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Calculus 2 : Integrals

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Example Questions

Example Question #14 : Volume Of A Solid

Evaluate the following trigonometric integral:

$\int \sin^{6}x \, \cos^{3}x\,dx$

Possible Answers:

$\frac{1}{9}\sin ^{9}x-\frac{1}{7}\cos^{7}x+C$

$\frac{1}{7}\sin ^{7}x-\frac{1}{9}\sin ^{9}x+C$

$\frac{1}{7}\cos ^{7}x-\frac{1}{9}\cos ^{9}x+C$

$\frac{1}{3}\cos ^{3}x-\frac{1}{5}\cos^{5}x+C$

$\frac{1}{3}\sin ^{3}x-\frac{1}{5}\sin ^{5}x+C$

Correct answer:

$\frac{1}{7}\sin ^{7}x-\frac{1}{9}\sin ^{9}x+C$

Explanation:

To solve this, first factor out one cosine and write the rest in terms of sine:

$\int \sin^{6}x \, \cos^{3}x\,dx=\int \sin^{6}x \, \cos^{2}x\,\cos x\, dx=\int \sin^{6}x (1-\sin^{2}x)\cos x\, dx$

From here, we can do a substitution where u=sin(x) and du=cos(x)dx:

$\int \sin^{6}x (1-\sin^{2}x)\cos x\, dx=\int u^{6}(1-u^{2})du=\int(u^{6}-u^{8})du=\frac{u^{7}}{7}-\frac{u^{9}}{9}+C$

Finally, re-substitute sin(x) for u:

$\frac{u^{7}}{7}-\frac{u^{9}}{9}+C=\frac{\sin^{7}x}{7}-\frac{\sin^{9}x}{9}+C$

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Example Question #1 : Volume Of Cross Sections And Area Of Region

Using the method of cylindrical disks, find the volume of the region revolved around the x-axis of the graph of

f(x)=-x^2+10x+3

on the interval

[0,10]

Possible Answers:

$\frac{13270}{3}\pi$ units cubed

$10000\pi$ units cubed

$20000\pi$ units cubed

$15000\pi$ units cubed

Correct answer:

$\frac{13270}{3}\pi$ units cubed

Explanation:

The formula for volume of the region revolved around the x-axis is given as

$V=\pi \int_{a}^{b} r^2\, dx$

where r=f(x)

As such

$V=\pi \int_{0}^{10} (-x^2+10x+3)^2\,dx = \pi \int_{0}^{10}x^4-20x^3+94x^2+60x+9\,dx$

When taking the integral, we use the inverse power rule which states

$\int x^n\,dx = \frac{x^{n+1}}{n+1}$

Applying this rule term by term we get

$= \pi \int_{0}^{10}x^4-20x^3+94x^2+60x+9\,dx = \pi\left ( \frac{x^5}{5} - \frac{20x^4}{4} + \frac{94x^3}{3} + \frac{60x^2}{2} + 9x\right |_{0}^{10}$

And by the corollary of the Fundamental Theorem of Calculus

$\pi\left ( \frac{x^5}{5} - \frac{20x^4}{4} + \frac{94x^3}{3} + \frac{60x^2}{2} + 9x\right |_{0}^{10}=\pi \left ( \left [ \frac{(10)^5}{5}-\frac{20(10)^4}{4}+\frac{94(10)^3}{3}+\frac{60(10)^2}{2}+9(10) \right ]-0 \right )$

$=\frac{13270}{3}\pi$

As such the volume is

$\frac{13270}{3}\pi$ units cubed

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Example Question #1 : Volume Of Cross Sections And Area Of Region

Find the volume of the solid generated by rotating about the y-axis the region under the curve $y =\sqrt{x}$ , from to .

Possible Answers:

$\frac{256}{7}\pi^2$

$\frac{128}{5}\pi$

$\frac{\pi}{2}$

$\frac{64}{3}\pi$

None of the other answers

Correct answer:

$\frac{128}{5}\pi$

Explanation:

Since we are revolving a function of around the y-axis, we will use the method of cylindrical shells to find the volume.

Using the formula for cylindrical shells, we have

$\int_{0}^{4}2\pi x (\sqrt{x})dx$

$=\int_{0}^{4}2\pi x^{3/2}dx$

$=[2\pi x^{5/2}]^{4}_{0}$

$=\frac{128}{5}\pi$ .

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Example Question #2051 : Calculus Ii

Find the volume of the solid region swept out by the area bounded by the the functions $\small \small \small y = \sqrt{3x}$ and $\small \small y = \frac{1}{4}x^2 +1$ about the $\small x$ -axis.

Possible Answers:

$\small V=\frac{7}{2}\pi (\sqrt[3]{7})^2$

$V=\frac{9}{10}\pi(\sqrt[3]{3})^2$

$\small V=\frac{17}{2}\pi (\sqrt[3]{3})$

$\small V=\frac{15}{2}\pi (\sqrt[3]{23})$

$\small V=\frac{7}{2}\pi (\sqrt[3]{233})$

Correct answer:

$V=\frac{9}{10}\pi(\sqrt[3]{3})^2$

Explanation:

Find the volume of the solid region swept out by the area bounded by the the functions $\small \small \small y = \sqrt{3x}$ and $\small \small \small y = x^2$ about the $\small x$ -axis.

The cross sectional area of the solid can be written as a function of $\small x$ by first finding the cross-sectional areas swept out by each of the individual functions as they rotate about the x-axis, and then subtracting the inner area.

Visualize a line drawn perpendicularly to the x-axis at some point meeting the function $\small \small \small y = \sqrt{3x}$ . As we rotate about the x-axis, we sweep out a circular cross-sectional area with a radius equal to the function value $\small \small \small y = \sqrt{3x}$ at the corresponding value of $\small x$ . The area of this disk is therefore:

$\small \small \small \small A_1(x)=\pi (\sqrt{3x})^2=\pi(3x)=3\pi x$

Similarly for the function $\small \small \small y = x^2$ , the cross-sectional area is:

$\small \small A_2(x)=\pi (x^2)^2=\pi x^4$

Problem 6 plots for finding the volume of a solid

The plot of the two functions helps visualize the geometry. The parabola is clearly the "inner" radius, and therefore we must subtract the areas $\small A_2(x)$ from $\small A_1(x)$ to find the cross-sectional area of the solid:

$\small \small A(x)=A_1(x)-A_2(x)$

$\small \small A(x)=3\pi x-\pi x^4$

Now we must find the limits of integration by finding the intersection points.

$\small \sqrt{3x}=x^2$

Solve for $\small x$

$\small \small 3x=x^4\: \: \: \rightarrow\: \: \:3=x^3 \: \: \:\rightarrow\: \: \: x=\sqrt[3]{3}$

The other solution is $\small x = 0$ . Now simply integrate the cross sectional area $\small A(x)$ over the interval $\small [0,\sqrt[3]{3} ]$ to find the volume of the solid,

$\small V=\small \int_0^{\sqrt[3]3} (3\pi x-\pi x^4)dx$

$\small \small \small \small \small \small V=\left[\frac{3\pi}{2} x^2 - \frac{\pi}{5} x^5\right ]_{0}^{\sqrt[3]{3}}\: \: \:$

$\small V=\frac{3}{2}\pi (\sqrt[3]{3})^2 - \frac{1}{5}\pi (\sqrt[3]{3})^5$

To simplify further, write the terms with a common denominator and factor.

$\small V=\frac{15}{10}\pi (\sqrt[3]{3})^2 - \frac{2}{10}\pi (\sqrt[3]{3})^5$

$V=\frac{1}{10}\pi(\sqrt[3]{3})^2 \left[15-2(\sqrt[3]{3})^3\right]$

$V=\frac{1}{10}\pi(\sqrt[3]{3})^2 \left[15-6\right]$

$V=\frac{9}{10}\pi(\sqrt[3]{3})^2$

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Example Question #16 : Volume Of A Solid

Find the volume of solid of revolution for the given function:

$y(x)=x^2+1; \:\:x\in [0,2]$ .

Possible Answers:

$\frac{206\pi}{16}$

$\frac{32\pi}{5}$

$\frac{16\pi}{3}$

$\frac{305\pi}{8}$

Correct answer:

$\frac{206\pi}{16}$

Explanation:

The volume of a solid of revolution for a given function is found using the disk method. Answer is obtained upon evaluating the following integral:

$\\\int_{0}^{2}\pi(x^2+1)^2\:dx\\\\=\pi\int_{0}^{2}x^4+2x^2+1\:dx\\\\=\left.\begin{matrix} \pi(\frac{x^5}{5}+\frac{2x^3}{3}+x) \end{matrix}\right|_{0}^{2}\\\\=\frac{206\pi}{15}$

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Example Question #301 : Integrals

Approximate the length of the curve of $f (x) = \ln (x + 1)$ on the interval [0,2] . Use Simpson's Parabolic Rule with n = 4 to make your estimate to the nearest thousandth.

Possible Answers:

2.442

2.303

2.486

2.198

2.667

Correct answer:

2.303

Explanation:

The length of the curve of f(x) on the interval [a,b] can be determined by evaluating the integral

$\int_{a}^{b} \sqrt{1 + f'(x) ^{2}} \; \mathrm{d}x$ .

$f (x) = \ln (x + 1)$ , so

$f'(x) = \frac{1}{x+1}$ , and the integral to be approximated is

$\int_{0}^{2} \sqrt{1 + \left ( \frac{1}{x+1} \right ) ^{2}} \; \mathrm{d}x$

$= \int_{0}^{2} \sqrt{\frac{x^{2}+2x+1}{x^{2}+2x+1} + \frac{1}{x^{2}+2x+1} } \; \mathrm{d}x$

$= \int_{0}^{2} \sqrt{\frac{x^{2}+2x+2}{x^{2}+2x+1} } \; \mathrm{d}x$

We divide the interval [0,2] into four subintervals of width $\Delta x = 2 \div 4 = 0.5$ each, so

$x_{0} = 0, x_{1} = 0.5, x_{2} = 1, x_{3} = 1.5, x_{4} = 2$

By Simpson's rule, we can estimate the integral by evaluating

$\frac{\Delta x}{3} \left ( g (x_{0}) + 4g(x_{1}) + 2g(x_{2}) + 4 g (x_{3}) + g (x_{4}) \right )$

$= \frac{0.5 }{3} \left ( g (0) + 4g(0.5) + 2g(1) + 4 g (1.5) + g (2) \right )$

where $g(x) = \sqrt{\frac{x^{2}+2x+2}{x^{2}+2x+1}$

We evaluate at these points, then substitute:

$g(0) = \sqrt{\frac{0^{2}+2 \cdot 0+2}{0^{2}+2 \cdot 0+1} } \approx 1.4142$

$g(0.5) = \sqrt{\frac{0.5^{2}+2 \cdot 0.5+2}{0.5^{2}+2 \cdot 0.5+1} } \approx 1.2019$

$g(1) = \sqrt{\frac{1^{2}+2 \cdot 1+2}{1^{2}+2 \cdot 1+1} } \approx 1.1180$

$g(1.5) = \sqrt{\frac{1.5^{2}+2 \cdot 1.5+2}{1.5^{2}+2 \cdot 1.5+1} } \approx 1.0770$

$g(2) = \sqrt{\frac{2^{2}+2 \cdot 2+2}{2^{2}+2 \cdot 2+1} } \approx 1.0541$

The approximation is therefore

$\approx \frac{0.5 }{3} \left ( 1.4142 +4 \cdot 1.2019 + 2\cdot 1.1180 +4 \cdot 1.0770 +1.0541 \right )$

$\approx \frac{0.5 }{3} \left ( 1.4142 +4.8076 + 2.2360 +4.3080 +1.0541 \right )$

$\approx \frac{0.5 }{3} \left (13.8199 \right ) \approx 2.303$

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Example Question #3 : Length Of Curve, Distance Traveled, Accumulated Change, Motion Of Curve

Give the arclength of the graph of the function $f(x) = \frac{2}{3} x \sqrt{x}$ on the interval $\left [ 0,3 \right ]$ .

Possible Answers:

$\frac{16}{3}$

$\frac{11}{3}$

$\frac{14}{3}$

Correct answer:

$\frac{14}{3}$

Explanation:

The length of the curve of f(x) on the interval [a,b] can be determined by evaluating the integral

$\int_{a}^{b} \sqrt{1 + f'(x) ^{2}} \; \mathrm{d}x$ .

$f(x) = \frac{2}{3} x \sqrt{x} = \frac{2}{3} x^{\frac{3}{2}}$

$f(x) = \frac{2}{3} \cdot \frac{3}{2} x^{\frac{1}{2}} = \sqrt{x}$ .

The above integral becomes

$\int_{0}^{3} \sqrt{1 + (\sqrt{x}) ^{2}} \; \mathrm{d}x$

$\int_{0}^{3} \sqrt{x+1} \; \mathrm{d}x$

Substitute u = x + 1 . Then $\frac{\mathrm{d} u}{\mathrm{d} x} = 1$ , $\mathrm{d} u = \mathrm{d} x$ , and the integral becomes

$\int_{1}^{4} \sqrt{u} \; \mathrm{d}u = \int_{1}^{4} u^{\frac{1}{2}} \; \mathrm{d}u$

$= \left.\begin{matrix} \frac{u^{\frac{3}{2}}}{\frac{3}{2}} \\ \\ \end{matrix}\right| \begin{matrix} 4\\ \\ 1 \end{matrix}$

$= \left.\begin{matrix} \frac{2u \sqrt{u}}{3} \\ \\ \end{matrix}\right| \begin{matrix} 4\\ \\ 1 \end{matrix}$

$= \frac{2 \cdot 4 \sqrt{4}}{3} - \frac{2 \cdot 1 \sqrt{1}}{3}$

$= \frac{2 \cdot 4 \cdot 2 }{3} - \frac{2 \cdot 1 \cdot 1}{3} = \frac{16}{3} - \frac{2}{3} =\frac{14}{3}$

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Example Question #1 : Length Of Curve, Distance Traveled, Accumulated Change, Motion Of Curve

Give the arclength of the graph of the function $f(x) = \frac{1}{2} \ln \left ( \cos 2 x \right )$ on the interval $\left [ 0, \frac{\pi}{8} \right ]$ .

Possible Answers:

$\frac{1 }{2}+\frac{1 }{2} \sqrt{2}$

$1 + \sqrt{2}$

$1+\frac{1 }{2} \ln 2$

$\frac{1}{2} \ln \left ( 1 + \sqrt{2} \right )$

$\ln \left ( 1 + \sqrt{2} \right )$

Correct answer:

$\frac{1}{2} \ln \left ( 1 + \sqrt{2} \right )$

Explanation:

The length of the curve of f(x) on the interval [a,b] can be determined by evaluating the integral

$\int_{a}^{b} \sqrt{1 + f'(x) ^{2}} \; \mathrm{d}x$ .

$f(x) = \frac{1}{2} \ln \left ( \cos 2 x \right )$ , so

$f'(x) = \frac{1}{2} \cdot \frac{\frac{\mathrm{d} }{\mathrm{d} x}\cos 2 x}{\cos 2 x} = \frac{1}{2} \cdot \frac{-2 \sin 2x}{\cos 2 x} = - \tan 2x$

The integral becomes

$\int_{0}^{ \frac{\pi}{8}} \sqrt{1 + \left ( - \tan 2x \right )^{2}} \; \mathrm{d}x$

$= \int_{0}^{ \frac{\pi}{8}} \sqrt{1 + \tan ^{2} 2x } \; \mathrm{d}x$

$= \int_{0}^{ \frac{\pi}{8}} \sqrt{\sec ^{2} 2x } \; \mathrm{d}x$

$= \int_{0}^{ \frac{\pi}{8}} \sec 2x \; \mathrm{d}x$

Use substitution - set u = 2x . Then $\frac{\mathrm{d} u}{\mathrm{d} x} = 2$ , and $\mathrm{d} x = \frac{1}{2}\mathrm{d} u$ . The bounds of integration become $2 \cdot 0 = 0$ and $2 \cdot \frac{\pi}{8} = \frac{\pi}{4}$ , and the integral becomes

$\frac{1}{2} \int_{0}^{ \frac{\pi}{4}} \sec u \; \mathrm{d}u$

$=\left.\begin{matrix}\frac{1}{2} \ln | \sec u + \tan u | \\ \\ \end{matrix}\right|\begin{matrix} \frac{\pi}{4}\\ \\ 0 \end{matrix}$

$=\frac{1}{2} \ln \left | \sec \frac{\pi}{4} + \tan \frac{\pi}{4} \right | - \frac{1}{2} \ln | \sec 0 + \tan 0 |$

$=\frac{1}{2} \ln \left | \sqrt{2} + 1 \right | -\frac{1}{2} \ln | 1 + 0 |$

$=\frac{1}{2} \ln \left ( 1 + \sqrt{2} \right ) -\frac{1}{2} \ln 1$

$=\frac{1}{2} \ln \left ( 1 + \sqrt{2} \right ) - 0$

$=\frac{1}{2} \ln \left (1 + \sqrt{2} \right )$

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Example Question #302 : Integrals

Give the arclength of the graph of the function $f(x) = \frac{1}{3} \left ( \cosh 3 x \right )$ on the interval $\left [ 0, 2 \right ]$ .

Possible Answers:

$\frac{ e^{12} - 1 }{6e^{6}}$

$\frac{ e^{6} +1 }{6e^{3}}$

$\frac{ e^{12} + 1 }{6e^{6}}$

$2e^{6}$

$\frac{ e^{6} - 1 }{6e^{3}}$

Correct answer:

$\frac{ e^{12} - 1 }{6e^{6}}$

Explanation:

The length of the curve of f(x) on the interval [a,b] can be determined by evaluating the integral

$\int_{a}^{b} \sqrt{1 + f'(x) ^{2}} \; \mathrm{d}x$ .

$f(x) = \frac{1}{3} \left ( \cosh 3 x \right )$ , so

$f'(x) = \frac{1}{3} \cdot 3 \sinh 3 x = \sinh 3 x$

and the integral becomes

$\int_{0}^{2} \sqrt{1 + \sinh^{2} 3 x } \; \mathrm{d}x$

$= \int_{0}^{2} \sqrt{ \cosh^{2} 3 x } \; \mathrm{d}x$

$= \int_{0}^{2} \cosh 3 x \; \mathrm{d}x$

Use substitution - set u =3x . Then $\frac{\mathrm{d} u}{\mathrm{d} x} = 3$ , and $\mathrm{d} x = \frac{1}{3}\mathrm{d} u$ . The bounds of integration become 0 and 6; the integral becomes

$= \int_{0}^{6} \frac{1}{3} \cosh u \; \mathrm{d}u$

$=\frac{1}{3} \int_{0}^{6} \cosh u \; \mathrm{d}u$

$= \frac{1}{3} \cdot \left.\begin{matrix} \sinh u\\ \\ \end{matrix}\right|\left.\begin{matrix} 6\\ \\ 0 \end{matrix}\right$

$=\frac{1}{3}\left ( \sinh 6 - \sinh 0 \right ) = \frac{1}{3} \sinh 6 =\frac{1}{3 } \cdot \frac{ e^{6} - e^{-6}}{2} = \frac{ \left (e^{6} - e^{-6} \right )e^{6}}{6e^{6}} = \frac{ e^{12} - 1 }{6e^{6}}$

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Example Question #303 : Integrals

What is the average value of the function $f(x) = \tan x$ on the interval $\left [\frac{ \pi }{6} , \frac{ \pi }{3}\right ]$ ?

Possible Answers:

$\frac{ \pi \ln 3}{6}$

$\frac{6 \ln 3}{\pi}$

$\frac{3 \ln 3}{\pi}$

$\frac{ \pi \ln 3}{3}$

$\frac{ \ln 3}{\pi}$

Correct answer:

$\frac{3 \ln 3}{\pi}$

Explanation:

The average value of the function $f(x) = \tan x$ on the interval $\left [\frac{ \pi }{6} , \frac{ \pi }{3}\right ]$ is equal to

$\frac{1}{ \frac{ \pi}{3}-( \frac{ \pi}{6})} \int_{ \frac{ \pi}{6}}^{ \frac{ \pi}{3}}\tan x \cdot \mathrm{d}x$

$= \frac{1}{ \frac{ \pi}{6}} \int_{ \frac{ \pi}{6}}^{ \frac{ \pi}{3}}\tan x \cdot \mathrm{d}x$

$= \frac{6}{ \pi} \int_{ \frac{ \pi}{6}}^{ \frac{ \pi}{3}}\tan x \cdot \mathrm{d}x$

$= \frac{6}{ \pi} \cdot \left.\begin{matrix} \ln\left | \sec x \right |\\ \\ \end{matrix} \right | \begin{matrix} \frac{\pi}{3}\\ \\ \frac{ \pi}{6} \end{matrix}$

$= \frac{6}{ \pi} \left ( \ln\left | \sec \frac{ \pi}{3} \right | - \ln\left | \sec \frac{ \pi}{6} \right | \right )$

$= \frac{6}{ \pi} \left ( \ln 2 - \ln \frac{2\sqrt{3}}{3} \right )$

$= \frac{6}{ \pi} \ln \left (2 \div \frac{2\sqrt{3}}{3} \right )$

$= \frac{6}{ \pi} \ln \sqrt{3}= \frac{6}{ \pi} \cdot \frac{1}{2} \cdot \ln 3 = \frac{3 \ln 3}{\pi}$

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Calculus 2 : Integrals

Study concepts, example questions & explanations for Calculus 2

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Example Questions

Example Question #14 : Volume Of A Solid

Example Question #1 : Volume Of Cross Sections And Area Of Region

Example Question #1 : Volume Of Cross Sections And Area Of Region

Example Question #2051 : Calculus Ii

Example Question #16 : Volume Of A Solid

Example Question #301 : Integrals

Example Question #3 : Length Of Curve, Distance Traveled, Accumulated Change, Motion Of Curve

Example Question #1 : Length Of Curve, Distance Traveled, Accumulated Change, Motion Of Curve

Example Question #302 : Integrals

Example Question #303 : Integrals

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