave downward on the interval \( \left( {-\infty ,\frac{1}{2}}\right) \) and concave upward on the interval \( \left( {\frac{1}{2},\infty }\right) \) .
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Yes
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If \( f\left( x\right) = {x}^{4} - {x}^{3} \), then
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\[ {f}^{\prime }\left( x\right) = 4{x}^{3} - 3{x}^{2} = {x}^{2}\left( {{4x} - 3}\right) \] and \[ {f}^{\prime \prime }\left( x\right) = {12}{x}^{2} - {6x} = {6x}\left( {{2x} - 1}\right) . \] Hence \( f \) has stationary points \( x = 0 \) and \( x = \frac{3}{4} \) . Since \[ {f}^{\prime \prime }\left( 0\right) = 0 \] and \[ {f}^{\prime \prime }\left( \frac{3}{4}\right) = \frac{9}{4} > 0 \] we see that \( f \) has a local minimum at \( x = \frac{3}{4} \) . Although the second derivative test tells us nothing about the nature of the critical point \( x = 0 \), we know, since \( f \) has a local minimum at \( x = \frac{3}{4} \), that \( f \) is decreasing on \( \left( {0,\frac{3}{4}}\right) \) and increasing on \( \left( {\frac{3}{4},\infty }\right) \) . Moreover, since \( {4x} - 3 < 0 \) for all \( x < 0 \), it follows that \( {f}^{\prime }\left( x\right) < 0 \) for all \( x < 0 \), and so \( f \) is also decreasing on \( \left( {-\infty ,0}\right) \) . Hence \( f \) has neither a local maximum nor a local minimum at \( x = 0 \) . Finally, since \( {f}^{\prime \prime }\left( x\right) < 0 \) for \( 0 < x < \frac{1}{2} \) and \( {f}^{\prime \prime }\left( x\right) > 0 \) for all other \( x \), we see that the graph of \( f \) is concave downward on the interval \( \left( {0,\frac{1}{2}}\right) \) and concave upward on the intervals \( \left( {-\infty ,0}\right) \) and \( \left( {\frac{1}{2},\infty }\right) \) . See Figure 1.12.2.
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Yes
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If \( f\left( x\right) = 3{x}^{2} \), then \( F\left( x\right) = {x}^{3} \) is an integral of \( f \) on \( \left( {-\infty ,\infty }\right) \) since \( {F}^{\prime }\left( x\right) = 3{x}^{2} \) for all \( x \).
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However, note that \( F \) is not the only integral of \( f \) : for other examples, both \( G\left( x\right) = {x}^{3} + 4 \) and \( H\left( x\right) = {x}^{3} + {15} \) are integrals of \( f \) as well. Indeed, since the derivative of a constant is 0 the function \( L\left( x\right) = {x}^{3} + c \) is an integral of \( f \) for any constant \( c \) .
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Yes
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Since\n\n\[ \frac{d}{dx}\left( {\frac{3}{2}{x}^{2} + {4x}}\right) = {3x} + 4 \]\n\nany integral of \( f\left( x\right) = {3x} + 4 \) must be of the form
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\n\n\[ F\left( x\right) = \frac{3}{2}{x}^{2} + {4x} + c \]\n\nfor some constant \( c \) .
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Yes
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Since\n\n\[ \frac{d}{dx}\left( {4{x}^{3} - \sin \left( x\right) }\right) = {12}{x}^{2} - \cos \left( x\right) ,\]
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it follows that\n\n\[ \int \left( {{12}{x}^{2} - \cos \left( x\right) }\right) {dx} = 4{x}^{3} - \sin \left( x\right) + c, \]
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Yes
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Suppose we wish to find the integral \( F\\left( x\\right) \) of \( f\\left( x\\right) = 5{x}^{2} - 7 \) for which \( F\\left( 1\\right) = {10} \).
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Now\n\n\\[ \n\\int \\left( {5{x}^{2} - 7}\\right) {dx} = \\frac{5}{3}{x}^{3} - {7x} + c \n\\]\n\nso\n\n\\[ \nF\\left( x\\right) = \\frac{5}{3}{x}^{3} - {7x} + c \n\\]\n\nfor some constant \( c \). Now we want\n\n\\[ \n{10} = F\\left( 1\\right) = \\frac{5}{3} - 7 + c \n\\]\n\nso we must have\n\n\\[ \nc = {10} + 7 - \\frac{5}{3} = \\frac{46}{3}. \n\\]\n\nHence the desired integral is\n\n\\[ \nF\\left( x\\right) = \\frac{5}{3}{x}^{3} - {7x} + \\frac{46}{3}. \n\\]
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Yes
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Suppose the velocity of an object oscillating at the end of a spring is\n\n\[ v\left( t\right) = - {20}\sin \left( {5t}\right) \text{centimeters/second.} \]\n\nIf \( x\left( t\right) \) is the position of the object at time \( t \), then
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\[ x\left( t\right) = - \int {20}\sin \left( {5t}\right) {dt} = 4\cos \left( {5t}\right) + c\text{ centimeters } \]\n\nfor some constant \( c \) . If in addition we know that the object was initially 4 centimeters from the origin, that is, that \( x\left( 0\right) = 4 \), then we would have\n\n\[ 4 = x\left( 0\right) = 4 + c. \]\n\nHence we would have \( c = 0 \), and so\n\n\[ x\left( t\right) = 4\cos \left( {5t}\right) \text{ centimeters } \]\n\ncompletely specifies the position of the object at time \( t \) .
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Yes
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In an earlier example, we had \( v\left( t\right) = - {20}\sin \left( {5t}\right) \) centimeters per second and \( x\left( 0\right) = 4 \) centimeters. To approximate \( x\left( 2\right) \), we will divide \( \left\lbrack {0,2}\right\rbrack \) into four equal subintervals, each of length 0.5.
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That is, we will take\n\n\[ \n{t}_{0} = {0.0},{t}_{1} = {0.5},{t}_{2} = 1,{t}_{3} = {1.5},{t}_{4} = 2\text{,}\n\]\n\nand\n\n\[ \n\Delta {t}_{1} = {0.5},\Delta {t}_{2} = {0.5},\Delta {t}_{3} = {0.5},\Delta {t}_{4} = {0.5}\text{.}\n\]\n\nGood choices for points to evaluate \( v\left( t\right) \) are the midpoints of the subintervals. In this case, that means we should take\n\n\[ \n{t}_{1}^{ * } = {0.25},{t}_{2}^{ * } = {0.75},{t}_{3}^{ * } = {1.25},{t}_{4}^{ * } = {1.75}\text{.}\n\]\n\nThen we have\n\n\[ \nx\left( 2\right) \approx x\left( 0\right) + v\left( {0.25}\right) \Delta {t}_{1} + v\left( {0.75}\right) \Delta {t}_{2} + v\left( {1.25}\right) \Delta {t}_{3} + v\left( {1.75}\right) \Delta {t}_{4}\n\]\n\n\[ \n= 4 - {20}\sin \left( {1.25}\right) \left( {0.5}\right) - {20}\sin \left( {3.75}\right) \left( {0.5}\right) - {20}\sin \left( {6.25}\right) \left( {0.5}\right)\n\]\n\n\[ \n- {20}\sin \left( {8.75}\right) \left( {0.5}\right)\n\]\n\n\[ \n\approx - {5.6897}\n\]\n\nNote that, from our earlier work, we know that the exact answer is\n\n\[ \nx\left( 2\right) = 4\cos \left( {10}\right) \approx - {3.3563}.\n\]
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Yes
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From the observation that\n\n\[ f\left( x\right) = \frac{1}{1 + {x}^{2}} \]\n\nis increasing on \( \left( {-\infty ,0\rbrack \text{and decreasing on}\lbrack 0,\infty }\right) \), it is easy to see that\n\n\[ \frac{1}{2} \leq \frac{1}{1 + {x}^{2}} \leq 1 \]\n\nfor all \( x \) in \( \left\lbrack {-1,1}\right\rbrack \) . Hence\n\n\[ 1 \leq {\int }_{-1}^{1}\frac{1}{1 + {x}^{2}}{dx} \leq 2 \]
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We will eventually see, in Example 2.6.20, that\n\n\[ {\int }_{-1}^{1}\frac{1}{1 + {x}^{2}}{dx} = \frac{\pi }{2} \approx {1.5708}. \]
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No
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If \( \alpha \) is any infinitesimal, than \( \alpha \sim o\left( 1\right) \)
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since \( \frac{\alpha }{1} = \alpha \) is an infinitesimal.
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Yes
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If \( \epsilon \) is any nonzero infinitesimal, then \( {\epsilon }^{2} \sim o\left( \epsilon \right) \)
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\[ \frac{{\epsilon }^{2}}{\epsilon } = \epsilon \simeq 0 \]
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Yes
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Theorem 2.4.1. Suppose \( B \) is a function that for any real numbers \( a < b \) in an open interval \( I \) assigns a value \( B\left( {a, b}\right) \) and satisfies\n\n- for any \( a < c < b \) in \( I, B\left( {a, b}\right) = B\left( {a, c}\right) + B\left( {c, b}\right) \), and\n\n- for some continuous function \( h \) and any nonzero infinitesimal \( {dx} \),\n\n\[ B\left( {x, x + {dx}}\right) - h\left( x\right) {dx} \sim o\left( {dx}\right) \]\n\nfor any \( x \) in \( I \) .\n\nThen\n\n\[ B\left( {a, b}\right) = {\int }_{a}^{b}h\left( x\right) {dx} \]\n\nfor any real numbers \( a \) and \( b \) in \( I \) .
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We will look at several applications of definite integrals in the next section. For now, we note how this theorem provides a method for evaluating integrals. Namely, given a function \( f \) which is differentiable on an open interval \( I \), define, for every \( a < b \) in \( I \),\n\n\[ B\left( {a, b}\right) = f\left( b\right) - f\left( a\right) .\n\nThen, for any \( a, b \), and \( c \) in \( I \) with \( a < c < b \),\n\n\[ B\left( {a, b}\right) = f\left( b\right) - f\left( a\right) \]\n\n\[ = \left( {f\left( b\right) - f\left( c\right) }\right) + \left( {f\left( c\right) - f\left( a\right) }\right) \]\n\n\[ = B\left( {a, c}\right) + B\left( {c, b}\right) \text{.} \]\n\nMoreover, for any infinitesimal \( {dx} \) and any \( x \) in \( I \),\n\n\[ \frac{B\left( {x, x + {dx}}\right) }{dx} = \frac{f\left( {x + {dx}}\right) - f\left( x\right) }{dx} \simeq {f}^{\prime }\left( x\right) ,\]\n\nfrom which it follows that\n\n\[ \frac{B\left( {x, x + {dx}}\right) - {f}^{\prime }\left( x\right) {dx}}{dx} \]\n\n is an infinitesimal. Hence\n\n\[ B\left( {x,{dx}}\right) - {f}^{\prime }\left( x\right) {dx} \sim o\left( {dx}\right) ,\]\n\nand so it follows from Theorem 2.4.1 that\n\n\[ f\left( b\right) - f\left( a\right) = B\left( {a, b}\right) = {\int }_{a}^{b}{f}^{\prime }\left( x\right) {dx}. \]
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Yes
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To evaluate\n\n\[ \n{\int }_{0}^{1}{xdx} \n\]
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we first note that \( g\left( x\right) = x \) is the derivative of \( f\left( x\right) = \frac{1}{2}{x}^{2} \) . Hence, by Theorem 2.4.2,\n\n\[ \n{\int }_{0}^{1}{xdx} = f\left( 1\right) - f\left( 0\right) = \frac{1}{2} - 0 = \frac{1}{2} \n\]
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Yes
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Since\n\n\[ \int {x}^{2}{dx} = \frac{1}{3}{x}^{3} + c \]
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we have\n\n\[ {\int }_{1}^{2}{x}^{2}{dx} = {\left. \frac{1}{3}{x}^{3}\right| }_{1}^{2} = \frac{8}{3} - \frac{1}{3} = \frac{7}{3}. \]
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Yes
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Example 2.4.5. Since\n\n\\[ \n\\int - {20}\\sin \\left( {5x}\\right) {dx} = 4\\cos \\left( {5x}\\right) + c, \n\\]\n\nwe have\n\n\\[ \n{\\int }_{0}^{2\\pi } - {20}\\sin \\left( {5t}\\right) {dt} = {\\left. 4\\cos \\left( 5t\\right) \\right| }_{0}^{2\\pi } = 4 - 4 = 0. \n\\]
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\\[ \n{\\int }_{0}^{2\\pi } - {20}\\sin \\left( {5t}\\right) {dt} = {\\left. 4\\cos \\left( 5t\\right) \\right| }_{0}^{2\\pi } = 4 - 4 = 0. \n\\]
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Yes
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Let \( A \) be the area of the region \( R \) bounded by the curves with equations \( y = {x}^{2} \) and \( y = x + 2 \) . Note that these curves intersect when \( {x}^{2} = x + 2 \), that is when\n\n\[ 0 = {x}^{2} - x - 2 = \left( {x + 1}\right) \left( {x - 2}\right) . \]\n\nHence they intersect at the points \( \left( {-1,1}\right) \) and \( \left( {2,4}\right) \), and so \( R \) is the region in the plane bounded above by the curve \( y = x + 2 \), below by the curve \( y = {x}^{2} \), on the right by \( x = - 1 \), and on the left by \( x = 2 \) . See Figure 2.5.2. Thus we have\n\n\[ A = {\int }_{-1}^{2}\left( {x + 2 - {x}^{2}}\right) {dx} \]
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\[ = {\left. \left( \frac{1}{2}{x}^{2} + 2x - \frac{1}{3}{x}^{3}\right) \right| }_{-1}^{2} \]\n\n\[ = \left( {2 + 4 - \frac{8}{3}}\right) - \left( {\frac{1}{2} - 2 + \frac{1}{3}}\right) \]\n\n\[ = \frac{9}{2}\text{.} \]
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Yes
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If \( A \) is the area of the region \( R \) beneath the graph of \( y = \sin \left( x\right) \) over the interval \( \left\lbrack {0,\pi }\right\rbrack \)
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\[ A = {\int }_{0}^{\pi }\sin \left( x\right) {dx} = - {\left. \cos \left( x\right) \right| }_{0}^{\pi } = 1 + 1 = 2. \]
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Yes
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Example 2.5.4. The unit sphere \( S \), with center at the origin, is the set of all points \( \left( {x, y, z}\right) \) satisfying \( {x}^{2} + {y}^{2} + {z}^{2} = 1 \) (see Figure 2.5.7). For a fixed value of \( z \) between -1 and 1, the cross section \( R\left( z\right) \) of \( S \) perpendicular to the \( z \) -axis is the set of points \( \left( {x, y}\right) \) satisfying the equation \( {x}^{2} + {y}^{2} = 1 - {z}^{2} \) . That is, \( R\left( z\right) \) is a circle with radius \( \sqrt{1 - {z}^{2}} \) . Hence \( R\left( z\right) \) has area
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\[ A\left( z\right) = \pi \left( {1 - {z}^{2}}\right) \] If \( V \) is the volume of \( S \), it now follows that \[ V = {\int }_{-1}^{1}\pi \left( {1 - {z}^{2}}\right) {dx} \] \[ = {\left. \pi \left( z - \frac{1}{3}{z}^{3}\right) \right| }_{-1}^{1} \] \[ = \pi \left( {\frac{2}{3} + \frac{2}{3}}\right) \] \[ = \frac{4\pi }{3}\text{.} \]
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Yes
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Let \( T \) be the region bounded by the \( z \) -axis and the graph of \( z = {x}^{2} \) for \( 0 \leq x \leq 1 \) . Let \( B \) be the three-dimensional body created by rotating \( T \) about the \( z \) -axis. See Figure 2.5.8. If \( R\left( z\right) \) is a cross section of \( B \) perpendicular to the \( z \) -axis, then \( R\left( z\right) \) is a circle with radius \( \sqrt{z} \) . Thus, if \( A\left( z\right) \) is the area of \( R\left( z\right) \), we have\n\n\[ A\left( z\right) = {\pi z}. \]
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If \( V \) is the volume of \( B \), then\n\n\[ V = {\int }_{0}^{1}{\pi zdz} = {\left. \pi \frac{{z}^{2}}{2}\right| }_{0}^{1} = \frac{\pi }{2}. \]
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Yes
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Example 2.5.6. Let \( T \) be the region bounded by the graphs of \( z = {x}^{4} \) and \( x = {x}^{2} \) for \( 0 \leq x \leq 1 \) . Let \( B \) be the three-dimensional body created by rotating \( T \) about the \( z \) -axis. See Figure 2.5.9. If \( R\left( z\right) \) is a cross section of \( B \) perpendicular to the \( z \) -axis, then \( R\left( z\right) \) is the region between the circles with radii \( {z}^{\frac{1}{4}} \) and \( \sqrt{z} \) , an annulus. Hence if \( A\left( z\right) \) is the area of \( R\left( z\right) \), then
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\[ A\left( z\right) = \pi {\left( {z}^{\frac{1}{4}}\right) }^{2} - \pi {\left( \sqrt{z}\right) }^{2} = \pi \left( {\sqrt{z} - z}\right) . \] If \( V \) is the volume of \( B \), then \[ V = {\int }_{0}^{1}\pi \left( {\sqrt{z} - z}\right) {dz} = {\left. \pi \left( \frac{2}{3}{z}^{\frac{3}{2}} - \frac{1}{2}{z}^{2}\right) \right| }_{0}^{1} = \pi \left( {\frac{2}{3} - \frac{1}{2}}\right) = \frac{\pi }{6}. \]
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Yes
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Let \( C \) be the graph of \( f\left( x\right) = {x}^{\frac{3}{2}} \) over the interval \( \left\lbrack {0,1}\right\rbrack \) (see Figure 2.5.11) and let \( L \) be the length of \( C \) . Since \( {f}^{\prime }\left( x\right) = \frac{3}{2}\sqrt{x} \), we have\n\n\[ L = {\int }_{0}^{1}\sqrt{1 + \frac{9}{4}x}{dx} \]
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Now\n\[ \int \sqrt{x}{dx} = \frac{2}{3}{x}^{\frac{3}{2}} + c \]\nso we might expect an integral of \( \sqrt{1 + \frac{9}{4}} \) to be\n\n\[ \frac{2}{3}{\left( 1 + \frac{9}{4}x\right) }^{\frac{3}{2}} + c \]\n\nHowever,\n\n\[ \frac{d}{dx}\frac{2}{3}{\left( 1 + \frac{9}{4}x\right) }^{\frac{3}{2}} = \frac{9}{4}\sqrt{1 + \frac{9}{4}x} \]\n\nand so, dividing our original guess by \( \frac{9}{4} \), we have\n\n\[ \int \sqrt{1 + \frac{9}{4}x}{dx} = \frac{8}{27}{\left( 1 + \frac{9}{4}x\right) }^{\frac{3}{2}} + c, \]\n\nwhich may be verified by differentiation. Hence\n\n\[ L = {\int }_{0}^{1}\sqrt{1 + \frac{9}{4}x}{dx} \]\n\n\[ = {\left. \frac{8}{27}{\left( 1 + \frac{9}{4}x\right) }^{\frac{3}{2}}\right| }_{0}^{1} \]\n\n\[ = \frac{8}{27}\left( {\frac{{13}\sqrt{13}}{8} - 1}\right) \]\n\n\[ = \frac{{13}\sqrt{13} - 8}{27} \]\n\n\[ \approx {1.4397}\text{.} \]
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Yes
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Example 2.5.8. Let \( C \) be the graph of \( f\left( x\right) = {x}^{2} \) over the interval \( \left\lbrack {0,1}\right\rbrack \) (see Figure 2.5.12) and let \( L \) be the length of \( C \) . Since \( {f}^{\prime }\left( x\right) = {2x} \) ,
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\[ L = {\int }_{0}^{1}\sqrt{1 + 4{x}^{2}}{dx} \]\n\nHowever, we do not have the tools at this time to evaluate this definite integral exactly. Still, we may use (2.5.24) to find an approximation for \( L \) . For example, if we take \( N = {100} \) in (2.5.24), then \( {\Delta x} = {0.01} \) and\n\n\[ \Delta {y}_{i} = f\left( {0.01i}\right) - f\left( {{0.01}\left( {i - 1}\right) }\right) \]\n\n\[ = {\left( {0.01}i\right) }^{2} - {\left( {0.01}\left( i - 1\right) \right) }^{2} \]\n\n\[ = {0.0001}\left( {{i}^{2} - \left( {{i}^{2} - {2i} + 1}\right) }\right) \]\n\n\[ = {0.0001}\left( {{2i} - 1}\right) \]\n\n\( \left( {2.5.29}\right) \)\n\nfor \( i = 1,2,\ldots, N \), and so\n\n\[ L \approx \mathop{\sum }\limits_{{i = 1}}^{{100}}\sqrt{{\left( \Delta x\right) }^{2} + {\left( \Delta {y}_{i}\right) }^{2}} \approx {1.4789}. \]\n\n\n\nFigure 2.5.12: Graph of \( y = {x}^{2} \) over \( \left\lbrack {0,1}\right\rbrack \)
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No
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To evaluate\n\n\[ \int {2x}\sqrt{1 + {x}^{2}}{dx} \]
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let\n\n\[ u = 1 + {x}^{2} \]\n\n\[ {du} = {2xdx}\text{.} \]\n\nThen\n\n\[ \int {2x}\sqrt{1 + {x}^{2}}{dx} = \int \sqrt{u}{du} = \frac{2}{3}{u}^{\frac{3}{2}} + c = \frac{2}{3}{\left( 1 + {x}^{2}\right) }^{\frac{3}{2}} + c. \]
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Yes
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To evaluate\n\n\[ \int x\sin \left( {x}^{2}\right) {dx} \]
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let\n\n\[ u = {x}^{2} \]\n\n\[ {du} = {2xdx} \]\n\nNote that in this case we cannot make a direct substitution of \( u \) and \( {du} \) since \( {du} = {2xdx} \) does not appear as part of the integral. However, \( {du} \) differs from \( {xdx} \) by only a constant factor, and we may rewrite \( {du} = {2xdx} \) as\n\n\[ \frac{1}{2}{du} = {xdx} \]\n\nNow we may perform the change of variable:\n\n\[ \int x\sin \left( u\right) {dx} = \frac{1}{2}\int \sin \left( u\right) {du} = - \frac{1}{2}\cos \left( u\right) + c = - \frac{1}{2}\cos \left( {x}^{2}\right) + c. \]
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Yes
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Note that we could evaluate the integral\n\n\[ \int \cos \left( {4x}\right) {dx} \]
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using the substitution\n\n\[ u = {4x} \]\n\n\[ {du} = {4dx} \]\n\nwhich gives us\n\n\[ \int \cos \left( {4x}\right) {dx} = \frac{1}{4}\int \cos \left( u\right) {du} = \frac{1}{4}\sin \left( u\right) + c = \frac{1}{4}\sin \left( {4x}\right) + c. \]
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Yes
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To evaluate\n\n\[ \int {\cos }^{2}\left( {5x}\right) \sin \left( {5x}\right) {dx} \]
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let\n\n\[ u = \cos \left( {5x}\right) \]\n\n\[ {du} = - 5\sin \left( {5x}\right) {dx} \]\n\nThen\n\n\[ \int {\cos }^{2}\left( {5x}\right) \sin \left( {5x}\right) {dx} = - \frac{1}{5}\int {u}^{2}{du} = - \frac{1}{15}{u}^{3} + c = - \frac{1}{15}{\cos }^{3}\left( {5x}\right) + c. \]
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Yes
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To evaluate\n\n\[ \n{\int }_{0}^{1}\frac{x}{\sqrt{1 + {x}^{2}}}{dx} \n\]
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let\n\n\[ \nu = 1 + {x}^{2} \]\n\n\[ {du} = {2xdx}\text{.} \]\n\nNote that when \( x = 0, u = 1 \), and when \( x = 1, u = 2 \) . Hence\n\n\[ \n{\int }_{0}^{1}\frac{x}{\sqrt{1 + {x}^{2}}}{dx} = \frac{1}{2}{\int }_{1}^{2}\frac{1}{\sqrt{u}}{du} = {\left. \sqrt{u}\right| }_{1}^{2} = \sqrt{2} - 1. \n\]
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Yes
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To evaluate\n\n\[ \n{\int }_{0}^{\frac{\pi }{4}}{\cos }^{2}\left( {2x}\right) \sin \left( {2x}\right) {dx} \n\]
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let\n\n\[ \nu = \cos \left( {2x}\right) \]\n\n\[ {du} = - 2\sin \left( {2x}\right) {dx} \]\n\nThen \( u = 1 \) when \( x = 0 \) and \( u = 0 \) when \( x = \frac{\pi }{4} \), so\n\n\[ \n{\int }_{0}^{\frac{\pi }{4}}{\cos }^{2}\left( {2x}\right) \sin \left( {2x}\right) {dx} = - \frac{1}{2}{\int }_{1}^{0}{u}^{2}{du}. \n\]\n\nNote that, after making the change of variable, the upper limit of integration is less than the lower limit of integration, a situation not covered by our definition of the definite integral or our statement of the fundamental theorem of calculus. However, the result on substitutions above shows that we will obtain the correct result if we apply the fundamental theorem as usual. Moreover, this points toward an extension of our definition: if \( b < a \), then we should have\n\n\[ \n{\int }_{a}^{b}f\left( x\right) {dx} = - {\int }_{b}^{a}f\left( x\right) {dx} \n\]\n\n(2.6.6)\n\nwhich is consistent with both the fundamental theorem of calculus and with the definition of the definite integral (since, if \( b < a,{dx} = \frac{b - a}{N} < 0 \) for any positive infinite integer \( N \) ). With this, we may finish the evaluation:\n\n\[ \n{\int }_{0}^{\frac{\pi }{4}}{\cos }^{2}\left( {2x}\right) \sin \left( {2x}\right) {dx} = - \frac{1}{2}{\int }_{1}^{0}{u}^{2}{du} = {\left. \frac{1}{2}{\int }_{0}^{1}{u}^{2}du = \frac{{u}^{3}}{6}\right| }_{0}^{1} = \frac{1}{6}. \n\]
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Yes
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Consider the integral\n\n\[ \int x\cos \left( x\right) {dx} \]
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If we let \( u = x \) and \( {dv} = \cos \left( x\right) {dx} \), then \( {du} = {dx} \) and we may let \( v = \sin \left( x\right) \) . Note that we have some choice for \( v \) since the only requirement is that it is an integral of \( \cos \left( x\right) \) . Using (2.6.10), we have\n\n\[ \int x\sin \left( x\right) {dx} = {uv} - \int {vdu} = x\sin \left( x\right) - \int \sin \left( x\right) {dx} = x\sin \left( x\right) + \cos \left( x\right) + c. \]
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No
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To evaluate\n\n\[ \n{\int }_{0}^{\pi }{x}^{2}\sin \left( x\right) {dx} \n\]
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let\n\n\[ \nu = {x}^{2}\;{dv} = \sin \left( x\right) {dx} \]\n\n\[ {du} = {2xdx}\;v = - \cos \left( x\right) . \]\n\nThen, using (2.6.11),\n\n\[ {\int }_{0}^{\pi }{x}^{2}\sin \left( x\right) {dx} = - {\left. {x}^{2}\cos \left( x\right) \right| }_{0}^{\pi } + {\int }_{0}^{\pi }{2x}\cos \left( x\right) {dx} = {\pi }^{2} + {\int }_{0}^{\pi }{2x}\cos \left( x\right) {dx}. \]\n\nNote that the final integral is simpler than the integral with which we started, but still requires another integration by parts to finish the evaluation. Namely, if we now let\n\n\[ u = {2x}\;{dv} = \cos \left( x\right) \]\n\n\[ {du} = {2dx}\;v = \sin \left( x\right) , \]\n\nwe have\n\n\[ {\int }_{0}^{\pi }{x}^{2}\sin \left( x\right) {dx} = {\left. {\pi }^{2} + 2x\sin \left( x\right) \right| }_{0}^{\pi } - {\int }_{0}^{\pi }2\sin \left( x\right) {dx} \]\n\n\[ = {\pi }^{2} + \left( {0 - 0}\right) + {\left. 2\cos \left( x\right) \right| }_{0}^{\pi } \]\n\n\[ = {\pi }^{2} - 2 - 2 \]\n\n\[ = {\pi }^{2} - 4\text{.} \]
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Yes
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To evaluate\n\n\[ \n{\int }_{0}^{1}x\sqrt{1 + x}{dx} \n\]
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let\n\n\[ \nu = x \]\n\n\[ {dv} = \sqrt{1 + x}{dx} \]\n\n\[ {du} = {dx}\;v = \frac{2}{3}{\left( 1 + x\right) }^{\frac{3}{2}}. \]\n\nThen\n\n\[ {\int }_{0}^{1}x\sqrt{1 + x}{dx} = {\left. \frac{2}{3}x{\left( 1 + x\right) }^{\frac{3}{2}}\right| }_{0}^{1} - \frac{2}{3}{\int }_{0}^{1}{\left( 1 + x\right) }^{\frac{3}{2}}{dx} \]\n\n\[ = {\left. \frac{4\sqrt{2}}{3} - \frac{4}{15}{\left( 1 + x\right) }^{\frac{5}{2}}\right| }_{0}^{1} \]\n\n\[ = \frac{4\sqrt{2}}{3} - \frac{{16}\sqrt{2} - 4}{15} \]\n\n\[ = \frac{4\sqrt{2} + 4}{15}\text{. } \]\n
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Yes
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To evaluate the integral\n\n\[ \n{\int }_{0}^{\pi }{\sin }^{2}\left( x\right) {dx} \n\]
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we will use the half-angle formula:\n\n\[ \n{\sin }^{2}\left( x\right) = \frac{1 - \cos \left( {2x}\right) }{2}. \n\]\n\n\( \left( {2.6.12}\right) \)\n\nThen\n\n\[ \n{\int }_{0}^{\pi }{\sin }^{2}\left( x\right) {dx} = \frac{1}{2}{\int }_{0}^{\pi }\left( {1 - \cos \left( {2x}\right) }\right) {dx} \n\]\n\n\[ \n= {\left. \frac{1}{2}x\right| }_{0}^{\pi } - {\left. \frac{1}{4}\sin \left( 2x\right) \right| }_{0}^{\pi } \n\]\n\n\[ \n= \frac{\pi }{2} \n\]
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Yes
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Example 2.6.11. Using 2.6.13 twice, we have\n\n\\[ \n{\\int }_{0}^{\\pi }{\\cos }^{4}\\left( {3x}\\right) {dx} = {\\int }_{0}^{\\pi }{\\left( {\\cos }^{2}\\left( 3x\\right) \\right) }^{2}{dx} \n\\]\n
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\n\\[ \n= {\\int }_{0}^{\\pi }{\\left( \\frac{1}{2}\\left( 1 + \\cos \\left( 6x\\right) \\right) \\right) }^{2}{dx} \n\\]\n\n\\[ \n= \\frac{1}{4}{\\int }_{0}^{\\pi }\\left( {1 + 2\\cos \\left( {6x}\\right) + {\\cos }^{2}\\left( {6x}\\right) }\\right) {dx} \n\\]\n\n\\[ \n= {\\left. \\frac{1}{4}x\\right| }_{0}^{\\pi } + {\\left. \\frac{1}{12}\\sin \\left( 6x\\right) \\right| }_{0}^{\\pi } + \\frac{1}{8}{\\int }_{0}^{\\pi }\\left( {1 + \\cos \\left( {12x}\\right) }\\right) {dx} \n\\]\n\n\\[ \n= {\\left. \\frac{\\pi }{4} + \\frac{1}{8}x\\right| }_{0}^{\\pi } + {\\left. \\frac{1}{96}\\sin \\left( {12}x\\right) \\right| }_{0}^{\\pi } \n\\]\n\n\\[ \n= \\frac{3\\pi }{8}\\text{. } \n\\]\n
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Yes
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Suppose \( n \geq 2 \) is an integer and we wish to evaluate\n\n\[ \n{\int }_{0}^{\pi }{\sin }^{n}\left( x\right) {dx} \n\]
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We begin with an integration by parts: if we let\n\n\[ \nu = {\sin }^{n - 1}\left( x\right) \;{dv} = \sin \left( x\right) {dx} \]\n\n\[ {du} = \left( {n - 1}\right) {\sin }^{n - 2}\left( x\right) \cos \left( x\right) {dx}\;v = - \cos \left( x\right) ,\]\n\nthen\n\n\[ {\int }_{0}^{\pi }{\sin }^{n}\left( x\right) {dx} = - {\left. {\sin }^{n - 1}\left( x\right) \cos \left( x\right) \right| }_{0}^{\pi } + \left( {n - 1}\right) {\int }_{0}^{\pi }{\sin }^{n - 2}\left( x\right) {\cos }^{2}\left( x\right) {dx} \]\n\n\[ = \left( {n - 1}\right) {\int }_{0}^{\pi }{\sin }^{n - 2}\left( x\right) {\cos }^{2}\left( x\right) {dx}. \]\n\nNow \( {\cos }^{2}\left( x\right) = 1 - {\sin }^{2}\left( x\right) \), so we have\n\n\[ {\int }_{0}^{\pi }{\sin }^{n}\left( x\right) {dx} = \left( {n - 1}\right) {\int }_{0}^{\pi }{\sin }^{n - 2}\left( x\right) \left( {1 - {\sin }^{2}\left( x\right) }\right) {dx} \]\n\n\[ = \left( {n - 1}\right) {\int }_{0}^{\pi }{\sin }^{n - 2}\left( x\right) {dx} - \left( {n - 1}\right) {\int }_{0}^{\pi }{\sin }^{n}\left( x\right) {dx}. \]\n\nNotice that \( {\int }_{0}^{\pi }{\sin }^{n}\left( x\right) {dx} \) occurs on both sides of this equation. Hence we may solve for this quantity, first obtaining\n\n\[ n{\int }_{0}^{\pi }{\sin }^{n}\left( x\right) {dx} = \left( {n - 1}\right) {\int }_{0}^{\pi }{\sin }^{n - 2}\left( x\right) {dx} \]\n\nand then\n\n\[ {\int }_{0}^{\pi }{\sin }^{n}\left( x\right) {dx} = \frac{n - 1}{n}{\int }_{0}^{\pi }{\sin }^{n - 2}\left( x\right) {dx}. \]
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Yes
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Example 2.6.13. An alternative to using a reduction formula in the last example begins with noting that\n\n\[ \n{\int }_{0}^{\pi }{\sin }^{5}\left( x\right) {dx} = {\int }_{0}^{\pi }{\sin }^{4}\left( x\right) \sin \left( x\right) {dx} \n\]
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\[ \n= {\int }_{0}^{\pi }{\left( {\sin }^{2}\left( x\right) \right) }^{2}\sin \left( x\right) {dx} \n\]\n\n\[ \n= {\int }_{0}^{\pi }{\left( 1 - {\cos }^{2}\left( x\right) \right) }^{2}\sin \left( x\right) {dx} \n\]\n\n\[ \n= {\int }_{0}^{\pi }\left( {1 - 2{\cos }^{2}\left( x\right) + {\cos }^{4}\left( x\right) }\right) \sin \left( x\right) {dx}. \n\]\n\nThe latter integral may now be evaluated using the change of variable\n\n\[ \nu = \cos \left( x\right) \]\n\n\[ {du} = - \sin \left( x\right) {dx} \]\n\ngiving us\n\n\[ \n{\int }_{0}^{\pi }{\sin }^{5}\left( x\right) {dx} = - {\int }_{1}^{-1}\left( {1 - 2{u}^{2} + {u}^{4}}\right) {du} \n\]\n\n\[ \n= {\int }_{-1}^{1}\left( {1 - 2{u}^{2} + {u}^{4}}\right) {du} \n\]\n\n\[ \n= {\left. \left( u - \frac{2}{3}{u}^{3} + \frac{1}{5}{u}^{5}\right) \right| }_{-1}^{1} \n\]\n\n\[ \n= \left( {1 - \frac{2}{3} + \frac{1}{5}}\right) - \left( {-1 + \frac{2}{3} - \frac{1}{5}}\right) \n\]\n\n\[ \n= \frac{16}{15} \n\]\n\nas we saw above.
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Yes
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To evaluate\n\n\[ \n{\int }_{0}^{\pi }\sin \left( {2x}\right) \cos \left( {3x}\right) {dx} \n\]
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we first note that, using (2.6.18) with \( a = 2 \) and \( b = 3 \) ,\n\n\[ \n\sin \left( {2x}\right) \cos \left( {3x}\right) = \frac{1}{2}\left( {\sin \left( {5x}\right) + \sin \left( {-x}\right) }\right) = \frac{1}{2}\left( {\sin \left( {5x}\right) - \sin \left( x\right) }\right) .\n\]\n\nHence\n\n\[ \n{\int }_{0}^{\pi }\sin \left( {2x}\right) \cos \left( {3x}\right) {dx} = \frac{1}{2}{\int }_{0}^{\pi }\sin \left( {5x}\right) {dx} - \frac{1}{2}{\int }_{0}^{\pi }\sin \left( x\right) {dx} \n\]\n\n\[ \n= - {\left. \frac{1}{10}\cos \left( 5x\right) \right| }_{0}^{\pi } + {\left. \frac{1}{2}\cos \left( x\right) \right| }_{0}^{\pi } \n\]\n\n\[ \n= \left( {\frac{1}{10} + \frac{1}{10}}\right) + \left( {-\frac{1}{2} - \frac{1}{2}}\right) \n\]\n\n\[ \n= - \frac{4}{5}\text{. } \n\]
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Yes
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To evaluate\n\n\[ \n{\int }_{0}^{\pi }\sin \left( {3x}\right) \sin \left( {5x}\right) {dx} \n\]
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we first note that, using (2.6.22) with \( a = 3 \) and \( b = 5 \) ,\n\n\[ \n\sin \left( {3x}\right) \sin \left( {5x}\right) = \frac{1}{2}\left( {\cos \left( {-{2x}}\right) - \cos \left( {8x}\right) }\right) = \frac{1}{2}\left( {\cos \left( {2x}\right) - \cos \left( {8x}\right) }\right) .\n\]\n\nNote that we would have the same identity if we had chosen \( a = 5 \) and \( b = 3 \) . Then\n\n\[ \n{\int }_{0}^{\pi }\sin \left( {3x}\right) \sin \left( {5x}\right) {dx} = \frac{1}{2}{\int }_{0}^{\pi }\cos \left( {2x}\right) {dx} - \frac{1}{2}{\int }_{0}^{\pi }\cos \left( {8x}\right) {dx}\n\]\n\n\[ \n= {\left. \frac{1}{4}\sin \left( 2x\right) \right| }_{0}^{\pi } - {\left. \frac{1}{16}\sin \left( 8x\right) \right| }_{0}^{\pi }\n\]\n\n\[ \n= 0\text{.} \n\]
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Yes
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To evaluate\n\n\[ \n{\\int }_{0}^{\\frac{\\pi }{2}}\\cos \\left( {3x}\\right) \\cos \\left( {5x}\\right) {dx} \n\]
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we note that, using (2.6.24) with \( a = 3 \) and \( b = 5 \) ,\n\n\[ \n\\cos \\left( {3x}\\right) \\cos \\left( {5x}\\right) = \\frac{1}{2}\\left( {\\cos \\left( {8x}\\right) + \\cos \\left( {-{2x}}\\right) }\\right) = \\frac{1}{2}\\left( {\\cos \\left( {8x}\\right) + \\cos \\left( {2x}\\right) }\\right) .\n\]\n\nHence\n\n\[ \n{\\int }_{0}^{\\frac{\\pi }{2}}\\cos \\left( {3x}\\right) \\cos \\left( {5x}\\right) {dx} = \\frac{1}{2}{\\int }_{0}^{\\frac{\\pi }{2}}\\cos \\left( {8x}\\right) {dx} + \\frac{1}{2}{\\int }_{0}^{\\frac{\\pi }{2}}\\cos \\left( {2x}\\right) {dx}\n\]\n\n\[ \n= {\\left. \\frac{1}{16}\\sin \\left( 8x\\right) \\right| }_{0}^{\\frac{\\pi }{2}} - {\\left. \\frac{1}{4}\\sin \\left( 2x\\right) \\right| }_{0}^{\\frac{\\pi }{2}}\n\]\n\n\[ \n= 0\\text{.}\n\]
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Yes
|
Since the graph of \( y = \sqrt{1 - {x}^{2}} \) for \( 0 \leq x \leq 1 \) is one-quarter of the circle \( {x}^{2} + {y}^{2} = 1 \) (see Figure 2.6.1), we know that\n\n\[{\int }_{0}^{1}\sqrt{1 - {x}^{2}}{dx} = \frac{\pi }{4}\]
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We will now see how to use a change of variable to evaluate this integral using the fundamental theorem. The idea is to make use of the trigonometric identity \( 1 - {\sin }^{2}\left( z\right) = {\cos }^{2}\left( z\right) \) . That is, suppose we let \( x = \sin \left( z\right) \) for \( 0 \leq z \leq \frac{\pi }{2} \) . Then\n\n\[ \sqrt{1 - {x}^{2}} = \sqrt{1 - {\sin }^{2}\left( z\right) } = \sqrt{{\cos }^{2}\left( z\right) } = \left| {\cos \left( z\right) }\right| = \cos \left( z\right) ,\]\n\nwhere the final equality follows since \( \cos \left( z\right) \geq 0 \) for \( 0 \leq z \leq \frac{\pi }{2} \) . Now\n\n\[ {dx} = \cos \left( z\right) {dz} \]\n\nso we have\n\n\[ {\int }_{0}^{1}\sqrt{1 - {x}^{2}}{dx} = {\int }_{0}^{\frac{\pi }{2}}\cos \left( z\right) \cos \left( z\right) {dz} \]\n\n\[ = {\int }_{0}^{\frac{\pi }{2}}{\cos }^{2}\left( z\right) {dz} \]\n\n\[ = \frac{1}{2}{\int }_{0}^{\frac{\pi }{2}}\left( {1 + \cos \left( {2z}\right) }\right) {dz} \]\n\n\[ = {\left. \frac{1}{2}z\right| }_{0}^{\frac{\pi }{2}} + {\left. \frac{1}{4}\sin \left( 2z\right) \right| }_{0}^{\frac{\pi }{2}} \]\n\n\[ = \frac{\pi }{4} \]\n\nas we expected.
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Yes
|
Let \( C \) be the circle with equation \( {x}^{2} + {y}^{2} = 1 \) and let \( L \) be the length of the shorter arc of \( C \) between \( \left( {-\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}}\right) \) and \( \left( {\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}}\right) \) (see Figure 2.6.2). Since the circumference of \( C \) is \( {2\pi } \) and this arc is one-fourth of the circumference of \( C \), we should have \( L = \frac{\pi }{2} \) . We will now show that this agrees with (2.5.28), the formula we derived for computing arc length.
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Now \( y = \sqrt{1 - {x}^{2}} \), so\n\n\[ \frac{dy}{dx} = \frac{1}{2}{\left( 1 - {x}^{2}\right) }^{-\frac{1}{2}}\left( {-{2x}}\right) = - \frac{x}{\sqrt{1 - {x}^{2}}}. \]\n\nHence\n\n\[ \sqrt{1 + {\left( \frac{dy}{dx}\right) }^{2}} = \sqrt{1 + \frac{{x}^{2}}{1 - {x}^{2}}} = \sqrt{\frac{1 - {x}^{2} + {x}^{2}}{1 - {x}^{2}}} = \frac{1}{\sqrt{1 - {x}^{2}}}. \]\n\nHence, by (2.5.28),\n\n\[ L = {\int }_{-\frac{1}{\sqrt{2}}}^{\frac{1}{\sqrt{2}}}\frac{1}{\sqrt{1 - {x}^{2}}}{dx} \]\n\nIf we let\n\n\[ x = \sin \left( z\right) \]\n\n\[ {dx} = \cos \left( z\right) {dz} \]\n\nthen\n\n\[ L = {\int }_{-\frac{\pi }{4}}^{\frac{\pi }{4}}\frac{1}{\sqrt{1 - {\sin }^{2}\left( z\right) }}\cos \left( z\right) {dz} \]\n\n\[ = {\int }_{-\frac{\pi }{4}}^{\frac{\pi }{4}}\frac{\cos \left( z\right) }{\sqrt{{\cos }^{2}\left( z\right) }}{dz} \]\n\n\[ = {\int }_{-\frac{\pi }{4}}^{\frac{\pi }{4}}\frac{\cos \left( z\right) }{\cos \left( z\right) }{dz} \]\n\n\[ = {\int }_{-\frac{\pi }{4}}^{\frac{\pi }{4}}{dz} \]\n\n\[ = \frac{\pi }{2}\text{. } \]
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Yes
|
Example 2.6.20. For a simpler example of the change of variable used in the previous example, consider the integral\n\n\\[ \n{\\int }_{-1}^{1}\\frac{1}{1 + {x}^{2}}{dx} \n\\]\n\nthe area under the curve\n\n\\[ \ny = \\frac{1}{1 + {x}^{2}} \n\\]\n\nover the interval \\( \\left\\lbrack {-1,1}\\right\\rbrack \\) (see Figure 2.6.3).
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If we let\n\n\\[ \nx = \\tan \\left( z\\right) \n\\]\n\n\\[ \n{dx} = {\\sec }^{2}\\left( z\\right) {dz} \n\\]\n\nand note that \\( \\tan \\left( {-\\frac{\\pi }{4}}\\right) = - 1 \\) and \\( \\tan \\left( \\frac{\\pi }{4}\\right) = 1 \\), then\n\n\\[ \n{\\int }_{-1}^{1}\\frac{1}{1 + {x}^{2}}{dx} = {\\int }_{-\\frac{\\pi }{4}}^{\\frac{\\pi }{4}}\\frac{1}{1 + {\\tan }^{2}\\left( z\\right) }{\\sec }^{2}\\left( z\\right) {dz} \n\\]\n\n\\[ \n= {\\int }_{-\\frac{\\pi }{4}}^{\\frac{\\pi }{4}}\\frac{{\\sec }^{2}\\left( z\\right) }{{\\sec }^{2}\\left( z\\right) }{dz} \n\\]\n\n\\[ \n= {\\int }_{-\\frac{\\pi }{4}}^{\\frac{\\pi }{4}}{dz} \n\\]\n\n\\[ \n= \\frac{\\pi }{2} \n\\]
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Yes
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If \( f\left( t\right) = {e}^{5t} \), then \( f \) is the composition of \( h\left( t\right) = {5t} \) and \( g\left( u\right) = {e}^{u} \) . Hence, using the chain rule,
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\[ {f}^{\prime }\left( t\right) = {g}^{\prime }\left( {h\left( t\right) }\right) {h}^{\prime }\left( t\right) = {e}^{5t} \cdot 5 = 5{e}^{5t}. \]
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Yes
|
Example 2.7.2. If \( f\left( x\right) = 6{e}^{-{x}^{2}} \), then
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\[ {f}^{\prime }\left( x\right) = - {12x}{e}^{-{x}^{2}}. \]
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Yes
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\[ {\int }_{0}^{1}{e}^{-t}{dt} = - {\left. {e}^{-t}\right| }_{0}^{1} = - {e}^{-1} + {e}^{0} = 1 - {e}^{-1} \approx {0.6321}. \]
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\[ {\int }_{0}^{1}{e}^{-t}{dt} = - {\left. {e}^{-t}\right| }_{0}^{1} = - {e}^{-1} + {e}^{0} = 1 - {e}^{-1} \approx {0.6321}. \]
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Yes
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To evaluate\n\n\[ \int x{e}^{-{x}^{2}}{dx} \]
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we will use the change of variable\n\n\[ u = - {x}^{2} \]\n\n\[ {du} = - {2xdx} \]\n\nThen\n\n\[ \int x{e}^{-2}{dx} = - \frac{1}{2}\int {e}^{u}{du} = - \frac{1}{2}{e}^{u} + c = - \frac{1}{2}{e}^{-{x}^{2}} + c. \]
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No
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To evaluate\n\n\\[ \n\\int x{e}^{-{2x}}{dx} \n\\]
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we will use integration by parts:\n\n\\[ \nu = x\\;{dv} = {e}^{-{2x}}{dx} \n\\]\n\n\\[ \n{du} = {dx}\\;v = - \\frac{1}{2}{e}^{-{2x}}. \n\\]\n\nThen\n\n\\[ \n\\int x{e}^{-{2x}}{dx} = - \\frac{1}{2}x{e}^{-{2x}} + \\frac{1}{2}\\int {e}^{-{2x}}{dx} = - \\frac{1}{2}x{e}^{-{2x}} - \\frac{1}{4}{e}^{-{2x}} + c. \n\\]
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Yes
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