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Theorem 5.1.1 (The Product Rule) If \( f\\left( x\\right) \) and \( g\\left( x\\right) \) are differentiable, then\n\n\[ \n\\frac{d}{dx}f\\left( x\\right) g\\left( x\\right) = f\\left( x\\right) {g}^{\\prime }\\left( x\\right) + {f}^{\\prime }\\left( x\\right) g\\left( x\\right) .\n\] | Proof From the limit definition of the derivative, write\n\n\[ \n\\frac{d}{dx}\\left( {f\\left( x\\right) g\\left( x\\right) }\\right) = \\mathop{\\lim }\\limits_{{h \\rightarrow 0}}\\frac{f\\left( {x + h}\\right) g\\left( {x + h}\\right) - f\\left( x\\right) g\\left( x\\right) }{h}\n\]\n\nNow we use the exact same tri... | Yes |
Example 5.1.2 Let \( f\left( x\right) = \left( {{x}^{2} + 1}\right) \) and \( g\left( x\right) = \left( {{x}^{3} - {3x}}\right) \) . Compute:\n\n\[ \n\frac{d}{dx}f\left( x\right) g\left( x\right) \n\] | Solution Write\n\n\[ \n\frac{d}{dx}f\left( x\right) g\left( x\right) = f\left( x\right) {g}^{\prime }\left( x\right) + {f}^{\prime }\left( x\right) g\left( x\right) \n\]\n\n\[ \n= \left( {{x}^{2} + 1}\right) \left( {3{x}^{2} - 3}\right) + {2x}\left( {{x}^{3} - {3x}}\right) \text{.} \n\] | Yes |
Theorem 5.2.1 (The Quotient Rule) If \( f\\left( x\\right) \) and \( g\\left( x\\right) \) are differentiable, then\n\n\[ \n\\frac{d}{dx}\\frac{f\\left( x\\right) }{g\\left( x\\right) } = \\frac{{f}^{\\prime }\\left( x\\right) g\\left( x\\right) - f\\left( x\\right) {g}^{\\prime }\\left( x\\right) }{g{\\left( x\\right)... | Proof First note that if we knew how to compute\n\n\[ \n\\frac{d}{dx}\\frac{1}{g\\left( x\\right) }\n\]\n\nthen we could use the product rule to complete our proof. Write\n\n\[ \n\\frac{d}{dx}\\frac{1}{g\\left( x\\right) } = \\mathop{\\lim }\\limits_{{h \\rightarrow 0}}\\frac{\\frac{1}{g\\left( {x + h}\\right) } - \\fr... | Yes |
Example 5.2.2 Compute:\n\n\[ \frac{d}{dx}\frac{{x}^{2} + 1}{{x}^{3} - {3x}} \] | Solution Write\n\n\[ \frac{d}{dx}\frac{{x}^{2} + 1}{{x}^{3} - {3x}} = \frac{{2x}\left( {{x}^{3} - {3x}}\right) - \left( {{x}^{2} + 1}\right) \left( {3{x}^{2} - 3}\right) }{{\left( {x}^{3} - 3x\right) }^{2}} \]\n\n\[ = \frac{-{x}^{4} - 6{x}^{2} + 3}{{\left( {x}^{3} - 3x\right) }^{2}}\text{. } \] | Yes |
Example 5.2.3 Compute\n\n\\[ \n\\frac{d}{dx}\\frac{{625} - {x}^{2}}{\\sqrt{x}} \n\\]\n\nin two ways. First using the quotient rule and then using the product rule. | Solution First, we'll compute the derivative using the quotient rule. Write\n\n\\[ \n\\frac{d}{dx}\\frac{{625} - {x}^{2}}{\\sqrt{x}} = \\frac{\\left( {-{2x}}\\right) \\left( \\sqrt{x}\\right) - \\left( {{625} - {x}^{2}}\\right) \\left( {\\frac{1}{2}{x}^{-1/2}}\\right) }{x}.\n\\]\n\n\nSecond, we’ll compute the derivativ... | Yes |
Theorem 6.1.1 (Chain Rule) If \( f\left( x\right) \) and \( g\left( x\right) \) are differentiable, then\n\n\[ \frac{d}{dx}f\left( {g\left( x\right) }\right) = {f}^{\prime }\left( {g\left( x\right) }\right) {g}^{\prime }\left( x\right) . \] | Proof Let \( {g}_{0} \) be some \( x \) -value and consider the following:\n\n\[ {f}^{\prime }\left( {g}_{0}\right) = \mathop{\lim }\limits_{{h \rightarrow 0}}\frac{f\left( {{g}_{0} + h}\right) - f\left( {g}_{0}\right) }{h}. \]\n\nSet \( h = g - {g}_{0} \) and we have\n\n\[ {f}^{\prime }\left( {g}_{0}\right) = \mathop{... | No |
Compute:\n\[ \frac{d}{dx}{\left( 1 + 2x\right) }^{5} \] | Solution Set \( f\left( x\right) = {x}^{5} \) and \( g\left( x\right) = 1 + {2x} \), now\n\n\[ {f}^{\prime }\left( x\right) = 5{x}^{4}\;\text{ and }\;{g}^{\prime }\left( x\right) = 2. \]\n\nHence\n\n\[ \frac{d}{dx}{\left( 1 + 2x\right) }^{5} = \frac{d}{dx}f\left( {g\left( x\right) }\right) \]\n\n\[ = {f}^{\prime }\left... | Yes |
Example 6.1.3 Compute:\n\n\[ \frac{d}{dx}\sqrt{1 + \sqrt{x}} \] | Solution Set \( f\left( x\right) = \sqrt{x} \) and \( g\left( x\right) = 1 + x \) . Hence,\n\n\[ \sqrt{1 + \sqrt{x}} = f\left( {g\left( {f\left( x\right) }\right) }\right) \;\text{ and }\;\frac{d}{dx}f\left( {g\left( {f\left( x\right) }\right) }\right) = {f}^{\prime }\left( {g\left( {f\left( x\right) }\right) }\right) ... | Yes |
Compute:\n\n\\[ \n\\frac{d}{dx}\\frac{{x}^{3}}{{x}^{2} + 1} \n\\] | Solution Rewriting this as\n\n\\[ \n\\frac{d}{dx}{x}^{3}{\\left( {x}^{2} + 1\\right) }^{-1} \n\\]\n\nset \\( f\\left( x\\right) = {x}^{-1} \\) and \\( g\\left( x\\right) = {x}^{2} + 1 \\) . Now\n\n\\( {x}^{3}{\\left( {x}^{2} + 1\\right) }^{-1} = {x}^{3}f\\left( {g\\left( x\\right) }\\right) \\; \\) and \\( \\;\\frac{d}... | Yes |
Compute \( \frac{dy}{dx} \) for the curve defined by \( {x}^{3} + {y}^{3} = {9xy} \). | Solution Starting with\n\n\[ {x}^{3} + {y}^{3} = {9xy} \]\n\nwe apply the differential operator \( \frac{d}{dx} \) to both sides of the equation to obtain\n\n\[ \frac{d}{dx}\left( {{x}^{3} + {y}^{3}}\right) = \frac{d}{dx}{9xy} \]\n\nApplying the sum rule we see\n\n\[ \frac{d}{dx}{x}^{3} + \frac{d}{dx}{y}^{3} = \frac{d}... | Yes |
Theorem 6.2.2 (The Derivative of the Natural Logarithm)\n\n\[ \frac{d}{dx}\ln \left( x\right) = \frac{1}{x} \] | Proof Recall\n\n\[ \ln \left( x\right) = y\; \Leftrightarrow \;{e}^{y} = x. \]\n\nHence\n\n\[ {e}^{y} = x \]\n\n\[ \frac{d}{dx}{e}^{y} = \frac{d}{dx}x \]\nDifferentiate both sides.\n\n\[ {e}^{y}\frac{dy}{dx} = 1 \]\nImplicit differentiation.\n\n\[ \frac{dy}{dx} = \frac{1}{{e}^{y}} = \frac{1}{x} \]\n\nSince \( y = \ln \... | Yes |
Example 6.3.1 Compute:\n\n\\[ \n\\frac{d}{dx}\\frac{{x}^{9}{e}^{4x}}{\\sqrt{{x}^{2} + 4}} \n\\] | Solution While we could use the product and quotient rule to solve this prob-\n\n---\n\nRecall the properties of logarithms:\n\n- \\( {\\log }_{b}\\left( {xy}\\right) = {\\log }_{b}\\left( x\\right) + {\\log }_{b}\\left( y\\right) \\)\n\n- \\( {\\log }_{b}\\left( {x/y}\\right) = {\\log }_{b}\\left( x\\right) - {\\log }... | Yes |
Compute: \[ \frac{d}{dx}{x}^{x} \] | The function \( {x}^{x} \) is tricky to differentiate. We cannot use the power rule, as the exponent is not a constant. However, if we set \( f\left( x\right) = {x}^{x} \) we can write\n\n\[ \ln \left( {f\left( x\right) }\right) = \ln \left( {x}^{x}\right) \]\n\n\[ = x\ln \left( x\right) \text{.} \]\n\nDifferentiating ... | Yes |
Theorem 6.3.3 (Power Rule) For any real number \( n \) , \[ \frac{d}{dx}{x}^{n} = n{x}^{n - 1} \] | Proof We will use logarithmic differentiation. Set \( f\left( x\right) = {x}^{n} \) . Write \[ \ln \left( {f\left( x\right) }\right) = \ln \left( {x}^{n}\right) \] \[ = n\ln \left( x\right) \text{.} \] Now differentiate both sides, and solve for \( {f}^{\prime }\left( x\right) \) \[ \frac{{f}^{\prime }\left( x\right) }... | Yes |
Theorem 7.1.1 (The Derivative of \( \\sin \\left( x\\right) \\) )\n\n\[ \n\\frac{d}{dx}\\sin \\left( x\\right) = \\cos \\left( x\\right) \n\] | Proof Using the definition of the derivative, write\n\n\[ \n\\mathop{\\lim }\\limits_{{h \\rightarrow 0}}\\frac{\\cos \\left( h\\right) - 1}{h} = \\mathop{\\lim }\\limits_{{h \\rightarrow 0}}\\left( {\\frac{\\cos \\left( h\\right) - 1}{h} \\cdot \\frac{\\cos \\left( h\\right) + 1}{\\cos \\left( h\\right) + 1}}\\right) ... | No |
Theorem 7.1.2 (The Derivative of \( \cos \left( x\right) \) ) \( \cos \left( x\right) \) is positive when \( \sin \left( x\right) \) is increasing, and that \( \cos \left( x\right) \) is negative when \( \sin \left( x\right) \) is decreasing. | \[ \frac{d}{dx}\cos \left( x\right) = - \sin \left( x\right) \] Proof Recall that \[ \cos \left( x\right) = \sin \left( {x + \frac{\pi }{2}}\right) \] \[ \sin \left( x\right) = - \cos \left( {x + \frac{\pi }{2}}\right) . \] Now: \[ \frac{d}{dx}\cos \left( x\right) = \frac{d}{dx}\sin \left( {x + \frac{\pi }{2}}\right) \... | Yes |
Theorem 7.1.3 (The Derivative of \( \tan \left( x\right) \) ) | \[ \frac{d}{dx}\tan \left( x\right) = {\sec }^{2}\left( x\right) \] Proof We’ll rewrite \( \tan \left( x\right) \) as \( \frac{\sin \left( x\right) }{\cos \left( x\right) } \) and use the quotient rule. Write \[ \frac{d}{dx}\tan \left( x\right) = \frac{d}{dx}\frac{\sin \left( x\right) }{\cos \left( x\right) } \] \[ = \... | Yes |
Theorem 7.1.4 (The Derivative of \( \sec \left( x\right) \) ) | \[ \frac{d}{dx}\sec \left( x\right) = \sec \left( x\right) \tan \left( x\right) \] Proof We’ll rewrite \( \sec \left( x\right) \) as \( {\left( \cos \left( x\right) \right) }^{-1} \) and use the power rule and the chain rule. Write \[ \frac{d}{dx}\sec \left( x\right) = \frac{d}{dx}{\left( \cos \left( x\right) \right) }... | Yes |
Theorem 7.1.5 (The Derivatives of Trigonometric Functions)\n\n\[ \text{-}\frac{d}{dx}\sin \left( x\right) = \cos \left( x\right) \text{.} \] | \[ \text{-}\frac{d}{dx}\sin \left( x\right) = \cos \left( x\right) \text{.} \] | No |
Theorem 7.2.1 (The Derivative of \( \arcsin \left( y\right) \) ) | \[ \frac{d}{dy}\arcsin \left( y\right) = \frac{1}{\sqrt{1 - {y}^{2}}} \] Proof To start, note that the Inverse Function Theorem, Theorem 6.2.3 assures us that this derivative actually exists. Recall \[ \arcsin \left( y\right) = \vartheta \; \Rightarrow \;\sin \left( \vartheta \right) = y. \] Hence \[ \sin \left( \varth... | Yes |
Theorem 7.2.2 (The Derivative of \( \arccos \left( y\right) \) ) | \[ \frac{d}{dy}\arccos \left( y\right) = \frac{-1}{\sqrt{1 - {y}^{2}}} \] Proof To start, note that the Inverse Function Theorem, Theorem 6.2.3 assures us that this derivative actually exists. Recall \[ \arccos \left( y\right) = \vartheta \; \Rightarrow \;\cos \left( \vartheta \right) = y. \] Hence \[ \cos \left( \vart... | Yes |
Theorem 7.2.3 (The Derivative of \( \arctan \left( y\right) \) ) | \[ \frac{d}{dy}\arctan \left( y\right) = \frac{1}{1 + {y}^{2}} \] Proof To start, note that the Inverse Function Theorem, Theorem 6.2.3 assures us that this derivative actually exists. Recall \[ \arctan \left( y\right) = \vartheta \; \Rightarrow \;\tan \left( \vartheta \right) = y. \] Hence \[ \tan \left( \vartheta \ri... | Yes |
Theorem 7.2.4 (The Derivatives of Inverse Trigonometric Functions)\n\n\[ \text{-}\frac{d}{dy}\arcsin \left( y\right) = \frac{1}{\sqrt{1 - {y}^{2}}}\text{.} \] | \[ \text{-}\frac{d}{dy}\arcsin \left( y\right) = \frac{1}{\sqrt{1 - {y}^{2}}}\text{.} \] | Yes |
Theorem 8.1.1 (L’Hôpital’s Rule) Let \( f\left( x\right) \) and \( g\left( x\right) \) be functions that are differentiable near \( a \) . If\n\n\[ \mathop{\lim }\limits_{{x \rightarrow a}}f\left( x\right) = \mathop{\lim }\limits_{{x \rightarrow a}}g\left( x\right) = 0\;\text{ or } \pm \infty ,\]\n\nand \( \mathop{\lim... | This theorem is somewhat difficult to prove, in part because it incorporates so many different possibilities, so we will not prove it here. | No |
Compute \[ \mathop{\lim }\limits_{{x \rightarrow 0}}\frac{\sin \left( x\right) }{x} \] | Solution Set \( f\left( x\right) = \sin \left( x\right) \) and \( g\left( x\right) = x \) . Since both \( f\left( x\right) \) and \( g\left( x\right) \) are differentiable functions at 0 , and \[ \mathop{\lim }\limits_{{x \rightarrow 0}}f\left( x\right) = \mathop{\lim }\limits_{{x \rightarrow 0}}g\left( x\right) = 0 \]... | Yes |
Example 8.1.3 \( \\left( {\\infty /\\infty }\\right) \) Compute\n\n\[ \n\\mathop{\\lim }\\limits_{{x \\rightarrow \\pi /2 + }}\\frac{\\sec \\left( x\\right) }{\\tan \\left( x\\right) }\n\] | Solution Set \( f\\left( x\\right) = \\sec \\left( x\\right) \) and \( g\\left( x\\right) = \\tan \\left( x\\right) \) . Both \( f\\left( x\\right) \) and \( g\\left( x\\right) \) are differen-\ntiable near \( \\pi /2 \) . Additionally,\n\n\[ \n\\mathop{\\lim }\\limits_{{x \\rightarrow \\pi /2 + }}f\\left( x\\right) = ... | Yes |
Example 8.1.4 \( \\left( {0 \\cdot \\infty }\\right) \) Compute\n\n\[ \n\\mathop{\\lim }\\limits_{{x \\rightarrow 0 + }}x\\ln x \n\] | Solution This doesn’t appear to be suitable for l’Hôpital’s Rule. As \( x \) ap-\n\nproaches zero, \( \\ln x \) goes to \( - \\infty \), so the product looks like\n\n(something very small) \( \\cdot \) (something very large and negative).\n\nThis product could be anything-a careful analysis is required. Write\n\n\[ \nx... | Yes |
Example 8.1.5 ( \( \infty - \infty \) ) Compute\n\n\[ \mathop{\lim }\limits_{{x \rightarrow 0}}\left( {\cot \left( x\right) - \csc \left( x\right) }\right) \] | Solution Here we simply need to write each term as a fraction,\n\n\[ \mathop{\lim }\limits_{{x \rightarrow 0}}\left( {\cot \left( x\right) - \csc \left( x\right) }\right) = \mathop{\lim }\limits_{{x \rightarrow 0}}\left( {\frac{\cos \left( x\right) }{\sin \left( x\right) } - \frac{1}{\sin \left( x\right) }}\right) \]\n... | Yes |
Example 8.1.7 \( \\left( {1}^{\\infty }\\right) \) Compute\n\n\[ \n\\mathop{\\lim }\\limits_{{x \\rightarrow \\infty }}{\\left( 1 + \\frac{1}{x}\\right) }^{x}\n\] | Solution Write\n\n\[ \n\\mathop{\\lim }\\limits_{{x \\rightarrow \\infty }}{\\left( 1 + \\frac{1}{x}\\right) }^{x} = \\mathop{\\lim }\\limits_{{x \\rightarrow \\infty }}{e}^{x\\ln \\left( {1 + \\frac{1}{x}}\\right) }.\n\]\n\nSo now look at the limit of the exponent\n\n\[ \n\\mathop{\\lim }\\limits_{{x \\rightarrow \\in... | Yes |
Example 8.2.1 Suppose you drive a car on a 600 mile road trip. Your distance from home is recorded by the plot shown in Figure 8.2. What was your average velocity during hours 4-8 of your trip? | Solution Examining Figure 8.2, we see that we were around 240 miles from home at hour 4, and 360 miles from home at hour 8. Hence our average velocity was \[ \frac{{360} - {240}}{8 - 4} = \frac{120}{4} = {30}\text{ miles per hour. } \] | Yes |
Again suppose, you drive a car 600 mile road trip. Your distance from home is recorded by the plot shown in Figure 8.2. What was your instantaneous velocity 8 hours into your trip? | Solution Since the instantaneous rate of change is measured by the derivative, we need to find the slope of the tangent line to the curve. At 7 hours, the curve is growing at an essentially constant rate. In fact, the growth rate seems to be constant from \( \\left( {7,{300}}\\right) \) to \( \\left( {{10},{500}}\\righ... | Yes |
The Mostar bridge in Bosnia is 25 meters above the river Neretva. For fun, you decided to dive off the bridge. Your position \( t \) seconds after jumping off is\n\n\[ p\left( t\right) = - {4.9}{t}^{2} + {25} \]\n\nWhen do you hit the water? What is your instantaneous velocity as you enter the water? What is your avera... | Solution To find when you hit the water, you must solve\n\n\[ - {4.9}{t}^{2} + {25} = 0 \]\n\nWrite\n\n\[ - {4.9}{t}^{2} = - {25} \]\n\n\[ {t}^{2} \approx {5.1} \]\n\n\[ t \approx {2.26}\text{.} \]\n\nHence after approximately 2.26 seconds, you gracefully enter the river.\n\nYour instantaneous velocity is given by \( {... | Yes |
A certain bacterium divides into two cells every 20 minutes. The initial population of a culture is 120 cells. Find a formula for the population. What is the average growth rate during the first 4 hours? What is the instantaneous growth rate of the population at 4 hours? What rate is the population growing at 20 hours? | Solution Since we start with 120 cells, and this population doubles every 20 minutes, then the population doubles three times an hour. So the formula for the population is\n\n\[ p\left( t\right) = {120} \cdot {2}^{3t} \]\n\nwhere \( t \) is time measured in hours.\n\nNow, the average growth rate during the first 4 hour... | Yes |
A plane is flying directly away from you at \( {500}\mathrm{{mph}} \) at an altitude of 3 miles. How fast is the plane's distance from you increasing at the moment when the plane is flying over a point on the ground 4 miles from you? | Solution We’ll use our general strategy to solve this problem. To start, draw a picture.\n\n\n\nNext we need to find an equation. By the Pythagorean Theorem we know that\n\n\[ {p}^{2} + {3}^{2} = {s}^{2} \]\n\nNow we... | Yes |
Example 8.3.2 You are inflating a spherical balloon at the rate of \( 7{\mathrm{\;{cm}}}^{3}/\mathrm{{sec}} \) . How fast is its radius increasing when the radius is \( 4\mathrm{\;{cm}} \) ? | Solution To start, draw a picture.\n\n\n\nNext we need to find an equation. Thinking of the variables \( r \) and \( V \) as functions of time, they are related by the equation\n\n\[ V\left( t\right) = \frac{{4\pi }{... | Yes |
Water is poured into a conical container at the rate of 10 \( {\mathrm{{cm}}}^{3}/\mathrm{{sec}} \). The cone points directly down, and it has a height of \( {30}\mathrm{\;{cm}} \) and a base radius of \( {10}\mathrm{\;{cm}} \). How fast is the water level rising when the water is \( 4\mathrm{\;{cm}} \) deep? | Solution To start, draw a picture.\n\n\n\nNote, no attempt was made to draw this picture to scale, rather we want all of the relevant information to be available to the mathematician.\n\nNow we need to find an equati... | Yes |
A swing consists of a board at the end of a \( {10}\mathrm{{ft}} \) long rope. Think of the board as a point \( P \) at the end of the rope, and let \( Q \) be the point of attachment at the other end. Suppose that the swing is directly below \( Q \) at time \( t = 0 \), and is being pushed by someone who walks at \( 6... | Solution To start, draw a picture.\n\n\n\n\n\nNow we must find an equation. From the right triangle in our picture, we see\n\n\[ \sin \left( \vartheta \right) = x/{10}\text{.} \]\n\nWe can now differentiate the equat... | Yes |
Example 8.3.5 A road running north to south crosses a road going east to west at the point \( P \) . Cyclist \( A \) is riding north along the first road, and cyclist \( B \) is riding east along the second road. At a particular time, cyclist \( A \) is 3 kilometers to the north of \( P \) and traveling at \( {20}\math... | Solution We start the same way we always do, we draw a picture.\n\n\n\nHere \( a\left( t\right) \) is the distance of cyclist \( A \) north of \( P \) at time \( t \), and \( b\left( t\right) \) the distance of cycli... | Yes |
Find the (absolute) maximum and minimum values of \( f\left( x\right) = {x}^{2} \) on the interval \( \left\lbrack {-2,1}\right\rbrack \) . | Solution To start, write\n\[ \frac{d}{dx}{x}^{2} = {2x} \]\n\nThe critical point is at \( x = 0 \) . By the Extreme Value Theorem, Theorem 9.1.1, we must also consider the endpoints of the closed interval, \( x = - 2 \) and \( x = 1 \) . Check\n\n\[ f\left( {-2}\right) = 4,\;f\left( 0\right) = 0,\;f\left( 1\right) = 1.... | Yes |
Example 9.1.3 Find the (absolute) maximum and minimum values of the function \( f\left( x\right) = \left| {x - 2}\right| \) on the interval \( \left\lbrack {1,4}\right\rbrack \) . | Solution To start, rewrite \( f\left( x\right) \) as\n\n\[ f\left( x\right) = \sqrt{{\left( x - 2\right) }^{2}} \]\n\nnow\n\n\[ \frac{d}{dx}f\left( x\right) = \frac{2\left( {x - 2}\right) }{2\sqrt{{\left( x - 2\right) }^{2}}} = \frac{x - 2}{\left| x - 2\right| }.\]\n\nThe derivative \( {f}^{\prime }\left( x\right) \) i... | Yes |
Of all rectangles of area \( {100}{\mathrm{\;{cm}}}^{2} \), which has the smallest perimeter? | Solution First we draw a picture, see Figure 9.5. If \( x \) denotes one of the sides of the rectangle, then the adjacent side must be \( {100}/x \) .\n\nThe perimeter of this rectangle is given by\n\n\[ p\left( x\right) = {2x} + 2\frac{100}{x}. \]\n\nWe wish to minimize \( p\left( x\right) \) . Note, not all values of... | Yes |
Find the rectangle with largest area that fits inside the graph of the parabola \( y = {x}^{2} \) below the line \( y = a \), where \( a \) is an unspecified constant value, with the top side of the rectangle on the horizontal line \( y = a \) . | Solution We want to maximize value of \( A\left( x\right) \) . The lower right corner of the rectangle is at \( \left( {x,{x}^{2}}\right) \), and once this is chosen the rectangle is completely determined. Then the area is\n\n\[ A\left( x\right) = \left( {2x}\right) \left( {a - {x}^{2}}\right) = - 2{x}^{3} + {2ax}. \]\... | Yes |
You are making cylindrical containers to contain a given volume. Suppose that the top and bottom are made of a material that is \( N \) times as expensive (cost per unit area) as the material used for the lateral side of the cylinder. Find (in terms of \( N \) ) the ratio of height to base radius of the cylinder that m... | Solution First we draw a picture, see Figure 9.8. Now we can write an expression for the cost of materials:\n\n\[ C = {2\pi crh} + {2\pi }{r}^{2}{Nc}. \]\n\nSince we know that \( V = \pi {r}^{2}h \), we can use this relationship to eliminate \( h \) (we could eliminate \( r \), but it’s a little easier if we eliminate ... | Yes |
Suppose you want to reach a point \( A \) that is located across the sand from a nearby road, see Figure 9.9. Suppose that the road is straight, and \( b \) is the distance from \( A \) to the closest point \( C \) on the road. Let \( v \) be your speed on the road, and let \( w \), which is less than \( v \), be your ... | Solution Let \( x \) be the distance short of \( C \) where you turn off, the distance from \( B \) to \( C \) . We want to minimize the total travel time. Recall that when traveling at constant velocity, time is distance divided by velocity.\n\nYou travel the distance from \( D \) to \( B \) at speed \( v \), and then... | Yes |
Use a linear approximation of \( f\left( x\right) = \sqrt[3]{x} \) at \( x = {64} \) to approximate \( \sqrt[3]{50} \) . | Solution To start, write\n\n\[ \frac{d}{dx}f\left( x\right) = \frac{d}{dx}{x}^{1/3} = \frac{1}{3{x}^{2/3}}. \]\n\nso our linear approximation is\n\n\[ \ell \left( x\right) = \frac{1}{3 \cdot {64}^{2/3}}\left( {x - {64}}\right) + 4 \]\n\n\[ = \frac{1}{48}\left( {x - {64}}\right) + 4 \]\n\n\[ = \frac{x}{48} + \frac{8}{3}... | Yes |
Use a linear approximation of \( f\left( x\right) = \sin \left( x\right) \) at \( x = 0 \) to approximate \( \sin \left( {0.3}\right) \) . | To start, write\n\n\[ \frac{d}{dx}f\left( x\right) = \cos \left( x\right) \]\n\nso our linear approximation is\n\n\[ \ell \left( x\right) = \cos \left( 0\right) \cdot \left( {x - 0}\right) + 0 \]\n\n\[ = x\text{.} \]\n\nHence a linear approximation for \( \sin \left( x\right) \) at \( x = 0 \) is \( \ell \left( x\right... | Yes |
Use differentials to approximate \( \sqrt[3]{50} \) . | Solution Since \( {4}^{3} = {64} \) is a perfect cube near 50, we will set \( {dx} = - {14} \) . In this\ncase\n\[ \frac{dy}{dx} = {f}^{\prime }\left( x\right) = \frac{1}{3{x}^{2/3}} \]\nhence\n\[ {dy} = \frac{1}{3{x}^{2/3}} \cdot {dx} \]\n\[ = \frac{1}{3 \cdot {64}^{2/3}} \cdot \left( {-{14}}\right) \]\n\[ = \frac{-7}... | Yes |
Use differentials to approximate \( \sin \left( {0.3}\right) \) . | Solution Since \( \sin \left( 0\right) = 0 \), we will set \( {dx} = {0.3} \) . In this case\n\n\[ \frac{dy}{dx} = {f}^{\prime }\left( x\right) = \cos \left( x\right) \]\n\nhence\n\n\[ {dy} = \cos \left( 0\right) \cdot {dx} \]\n\n\[ = 1 \cdot \left( {0.3}\right) \]\n\n\[ = {0.3} \]\n\nNow \( f\left( {.3}\right) \approx... | Yes |
Use Newton's Method to approximate the solution to\n\n\[ \n{x}^{3} = {50} \n\]\nto two decimal places. | Solution To start, set \( f\left( x\right) = {x}^{3} - {50} \) . We will use Newton’s Method to approximate a solution to the equation\n\n\[ \nf\left( x\right) = {x}^{3} - {50} = 0. \n\]\n\nLet’s choose \( {a}_{0} = 4 \) as our first guess. Now compute\n\n\[ \n{f}^{\prime }\left( x\right) = 3{x}^{2}. \n\]\n\nAt this po... | Yes |
Use Newton’s Method to approximate \( \sqrt[3]{50} \) to two decimal places. | The \( \sqrt[3]{50} \) is simply a solution to the equation\n\n\[ \n{x}^{3} - {50} = 0\text{.}\n\]\n\nSince we did this in the previous example, we have found \( \sqrt[3]{50} \approx {3.68} \). | No |
Example 10.2.3 Suppose that the velocity in meters per second of a ball tossed from a height of 1 meter is given by\n\n\[ v\left( t\right) = - {9.8t} + 6 \]\n\nRounding to two decimals at each step, use Euler’s Method with \( h = {0.2} \) to approximate the height of the ball after 1 second. | Solution We simply need to make a table and use Euler's Method.\n\n<table><thead><tr><th>n</th><th>\( {t}_{n} \)</th><th>\( {y}_{n} \)</th></tr></thead><tr><td>0</td><td>0</td><td>1</td></tr><tr><td>1</td><td>0.2</td><td>2.2</td></tr><tr><td>2</td><td>0.4</td><td>3.01</td></tr><tr><td>3</td><td>0.6</td><td>3.42</td></t... | Yes |
Theorem 10.3.1 (Rolle’s Theorem) Suppose that \( f\left( x\right) \) is differentiable on the interval \( \left( {a, b}\right) \), is continuous on the interval \( \left\lbrack {a, b}\right\rbrack \), and \( f\left( a\right) = f\left( b\right) \). Then \[ {f}^{\prime }\left( c\right) = 0 \] for some \( a < c < b \). | Proof By the Extreme Value Theorem, Theorem 9.1.1, we know that \( f\left( x\right) \) has a maximum and minimum value on \( \left\lbrack {a, b}\right\rbrack \). If maximum and minimum both occur at the endpoints, then \( f\left( x\right) = f\left( a\right) = f\left( b\right) \) at every point in \( \left\lbrack {a, b}... | Yes |
Example 10.3.2 Suppose you toss a ball into the air and then catch it. Must the ball's vertical velocity have been zero at some point? | Solution If \( p\left( t\right) \) is the position of the ball at time \( t \), then we may apply Rolle’s Theorem to see at some time \( c,{p}^{\prime }\left( c\right) = 0 \) . Hence the velocity must be zero at some point. | Yes |
Theorem 10.3.3 (Mean Value Theorem) Suppose that \( f\left( x\right) \) has a derivative on the interval \( \left( {a, b}\right) \) and is continuous on the interval \( \left\lbrack {a, b}\right\rbrack \) . Then \[ {f}^{\prime }\left( c\right) = \frac{f\left( b\right) - f\left( a\right) }{b - a} \] for some \( a < c < ... | Proof Let \[ m = \frac{f\left( b\right) - f\left( a\right) }{b - a}, \] and consider a new function \( g\left( x\right) = f\left( x\right) - m\left( {x - a}\right) - f\left( a\right) \) . We know that \( g\left( x\right) \) has a derivative on \( \left\lbrack {a, b}\right\rbrack \), since \( {g}^{\prime }\left( x\right... | Yes |
Suppose you drive a car from toll booth on a toll road to another toll booth 30 miles away in half of an hour. Must you have been driving at 60 miles per hour at some point? | Solution If \( p\left( t\right) \) is the position of the car at time \( t \), and 0 hours is the starting time with \( 1/2 \) hours being the final time, the Mean Value Theorem states there is a time \( c \)\n\n\[ \n{p}^{\prime }\left( c\right) = \frac{{30} - 0}{1/2} = {60}\;\text{ where }0 < c < 1/2. \n\]\n\nSince th... | Yes |
Theorem 10.3.5 If \( {f}^{\prime }\left( x\right) = 0 \) for all \( x \) in an interval \( I \), then \( f\left( x\right) \) is constant on I. | Proof Let \( a < b \) be two points in \( I \) . By the Mean Value Theorem we know\n\n\[ \frac{f\left( b\right) - f\left( a\right) }{b - a} = {f}^{\prime }\left( c\right) \]\n\nfor some \( c \) in the interval \( \left( {a, b}\right) \) . Since \( {f}^{\prime }\left( c\right) = 0 \) we see that \( f\left( b\right) = f\... | Yes |
Suppose two different functions have the same derivative. What can you say about the relationship between the two functions? | Solution Set \( h\left( x\right) = f\left( x\right) - g\left( x\right) \), so \( {h}^{\prime }\left( x\right) = {f}^{\prime }\left( x\right) - {g}^{\prime }\left( x\right) \) . Now \( {h}^{\prime }\left( x\right) = 0 \) on the interval \( \left( {a, b}\right) \) . This means that \( h\left( x\right) = k \) where \( k \... | Yes |
Describe all functions whose derivative is \( \sin \left( x\right) \) . | Solution One such function is \( - \cos \left( x\right) \), so all such functions have the form \( - \cos \left( x\right) + k \), see Figure 10.12. | Yes |
Theorem 11.1.1 (Basic Antiderivatives) | <table><tr><td>- \( \int {kdx} = {kx} + C \) .- \( \int \cos \left( x\right) {dx} = \sin \left( x\right) + C \) .</td><td>- \( \int \sec \left( x\right) \tan \left( x\right) {dx} = \sec \left( x\right) + C \) .</td></tr><tr><td>- \( \int {x}^{n}{dx} = \frac{{x}^{n + 1}}{n + 1} + C \)\( \left( {n \neq - 1}\right) \) .- ... | Yes |
Compute\n\n\[ \int 3{x}^{7}{dx} \] | Solution By Theorem 11.1.1 and Theorem 11.1.2, we see that\n\n\[ \int 3{x}^{7}{dx} = 3\int {x}^{7}{dx} \]\n\n\[ = 3 \cdot \frac{{x}^{8}}{8} + C\text{. } \] | Yes |
Compute\n\n\[ \int \left( {{x}^{4} + 5{x}^{2} - \cos \left( x\right) }\right) {dx} \] | Solution Let's start by simplifying the problem using the sum rule for antiderivatives, Theorem 11.1.2.\n\n\[ \int \left( {{x}^{4} + 5{x}^{2} - \cos \left( x\right) }\right) {dx} = \int {x}^{4}{dx} + 5\int {x}^{2}{dx} - \int \cos \left( x\right) {dx}. \]\n\nNow we may integrate term-by-term to find\n\n\[ \int \left( {{... | Yes |
Compute\n\n\\[ \n\\int \\frac{{x}^{3}}{\\sqrt{{x}^{4} - 6}}{dx} \n\\] | Solution Start by rewriting the indefinite integral as\n\n\\[ \n\\int {x}^{3}{\\left( {x}^{4} - 6\\right) }^{-1/2}{dx} \n\\]\n\nNow start with a guess of\n\n\\[ \n\\int {x}^{3}{\\left( {x}^{4} - 6\\right) }^{-1/2}{dx} \\approx {\\left( {x}^{4} - 6\\right) }^{1/2}. \n\\]\n\nTake the derivative of your guess to see if it... | Yes |
Compute\n\n\[ \int x{e}^{x}{dx} \] | Solution We try to guess the antiderivative. Start with a guess of\n\n\[ \int x{e}^{x}{dx} \approx x{e}^{x} \]\n\nTake the derivative of your guess to see if it is correct:\n\n\[ \frac{d}{dx}x{e}^{x} = {e}^{x} + x{e}^{x} \]\n\nAh! So we need only subtract \( {e}^{x} \) from our original guess. We now find\n\n\[ \int x{... | Yes |
Compute\n\n\[ \int \frac{2{x}^{2}}{7{x}^{3} + 3}{dx} \] | Solution We’ll start with a guess of\n\n\[ \int \frac{2{x}^{2}}{7{x}^{3} + 3}{dx} \approx \ln \left( {7{x}^{3} + 3}\right) \]\n\nTake the derivative of your guess to see if it is correct:\n\n\[ \frac{d}{dx}\ln \left( {7{x}^{3} + 3}\right) = \frac{{21}{x}^{2}}{7{x}^{3} + 3} \]\n\nWe are only off by a factor of \( 2/{21}... | Yes |
Compute\n\n\\[ \int {x}^{4}\sin \left( {3{x}^{5} + 7}\right) {dx} \\] | Solution Here we simply try to guess the antiderivative. Start with a guess of\n\n\\[ \int {x}^{4}\sin \left( {3{x}^{5} + 7}\right) {dx} \\approx \\cos \\left( {3{x}^{5} + 7}\right) .\n\nTo see if your guess is correct, take the derivative of \\( \\cos \\left( {3{x}^{5} + 7}\right) \\) ,\n\n\\[ \\frac{d}{dx}\\cos \\lef... | Yes |
A ball is tossed into the air with an initial velocity of \( {15}\mathrm{\;m}/\mathrm{s} \) . What is the velocity of the ball after 1 second? How about after 2 seconds? | Solution Knowing that the acceleration due to gravity is \( - {9.8}\mathrm{\;m}/{\mathrm{s}}^{2} \), we write\n\n\[ {v}^{\prime }\left( t\right) = - {9.8}\text{.} \]\n\nTo solve this differential equation, take the antiderivative of both sides\n\n\[ \int {v}^{\prime }\left( t\right) {dt} = \int - {9.8dt} \]\n\n\[ v\lef... | Yes |
A ball is tossed into the air with an initial velocity of \( {15}\mathrm{\;m}/\mathrm{s} \) from a height of 2 meters. When does the ball hit the ground? | Solution Knowing that the acceleration due to gravity is \( - {9.8}\mathrm{\;m}/{\mathrm{s}}^{2} \), we write\n\n\[ \n{p}^{\prime \prime }\left( t\right) = - {9.8}\text{.} \n\]\n\nStart by taking the antiderivative of both sides of the equation\n\n\[ \n\int {p}^{\prime \prime }\left( t\right) {dt} = \int - {9.8dt} \n\]... | Yes |
A culture of yeast starts with 100 cells. After 160 minutes, there are 350 cells. Assuming that the growth rate of the yeast is proportional to the number of yeast cells present, estimate when the culture will have 1000 cells. | Since the growth rate of the yeast is proportional to the number of yeast cells present, we have the following differential equation\n\n\[ \n{p}^{\prime }\left( t\right) = {kp}\left( t\right) \n\]\n\nwhere \( p\left( t\right) \) is the population of the yeast culture and \( t \) is time measured in minutes. We know tha... | Yes |
The half-life of carbon-14 (the time it takes for half of an amount of carbon-14 to decay) is about 5730 years. If the rate of decay is proportional to the amount of carbon-14, and if we found a bone with 1/70th of the amount of carbon-14 we would expect to find in a living organism, approximately how old is the bone? | Solution Since the rate of decay of carbon-14 is proportional to the amount of carbon-14 present, we can model this situation with the differential equation\n\n\[ \n{f}^{\prime }\left( t\right) = {kf}\left( t\right) \n\]\n\nWe know that this differential equation is solved by the function defined by\n\n\[ \nf\left( t\r... | Yes |
Consider the differential equation\n\n\[ \n{f}^{\prime }\left( x\right) = {\left( f\left( x\right) \right) }^{2} - {6f}\left( x\right) + 8.\n\]\n\nSuppose you know that \( f\left( 1\right) = {3.8} \) . Rounding to two decimals at each step, use Euler’s Method with \( h = {0.2} \) to approximate \( f\left( 3\right) \). | Solution To solve this problem we'll use a variation on Euler's Method. We'll make a table following this format\n\n<table><thead><tr><th>n</th><th>\( {x}_{n} \)</th><th>\( {y}_{n} \)</th></tr></thead><tr><td>0</td><td>\( {x}_{0} \)</td><td>\( {y}_{0} \)</td></tr><tr><td>1</td><td></td><td>\( {y}_{0} + h \cdot \left( {... | No |
Consider the differential equation\n\n\[ \n{f}^{\prime }\left( x\right) = {\left( f\left( x\right) \right) }^{2} - {6f}\left( x\right) + 8.\n\]\n\nSuppose you know that \( f\left( 1\right) = 4 \) . Rounding to two decimals at each step, use Euler’s Method with \( h = {0.2} \) to approximate \( f\left( 3\right) \) . | Solution Again we’ll use a variation on Euler’s Method. Making the table as we did before, see Table 11.2. This time our solution is simply the function \( f\left( x\right) = 4 \) . Note, this does solve the differential equation as, given\n\n\[ \n{f}^{\prime }\left( x\right) = {\left( f\left( x\right) \right) }^{2} - ... | No |
Consider the differential equation\n\n\[ \n{f}^{\prime }\left( x\right) = {\left( f\left( x\right) \right) }^{2} - {6f}\left( x\right) + 8.\n\]\n\nSuppose you know that \( f\left( 1\right) = 2 \) . Rounding to two decimals at each step, use Euler’s Method with \( h = {0.2} \) to approximate \( f\left( 3\right) \) . | Solution Using the same variation on Euler's Method as before, see Table 11.3. This time our solution is simply the function \( f\left( x\right) = 2 \) . Note, this does solve the differential equation as, given\n\n\[ \n{f}^{\prime }\left( x\right) = {\left( f\left( x\right) \right) }^{2} - {6f}\left( x\right) + 8\n\]\... | No |
Example 12.1.1 Compute\n\n\[ \n{\\int }_{0}^{3}{xdx} \n\] | Solution The definite integral \( {\\int }_{0}^{3}{xdx} \) measures signed area of the shaded region shown in figure 12.1. Since this region is a triangle, we can use the formula for the area of the triangle to compute\n\n\[ \n{\\int }_{0}^{3}{xdx} = \\frac{1}{2}3 \\cdot 3 = 9/2 \n\] | Yes |
Compute \[ {\int }_{-1}^{3}\lfloor x\rfloor {dx} \] | Solution The definite integral \( {\int }_{-1}^{3}\lfloor x\rfloor {dx} \) measures signed area of the shaded region shown in figure 12.2. We see that\n\n\[ {\int }_{-1}^{3}\lfloor x\rfloor {dx} = {\int }_{-1}^{0}\lfloor x\rfloor {dx} + {\int }_{0}^{1}\lfloor x\rfloor {dx} + {\int }_{1}^{2}\lfloor x\rfloor {dx} + {\int... | Yes |
Consider the interval \( \left\lbrack {-1,1}\right\rbrack \), where is \( F\left( x\right) \) increasing? Where is \( F\left( x\right) \) decreasing? When does \( F\left( x\right) \) have a local extrema? | Solution We can see a plot of \( f\left( t\right) \) along with the signed area measured by the accumulation function in Figure 12.3. The accumulation function starts off at zero, and then is decreasing as it accumulates negatively signed area. However when \( x > 0, F\left( x\right) \) starts to accumulate positively ... | Yes |
Example 12.1.5 Consider the following accumulation function for \( f\left( x\right) = {x}^{3} \) .\n\n\[ F\left( x\right) = {\int }_{-1}^{x}{t}^{3}{dt} \]\n\nWhere is \( F\left( x\right) \) increasing? Where is \( F\left( x\right) \) decreasing? When does \( F\left( x\right) \) have a extrema? | Solution From our previous example, we know that \( F\left( x\right) \) is increasing on \( \left( {0,1}\right) \) . Since \( f\left( t\right) \) continues to be positive at \( t = 1 \) and beyond, \( F\left( x\right) \) is increasing on \( \left( {0,\infty }\right) \) . On the other hand, we know from our previous exa... | Yes |
Compute the left Riemann sum that approximates\n\n\\[ \n{\\int }_{1}^{2}\\left( {{x}^{2} - {2x} + 2}\\right) {dx} \n\\]\n\nusing four equally spaced partitions of the interval \\( \\left\\lbrack {1,2}\\right\\rbrack \\) . | Solution Start by setting \\( f\\left( x\\right) = {x}^{2} - {2x} + 2 \\) and examining Figure 12.10. Our partition of \\( \\left\\lbrack {1,2}\\right\\rbrack \\) is\n\n\\[ \n\\left\\lbrack {1,{1.25}}\\right\\rbrack \\cup \\left\\lbrack {{1.25},{1.5}}\\right\\rbrack \\cup \\left\\lbrack {{1.5},{1.75}}\\right\\rbrack \\... | Yes |
Compute\n\n\\[ \n{\\int }_{3}^{7}\\left( {{2x} - 1}\\right) {dx} \n\\]\n\nvia a left Riemann sum. | Solution Start by setting \\( f\\left( x\\right) = {2x} - 1 \\) and examining Figure 12.11. The interval \\( \\left\\lbrack {3,7}\\right\\rbrack \\) is divided into \\( n \\) subintervals each of width \\( \\left( {7 - 3}\\right) /n \\) . Our left Riemann sum is now\n\n\\[ \n\\mathop{\\sum }\\limits_{{i = 0}}^{{n - 1}}... | Yes |
Theorem 13.1.1 (Fundamental Theorem of Calculus-Version I)\n\nSuppose that \( f\left( x\right) \) is continuous on the real numbers and let\n\n\[ F\left( x\right) = {\int }_{a}^{x}f\left( t\right) {dt} \]\n\nThen \( {F}^{\prime }\left( x\right) = f\left( x\right) \) . | Proof Using the limit definition of the derivative we’ll compute \( {F}^{\prime }\left( x\right) \) . Write\n\n\[ {F}^{\prime }\left( x\right) = \mathop{\lim }\limits_{{h \rightarrow 0}}\frac{F\left( {x + h}\right) - F\left( x\right) }{h} \]\n\n\[ = \mathop{\lim }\limits_{{h \rightarrow 0}}\frac{1}{h}\left( {{\int }_{a... | Yes |
Theorem 13.1.2 (Fundamental Theorem of Calculus-Version II)\n\nSuppose that \( f\\left( x\\right) \) is continuous on the interval \( \\left\\lbrack {a, b}\\right\\rbrack \) . If \( F\\left( x\\right) \) is any antiderivative of \( f\\left( x\\right) \), then\n\n\[ \n{\\int }_{a}^{b}f\\left( x\\right) {dx} = {\\left. F... | Proof We know from Theorem 13.1.1\n\n\[ \nG\\left( x\\right) = {\\int }_{a}^{x}f\\left( t\\right) {dt} \n\]\n\nis an antiderivative of \( f\\left( x\\right) \), and therefore any antiderivative \( F\\left( x\\right) \) of \( f\\left( x\\right) \) is of\n\nHere the notation\n\n\[ \n{\\left. F\\left( x\\right) \\right| }... | Yes |
Compute\n\n\[ \n{\int }_{1}^{2}\left( {{x}^{9} + \frac{1}{x}}\right) {dx} \n\] | Solution Here we start by finding an antiderivative of\n\n\[ \n{x}^{9} + \frac{1}{x} \n\]\n\nThe correct choice is \( \frac{{x}^{10}}{10} + \ln \left( x\right) \), one could verify this by taking the derivative.\n\nHence\n\n\[ \n{\int }_{1}^{2}\left( {{x}^{9} + \frac{1}{x}}\right) {dx} = {\left. \left( \frac{{x}^{10}}{... | Yes |
Suppose that the velocity in meters per second of a ball tossed from a height of 1 meter is given by\n\n\[ v\left( t\right) = - {9.8t} + 6 \]\n\nWhat is the height of the ball after 1 second? | Solution Since the derivative of position is velocity, and we want to know the height (position) after one second, we need to compute\n\n\[ {\int }_{0}^{1} - {9.8t} + {6dt} = {\left. \left( -{4.9}{t}^{2} + 6t\right) \right| }_{0}^{1} \]\n\n\[ = - {4.9} + 6 - 0 \]\n\n\[ = {1.1}\text{.} \]\n\nHowever, since the ball was ... | Yes |
Find the area below \( f\left( x\right) = - {x}^{2} + {4x} + 3 \) and above \( g\left( x\right) = \) \( - {x}^{3} + 7{x}^{2} - {10x} + 5 \) over the interval \( 1 \leq x \leq 2 \) . | In Figure 13.2 we show the two curves together, with the desired area shaded.\n\nIt is clear from the figure that the area we want is the area under \( f\left( x\right) \) minus the area under \( g\left( x\right) \), which is to say\n\n\[ \n{\int }_{1}^{2}f\left( x\right) {dx} - {\int }_{1}^{2}g\left( x\right) {dx} = {... | Yes |
Find the area between \( f\left( x\right) = - {x}^{2} + {4x} \) and \( g\left( x\right) = {x}^{2} - {6x} + 5 \) over the interval \( 0 \leq x \leq 1 \) . | Solution The curves are shown in Figure 13.3. Generally we should interpret \ | No |
Find the area between \( f\left( x\right) = - {x}^{2} + {4x} \) and \( g\left( x\right) = {x}^{2} - {6x} + 5 \) . | Solution The curves are shown in Figure 13.4. Here we are not given a specific interval, so it must be the case that there is a \ | No |
Theorem 14.1.1 (Integral Substitution Formula) If \( u\left( x\right) \) is differentiable on the interval \( \left\lbrack {a, b}\right\rbrack \) and \( f\left( x\right) \) is differentiable on the interval \( \left\lbrack {u\left( a\right), u\left( b\right) }\right\rbrack \), then\n\n\[{\int }_{a}^{b}{f}^{\prime }\lef... | Proof First we recognize the chain rule\n\n\[{\int }_{a}^{b}{f}^{\prime }\left( {u\left( x\right) }\right) {u}^{\prime }\left( x\right) {dx} = {\int }_{a}^{b}{\left( f \circ u\right) }^{\prime }\left( x\right) {dx}.\]\n\nHere as is customary in calculus courses, we are abusing notation slightly, allowing \( u \) to bot... | Yes |
Compute\n\n\[ \n{\int }_{1}^{3}x\cos \left( {x}^{2}\right) {dx} \n\] | Solution A little thought reveals that if \( x\cos \left( {x}^{2}\right) \) is the derivative of some function, then it must have come from an application of the chain rule. Here we have \( x \) on the \ | No |
Compute\n\n\[ \n{\int }_{1}^{3}x\cos \left( {x}^{2}\right) {dx} \n\] | Solution Here we will set \( u = {x}^{2} \) . Now \( {du} = {2xdx} \), we are thinking in terms of differentials. Now we see\n\n\[ \n{\int }_{u\left( 1\right) }^{u\left( 3\right) }\frac{\cos \left( u\right) }{2}{du} = {\int }_{1}^{3}\frac{\cos \left( {x}^{2}\right) }{2}{2xdx}. \n\]\n\nAt this point, we can continue as ... | Yes |
Compute\n\n\[ \n{\int }_{1}^{3}x\cos \left( {x}^{2}\right) {dx} \n\] | Solution Here we start as we did before, setting \( u = {x}^{2} \) . Now \( {du} = {2xdx} \) ,\n\nagain thinking in terms of differentials. Now we see\n\n\[ \n\int \frac{\cos \left( u\right) }{2}{du} = \int \frac{\cos \left( {x}^{2}\right) }{2}{2xdx} \n\]\n\nHence\n\n\[ \n\int x\cos \left( {x}^{2}\right) {dx} = \frac{\... | Yes |
Compute\n\n\[ \int {x}^{4}{\left( {x}^{5} + 1\right) }^{99}{dx} \] | Solution Here we set \( u = {x}^{5} + 1 \) so \( {du} = 5{x}^{4}{dx} \), and \( f\left( u\right) = \frac{{u}^{99}}{5} \) . Now\n\n\[ \int {x}^{4}{\left( {x}^{5} + 1\right) }^{99}{dx} = \int \frac{{u}^{99}}{5}{du} \]\n\n\[ = \frac{{u}^{100}}{500} \]\n\nRecalling that \( u = {x}^{5} + 1 \), we have our final answer\n\n\[... | Yes |
Compute\n\n\[ \n{\int }_{2}^{3}\frac{1}{x\ln \left( x\right) }{dx} \n\] | Solution Let \( u = \ln \left( x\right) \) so \( {du} = \frac{1}{x}{dx} \) . Write\n\n\[ \n{\int }_{2}^{3}\frac{1}{x\ln \left( x\right) }{dx} = {\int }_{\ln \left( 2\right) }^{\ln \left( 3\right) }\frac{1}{u}{du} \n\]\n\n\[ \n= {\left. \ln \left( u\right) \right| }_{\ln \left( 2\right) }^{\ln \left( 3\right) } \n\]\n\n... | Yes |
Compute\n\n\\[ \n\\int {x}^{3}\\sqrt{1 - {x}^{2}}{dx} \n\\] | Solution Here it is not apparent that the chain rule is involved. However, if it was involved, perhaps a good guess for \\( u \\) would be\n\n\\[ \nu = 1 - {x}^{2} \n\\]\n\nin this case\n\n\\[ \n{du} = - {2xdx} \n\\]\n\nNow consider our indefinite integral\n\n\\[ \n\\int {x}^{3}\\sqrt{1 - {x}^{2}}{dx} \n\\]\n\nimmediat... | Yes |
Compute\n\n\[ \int {\sin }^{5}{xdx} \] | Solution Rewrite the function:\n\n\[ \int {\sin }^{5}{xdx} = \int \sin x{\sin }^{4}{xdx} = \int \sin x{\left( {\sin }^{2}x\right) }^{2}{dx} = \int \sin x{\left( 1 - {\cos }^{2}x\right) }^{2}{dx}. \]\n\nNow use \( u = \cos x,{du} = - \sin {xdx} \) :\n\n\[ \int \sin x{\left( 1 - {\cos }^{2}x\right) }^{2}{dx} = \int - {\l... | Yes |
Evaluate\n\n\[ \int {\sin }^{6}{xdx} \] | Solution Use \( {\sin }^{2}x = \left( {1 - \cos \left( {2x}\right) }\right) /2 \) to rewrite the function:\n\n\[ \int {\sin }^{6}{xdx} = \int {\left( {\sin }^{2}x\right) }^{3}{dx} = \int \frac{{\left( 1 - \cos 2x\right) }^{3}}{8}{dx} \]\n\n\[ = \frac{1}{8}\int 1 - 3\cos {2x} + 3{\cos }^{2}{2x} - {\cos }^{3}{2xdx}. \]\n... | Yes |
Compute\n\n\[ \int {\sin }^{2}x{\cos }^{2}{xdx} \] | Solution Use the formulas \( {\sin }^{2}x = \left( {1 - \cos \left( {2x}\right) }\right) /2 \) and \( {\cos }^{2}x = \left( {1 + \cos \left( {2x}\right) }\right) /2 \)\nto get:\n\n\[ \int {\sin }^{2}x{\cos }^{2}{xdx} = \int \frac{1 - \cos \left( {2x}\right) }{2} \cdot \frac{1 + \cos \left( {2x}\right) }{2}{dx}. \] | No |
Theorem 14.3.1 (Integration by Parts Formula) If \( f\left( x\right) g\left( x\right) \) is differentiable on the interval \( \left\lbrack {a, b}\right\rbrack \), then\n\n\[{\int }_{a}^{b}f\left( x\right) {g}^{\prime }\left( x\right) {dx} = {\left. f\left( x\right) g\left( x\right) \right| }_{a}^{b} - {\int }_{a}^{b}{f... | Proof First note by the product rule we have\n\n\[ \frac{d}{dx}f\left( x\right) g\left( x\right) = f\left( x\right) {g}^{\prime }\left( x\right) + {f}^{\prime }\left( x\right) g\left( x\right) . \]\n\nNow integrate both sides of the equation above\n\n\[ {\int }_{a}^{b}\frac{d}{dx}f\left( x\right) g\left( x\right) {dx} ... | Yes |
Compute\n\n\[ \int \ln \left( x\right) {dx} \] | Solution Let \( u = \ln \left( x\right) \) so \( {du} = 1/{xdx} \) . Hence, \( {dv} = {1dx} \) so \( v = x \) and so\n\n\[ \int \ln \left( x\right) {dx} = x\ln \left( x\right) - \int \frac{x}{x}{dx} \]\n\n\[ = x\ln \left( x\right) - x + C\text{.} \] | Yes |
Compute\n\n\[ \int x\sin \left( x\right) {dx} \] | Solution Let \( u = x \) so \( {du} = {dx} \). Hence, \( {dv} = \sin \left( x\right) {dx} \) so \( v = - \cos \left( x\right) \) and\n\n\[ \int x\sin \left( x\right) {dx} = - x\cos \left( x\right) - \int - \cos \left( x\right) {dx} \]\n\n\[ = - x\cos \left( x\right) + \int \cos \left( x\right) {dx} \]\n\n\[ = - x\cos \... | Yes |
Compute\n\n\\[ \int {x}^{2}\sin \left( x\right) {dx} \\] | Solution Let \( u = {x}^{2},{dv} = \sin \left( x\right) {dx} \) ; then \( {du} = {2xdx} \) and \( v = - \cos \left( x\right) \) . Now\n\n\\[ \int {x}^{2}\sin \left( x\right) {dx} = - {x}^{2}\cos \left( x\right) + \int {2x}\cos \left( x\right) {dx}. \\]\n\nThis is better than the original integral, but we need to do int... | Yes |
Find the volume of a pyramid with a square base that is 20 meters tall and 20 meters on a side at the base. | Solution As with most of our applications of integration, we begin by asking how we might approximate the volume. Since we can easily compute the volume of a box, we will use some \ | No |
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