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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 | \[ f\left( {1 + \epsilon }\right) = 3\left( {1 + \epsilon }\right) + 5 = 8 + {3\epsilon } \simeq 8 = f\left( 1\right) ,\]\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... | Yes |
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 | \[ s\left( {c + \epsilon }\right) = f\left( {c + \epsilon }\right) + g\left( {c + \epsilon }\right) \simeq f\left( c\right) + g\left( c\right) = s\left( c\right) ,\] and so \( s \) is also continuous at \( c \) . | Yes |
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 \( \left( {-\infty ,\infty }\right) \) . | 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 \( \left( {-\infty ,\infty }\right) \) . | No |
The function \( f\left( x\right) = {x}^{2} \) attains neither a maximum nor a minimum value on the interval \( \left( {0,1}\right) \) . | 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 \( 0 < x < a \) . | Yes |
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\\lef... | 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 \( f\\left( 0\\right) = 0 \) . | Yes |
If \( y = {x}^{2} \), then, for any nonzero infinitesimal \( {dx} \) | \[ {dy} = {\left( x + dx\right) }^{2} - {x}^{2} = \left( {{x}^{2} + {2xdx} + {\left( dx\right) }^{2}}\right) - {x}^{2} = \left( {{2x} + {dx}}\right) {dx}. \] Hence \[ \frac{dy}{dx} = {2x} + {dx} \simeq {2x} \] and so the derivative of \( y \) with respect to \( x \) is \[ \frac{dy}{dx} = {2x} \] | Yes |
If \( f\left( x\right) = {4x} \), then, for any nonzero infinitesimal \( {dx} \) | \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 ... | Yes |
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 | \[ {\left. \frac{dy}{dx}\right| }_{x = 3} = \left( 2\right) \left( 3\right) = 6 \] | Yes |
If \( f\left( x\right) = {x}^{\frac{2}{3}} \), then for any infinitesimal \( {dx} \) , | \[ f\left( {0 + {dx}}\right) - f\left( 0\right) = f\left( {dx}\right) = {\left( dx\right) }^{\frac{2}{3}}, \] 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} = \f... | Yes |
If \( f\left( x\right) = \sqrt{x} \), then, as we have seen above, for any \( x > 0 \) and any nonzero infinitesimal \( {dx} \), | \[ f\left( {x + {dx}}\right) - f\left( x\right) = \sqrt{x + {dx}} - \sqrt{x} \] \[ = \left( {\sqrt{x + {dx}} - \sqrt{x}}\right) \frac{\sqrt{x + {dx}} + \sqrt{x}}{\sqrt{x + {dx}} + \sqrt{x}} \] \[ = \frac{\left( {x + {dx}}\right) - x}{\sqrt{x + {dx}} + \sqrt{x}} \] \[ = \frac{dx}{\sqrt{x + {dx}} + \sqrt{x}}. \] It now f... | Yes |
The function \( f\left( x\right) = \sqrt{x} \) is differentiable on \( \left( {0,\infty }\right) \) . | 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 \( {dx} \). | Yes |
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 | \[ {s}^{\prime }\left( x\right) = {f}^{\prime }\left( x\right) + {g}^{\prime }\left( x\right) \] | Yes |
If \( y = {x}^{2} + \sqrt{x} \), then, using our results from the previous section, | \[ \frac{dy}{dx} = \frac{d}{dx}\left( {x}^{2}\right) + \frac{d}{dx}\left( \sqrt{x}\right) = {2x} + \frac{1}{2\sqrt{x}}. \] | Yes |
Theorem 1.7.3. If \( c \) is a real constant, \( f \) is differentiable, and \( g\left( x\right) = {cf}\left( x\right) \) , then | \[ {g}^{\prime }\left( x\right) = c{f}^{\prime }\left( x\right) \] | Yes |
Example 1.7.3. If \( y = 5{x}^{2} \), then | \n\[
\frac{dy}{dx} = 5\frac{d}{dx}\left( {x}^{2}\right) = 5\left( {2x}\right) = {10x}.
\] | Yes |
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 | \[ {p}^{\prime }\left( x\right) = f\left( x\right) {g}^{\prime }\left( x\right) + g\left( x\right) {f}^{\prime }\left( x\right) . \] | Yes |
We may use the product rule to find a formula for the derivative of a positive integer power of \( x \) . | 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 t... | Yes |
When \( n = {34} \), the power rule shows that | \[ \frac{d}{dx}{x}^{34} = {34}{x}^{33} \] | Yes |
Example 1.7.6. If \( f\left( x\right) = {14}{x}^{5} \), then, combining the power rule with our result for constant multiples, | \[ {f}^{\prime }\left( x\right) = {14}\left( {5{x}^{4}}\right) = {70}{x}^{4}. \] | Yes |
If \( n < 0 \) is an integer, then | \( \frac{d}{dx}{x}^{n} = \frac{d}{dx}\left( \frac{1}{{x}^{-n}}\right) = - \frac{1}{{x}^{-{2n}}} \cdot \left( {-n{x}^{-n - 1}}\right) = n{x}^{n - 1}. \) | Yes |
If \( f\left( x\right) = 3{x}^{2} - \frac{5}{{x}^{7}} \) then find \( f'\left( x\right) \). | then \( f\left( x\right) = 3{x}^{2} - 5{x}^{-7} \), and so \[ {f}^{\prime }\left( x\right) = {6x} + {35}{x}^{-8} = {6x} + \frac{35}{{x}^{8}}. \] | Yes |
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... | \[ {q}^{\prime }\left( x\right) = \frac{g\left( x\right) {f}^{\prime }\left( x\right) - f\left( x\right) {g}^{\prime }\left( x\right) }{{\left( g\left( x\right) \right) }^{2}}. \] | Yes |
If\n\n\[ f\left( x\right) = \frac{3{x}^{2} - {6x} + 4}{{x}^{2} + 1} \]\n\nthen\n\n\[ {f}^{\prime }\left( x\right) = \frac{\left( {{x}^{2} + 1}\right) \left( {{6x} - 6}\right) - \left( {3{x}^{2} - {6x} + 4}\right) \left( {2x}\right) }{{\left( {x}^{2} + 1\right) }^{2}} \] | \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\[ = \frac{6{x}^{2} - {2x} - 6}{{\left( {x}^{2} + 1\right) }^{2}}. \] | Yes |
Example 1.7.12. We may use either 1.7.30 or 1.7.37 to differentiate\n\n\[ y = \frac{5}{{x}^{2} + 1}. \] | In either case, we obtain\n\n\[ \frac{dy}{dx} = - \frac{5}{{\left( {x}^{2} + 1\right) }^{2}}\frac{d}{dx}\left( {{x}^{2} + 1}\right) = - \frac{10x}{{\left( {x}^{2} + 1\right) }^{2}}. \] | Yes |
Theorem 1.7.9. If \( y \) is a differentiable function of \( u \) and \( u \) is a differentiable function of \( x \), then\n\n\[ \frac{dy}{dx} = \frac{dy}{du}\frac{du}{dx} \] | 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}^... | Yes |
If \( y = 3{u}^{2} \) and \( u = {2x} + 1 \), then | \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,
... | Yes |
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 \( g\left( x\right) = {x}^{2} + 1 \). | 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} ... | Yes |
Example 1.7.16. With \( n = {10} \) and \( g\left( x\right) = {x}^{2} + 3 \), we have | \[ \frac{d}{dx}{\left( {x}^{2} + 3\right) }^{10} = {10}{\left( {x}^{2} + 3\right) }^{9}\left( {2x}\right) = {20x}{\left( {x}^{2} + 3\right) }^{9}. \] | Yes |
If \( f\left( x\right) = \frac{15}{{\left( {x}^{4} + 5\right) }^{2}} \), then what is \( f'(x) \)? | then we may apply the previous result with \( n = - 2 \) and \( g\left( x\right) = {x}^{4} + 5 \) to obtain\n\n\[ {f}^{\prime }\left( x\right) = - {30}{\left( {x}^{4} + 5\right) }^{-3}\left( {4{x}^{3}}\right) = - \frac{{120}{x}^{3}}{{\left( {x}^{4} + 5\right) }^{3}}. \] | Yes |
Example 1.7.18. With \( r = \frac{1}{2} \) in the previous theorem, we have | \[ \frac{d}{dx}\sqrt{x} = \frac{1}{2}{x}^{-\frac{1}{2}} = \frac{1}{2\sqrt{x}} \] in agreement with our earlier direct computation. | Yes |
If \( y = {x}^{\frac{2}{3}} \), then | \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 . | Yes |
If \( f\left( x\right) = \sqrt{{x}^{2} + 1} \) | \[ {f}^{\prime }\left( x\right) = \frac{1}{2}{\left( {x}^{2} + 1\right) }^{-\frac{1}{2}}\left( {2x}\right) = \frac{x}{\sqrt{{x}^{2} + 1}}. \] | Yes |
Example 1.7.21. If\n\n\\[ \n g\\left( t\\right) = \\frac{1}{{t}^{4} + 5} \n\\]\n\nthen | \\[ \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\\] | Yes |
Using the chain rule, | \[ \frac{d}{dx}\cos \left( {4x}\right) = - \sin \left( {4x}\right) \frac{d}{dt}\left( {4x}\right) = - 4\sin \left( {4x}\right) . \] | Yes |
If \( f\left( t\right) = {\sin }^{2}\left( t\right) \), then, again using the chain rule, | \[ {f}^{\prime }\left( t\right) = 2\sin \left( t\right) \frac{d}{dt}\sin \left( t\right) = 2\sin \left( t\right) \cos \left( t\right) . \] | Yes |
If \( g\left( x\right) = \cos \left( {x}^{2}\right) \), then | \[ {g}^{\prime }\left( x\right) = - \sin \left( {x}^{2}\right) \left( {2x}\right) = - {2x}\cos \left( {x}^{2}\right) . \] | Yes |
Example 1.7.25. If \( f\left( x\right) = {\sin }^{3}\left( {4x}\right) \), then, using the chain rule twice, | \[ {f}^{\prime }\left( x\right) = 3{\sin }^{2}\left( {4x}\right) \frac{d}{dx}\sin \left( {4x}\right) = {12}{\sin }^{2}\left( {4x}\right) \cos \left( {4x}\right) . \] | Yes |
If \( f\left( x\right) = {x}^{5} - 6{x}^{2} + 5 \), then | \[ {f}^{\prime }\left( x\right) = 5{x}^{4} - {12x} \]\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\[ y = \frac{101}{6}\left( {x + \frac{1}{2}}\right) + \frac{111}{32}. \] | Yes |
Theorem 1.9.1. If \( f \) is differentiable on \( \left( {a, b}\right) \) and attains a maximum, or a minimum, value at \( c \), then \( {f}^{\prime }\left( c\right) = 0 \) . | 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 no... | 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 \( {f}^{\prime }\left( c\right) = 0 \) . | 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 d... | 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) ... | 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... | Yes |
Let \( f\left( x\right) = 2{x}^{3} - 3{x}^{2} - {12x} + 1 \) . Then | \[ {f}^{\prime }\left( x\right) = 6{x}^{2} - {6x} - {12} = 6\left( {{x}^{2} - x - 2}\right) = 6\left( {x - 2}\right) \left( {x + 1}\right) . \] 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 <... | Yes |
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, | \[ \cos \left( x\right) < - \frac{1}{2} \] 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 \... | Yes |
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... | 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... | Yes |
Consider the function \( g\left( t\right) = t - 2\cos \left( t\right) \) defined on the interval \( \left\lbrack {0,{2\pi }}\right\rbrack \) . | \[ {g}^{\prime }\left( t\right) = 1 + 2\sin \left( t\right) \] 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 stati... | Yes |
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)... | 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} =... | Yes |
Consider the problem of finding the extreme values of\n\n\[ y = 4{x}^{2} + \frac{1}{x} \]\n\non the interval \( \left( {0,\infty }\right) \) . | 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, \( 0 < x < \frac{1}{2} \). S... | Yes |
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 ... | 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}^{... | Yes |
Using \( a = 2 \) and \( b = 1 \) ,(1.11.1) becomes, after multiplying both sides of the equation by 4 ,\n\n\[ \n{x}^{2} + 4{y}^{2} = 4 \n\] | 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 \( y \neq 0 \) (corre... | Yes |
Consider the hyperbola \( H \) with equation\n\n\[ \n{x}^{2} - {4xy} + {y}^{2} = 4 \n\] | 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\nFo... | Yes |
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 ... | 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... | Yes |
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 af... | 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}... | Yes |
If \( y = 4{x}^{5} - 3{x}^{2} + 4 \), then | \[ \frac{dy}{dx} = {20}{x}^{4} - {6x} \] and so \[ \frac{{d}^{2}y}{d{x}^{2}} = {80}{x}^{3} - 6 \] | Yes |
If\n\n\[ f\\left( x\\right) = \\frac{1}{x} \]\n\nthen\n\n\[ {f}^{\\prime }\\left( x\\right) = - \\frac{1}{{x}^{2}} \] | \[ {f}^{\\prime \\prime }\\left( x\\right) = \\frac{2}{{x}^{3}} \]\n\n\[ {f}^{\\prime \\prime \\prime }\\left( x\\right) = - \\frac{6}{{x}^{4}} \]\n\nand\n\n\[ {f}^{\\left( 4\\right) }\\left( x\\right) = \\frac{24}{{x}^{5}}. \] | No |
If \( f\left( x\right) = 2{x}^{3} - 3{x}^{2} - {12x} + 1 \), then | \[ {f}^{\prime }\left( x\right) = 6{x}^{2} - {6x} - {12} \] 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 downwa... | Yes |
If \( f\left( x\right) = {x}^{4} - {x}^{3} \), then | \[ {f}^{\prime }\left( x\right) = 4{x}^{3} - 3{x}^{2} = {x}^{2}\left( {{4x} - 3}\right) \] 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 \[ {f}^{\prime \prime }\left( 0\right) = 0 \] a... | Yes |
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 \( x \). | 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... | Yes |
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 | \n\n\[ F\left( x\right) = \frac{3}{2}{x}^{2} + {4x} + c \]\n\nfor some constant \( c \) . | Yes |
Since\n\n\[ \frac{d}{dx}\left( {4{x}^{3} - \sin \left( x\right) }\right) = {12}{x}^{2} - \cos \left( x\right) ,\] | it follows that\n\n\[ \int \left( {{12}{x}^{2} - \cos \left( x\right) }\right) {dx} = 4{x}^{3} - \sin \left( x\right) + c, \] | Yes |
Suppose we wish to find the integral \( F\\left( x\\right) \) of \( f\\left( x\\right) = 5{x}^{2} - 7 \) for which \( F\\left( 1\\right) = {10} \). | 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 - \\fr... | Yes |
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 | \[ x\left( t\right) = - \int {20}\sin \left( {5t}\right) {dt} = 4\cos \left( {5t}\right) + c\text{ centimeters } \]\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) =... | Yes |
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. | 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 ... | Yes |
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 \( \left\lbrack {-1,1}\right\rbrack \) . ... | We will eventually see, in Example 2.6.20, that\n\n\[ {\int }_{-1}^{1}\frac{1}{1 + {x}^{2}}{dx} = \frac{\pi }{2} \approx {1.5708}. \] | No |
If \( \alpha \) is any infinitesimal, than \( \alpha \sim o\left( 1\right) \) | since \( \frac{\alpha }{1} = \alpha \) is an infinitesimal. | Yes |
If \( \epsilon \) is any nonzero infinitesimal, then \( {\epsilon }^{2} \sim o\left( \epsilon \right) \) | \[ \frac{{\epsilon }^{2}}{\epsilon } = \epsilon \simeq 0 \] | Yes |
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 \... | 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\le... | Yes |
To evaluate\n\n\[ \n{\int }_{0}^{1}{xdx} \n\] | 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\[ \n{\int }_{0}^{1}{xdx} = f\left( 1\right) - f\left( 0\right) = \frac{1}{2} - 0 = \frac{1}{2} \n\] | Yes |
Since\n\n\[ \int {x}^{2}{dx} = \frac{1}{3}{x}^{3} + c \] | we have\n\n\[ {\int }_{1}^{2}{x}^{2}{dx} = {\left. \frac{1}{3}{x}^{3}\right| }_{1}^{2} = \frac{8}{3} - \frac{1}{3} = \frac{7}{3}. \] | Yes |
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\\] | \\[ \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\\] | Yes |
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... | \[ = {\left. \left( \frac{1}{2}{x}^{2} + 2x - \frac{1}{3}{x}^{3}\right) \right| }_{-1}^{2} \]\n\n\[ = \left( {2 + 4 - \frac{8}{3}}\right) - \left( {\frac{1}{2} - 2 + \frac{1}{3}}\right) \]\n\n\[ = \frac{9}{2}\text{.} \] | Yes |
If \( A \) is the area of the region \( R \) beneath the graph of \( y = \sin \left( x\right) \) over the interval \( \left\lbrack {0,\pi }\right\rbrack \) | \[ A = {\int }_{0}^{\pi }\sin \left( x\right) {dx} = - {\left. \cos \left( x\right) \right| }_{0}^{\pi } = 1 + 1 = 2. \] | Yes |
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 \( z \) -... | \[ A\left( z\right) = \pi \left( {1 - {z}^{2}}\right) \] 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\... | Yes |
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 \( z \) -axi... | If \( V \) is the volume of \( B \), then\n\n\[ V = {\int }_{0}^{1}{\pi zdz} = {\left. \pi \frac{{z}^{2}}{2}\right| }_{0}^{1} = \frac{\pi }{2}. \] | Yes |
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 ... | \[ A\left( z\right) = \pi {\left( {z}^{\frac{1}{4}}\right) }^{2} - \pi {\left( \sqrt{z}\right) }^{2} = \pi \left( {\sqrt{z} - z}\right) . \] 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)... | Yes |
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\[ L = {\int }_{0}^{1}\sqrt{1 + \frac{9}{4}x}{dx} \] | 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 + \f... | Yes |
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 \( {f}^{\prime }\left( x\right) = {2x} \) , | \[ L = {\int }_{0}^{1}\sqrt{1 + 4{x}^{2}}{dx} \]\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} = ... | No |
To evaluate\n\n\[ \int {2x}\sqrt{1 + {x}^{2}}{dx} \] | let\n\n\[ u = 1 + {x}^{2} \]\n\n\[ {du} = {2xdx}\text{.} \]\n\nThen\n\n\[ \int {2x}\sqrt{1 + {x}^{2}}{dx} = \int \sqrt{u}{du} = \frac{2}{3}{u}^{\frac{3}{2}} + c = \frac{2}{3}{\left( 1 + {x}^{2}\right) }^{\frac{3}{2}} + c. \] | Yes |
To evaluate\n\n\[ \int x\sin \left( {x}^{2}\right) {dx} \] | 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\... | Yes |
Note that we could evaluate the integral\n\n\[ \int \cos \left( {4x}\right) {dx} \] | using the substitution\n\n\[ u = {4x} \]\n\n\[ {du} = {4dx} \]\n\nwhich gives us\n\n\[ \int \cos \left( {4x}\right) {dx} = \frac{1}{4}\int \cos \left( u\right) {du} = \frac{1}{4}\sin \left( u\right) + c = \frac{1}{4}\sin \left( {4x}\right) + c. \] | Yes |
To evaluate\n\n\[ \int {\cos }^{2}\left( {5x}\right) \sin \left( {5x}\right) {dx} \] | let\n\n\[ u = \cos \left( {5x}\right) \]\n\n\[ {du} = - 5\sin \left( {5x}\right) {dx} \]\n\nThen\n\n\[ \int {\cos }^{2}\left( {5x}\right) \sin \left( {5x}\right) {dx} = - \frac{1}{5}\int {u}^{2}{du} = - \frac{1}{15}{u}^{3} + c = - \frac{1}{15}{\cos }^{3}\left( {5x}\right) + c. \] | Yes |
To evaluate\n\n\[ \n{\int }_{0}^{1}\frac{x}{\sqrt{1 + {x}^{2}}}{dx} \n\] | 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\[ \n{\int }_{0}^{1}\frac{x}{\sqrt{1 + {x}^{2}}}{dx} = \frac{1}{2}{\int }_{1}^{2}\frac{1}{\sqrt{u}}{du} = {\left. \sqrt{u}\right| }_{1}^{2} = \sqrt{2} - 1. \n\] | Yes |
To evaluate\n\n\[ \n{\int }_{0}^{\frac{\pi }{4}}{\cos }^{2}\left( {2x}\right) \sin \left( {2x}\right) {dx} \n\] | 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\[ \n{\int }_{0}^{\frac{\pi }{4}}{\cos }^{2}\left( {2x}\right) \sin \left( {2x}\right) {dx} = - \frac{1}{2}{\int }_{1}^{0}{u}^{2}{du}. \n\]... | Yes |
Consider the integral\n\n\[ \int x\cos \left( x\right) {dx} \] | 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)... | No |
To evaluate\n\n\[ \n{\int }_{0}^{\pi }{x}^{2}\sin \left( x\right) {dx} \n\] | 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} ... | Yes |
To evaluate\n\n\[ \n{\int }_{0}^{1}x\sqrt{1 + x}{dx} \n\] | 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}... | Yes |
To evaluate the integral\n\n\[ \n{\int }_{0}^{\pi }{\sin }^{2}\left( x\right) {dx} \n\] | 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. ... | Yes |
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 | \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 } + ... | Yes |
Suppose \( n \geq 2 \) is an integer and we wish to evaluate\n\n\[ \n{\int }_{0}^{\pi }{\sin }^{n}\left( x\right) {dx} \n\] | 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} = ... | Yes |
Example 2.6.13. An alternative to using a reduction formula in the last example begins with noting that\n\n\[ \n{\int }_{0}^{\pi }{\sin }^{5}\left( x\right) {dx} = {\int }_{0}^{\pi }{\sin }^{4}\left( x\right) \sin \left( x\right) {dx} \n\] | \[ \n= {\int }_{0}^{\pi }{\left( {\sin }^{2}\left( x\right) \right) }^{2}\sin \left( x\right) {dx} \n\]\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)... | Yes |
To evaluate\n\n\[ \n{\int }_{0}^{\pi }\sin \left( {2x}\right) \cos \left( {3x}\right) {dx} \n\] | 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}^{\... | Yes |
To evaluate\n\n\[ \n{\int }_{0}^{\pi }\sin \left( {3x}\right) \sin \left( {5x}\right) {dx} \n\] | 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 hav... | Yes |
To evaluate\n\n\[ \n{\\int }_{0}^{\\frac{\\pi }{2}}\\cos \\left( {3x}\\right) \\cos \\left( {5x}\\right) {dx} \n\] | 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\nHenc... | 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\[{\int }_{0}^{1}\sqrt{1 - {x}^{2}}{dx} = \frac{\pi }{4}\] | 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 \( 0 \leq z \leq \frac{\pi }{2} \)... | 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 \... | 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}^{... | 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 Figu... | 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{\\p... | Yes |
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, | \[ {f}^{\prime }\left( t\right) = {g}^{\prime }\left( {h\left( t\right) }\right) {h}^{\prime }\left( t\right) = {e}^{5t} \cdot 5 = 5{e}^{5t}. \] | Yes |
Example 2.7.2. If \( f\left( x\right) = 6{e}^{-{x}^{2}} \), then | \[ {f}^{\prime }\left( x\right) = - {12x}{e}^{-{x}^{2}}. \] | Yes |
\[ {\int }_{0}^{1}{e}^{-t}{dt} = - {\left. {e}^{-t}\right| }_{0}^{1} = - {e}^{-1} + {e}^{0} = 1 - {e}^{-1} \approx {0.6321}. \] | \[ {\int }_{0}^{1}{e}^{-t}{dt} = - {\left. {e}^{-t}\right| }_{0}^{1} = - {e}^{-1} + {e}^{0} = 1 - {e}^{-1} \approx {0.6321}. \] | Yes |
To evaluate\n\n\[ \int x{e}^{-{x}^{2}}{dx} \] | we will use the change of variable\n\n\[ u = - {x}^{2} \]\n\n\[ {du} = - {2xdx} \]\n\nThen\n\n\[ \int x{e}^{-2}{dx} = - \frac{1}{2}\int {e}^{u}{du} = - \frac{1}{2}{e}^{u} + c = - \frac{1}{2}{e}^{-{x}^{2}} + c. \] | No |
To evaluate\n\n\\[ \n\\int x{e}^{-{2x}}{dx} \n\\] | 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\\] | Yes |
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