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Calculus 2 : Calculus II

Study concepts, example questions & explanations for Calculus 2

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9 Diagnostic Tests 308 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #1 : Finding Integrals

Evaluate:

$\int_{0}^{1} \ln (x^2 + 4x + 4) \; \mathrm{d}x$

Possible Answers:

$= -2 + \ln \frac{27}{4}$

The integral is undefined.

$= -1 + \ln \frac{729}{16}$

$= -2 + \ln \frac{729}{16}$

$= -1 + \ln \frac{27}{4}$

Correct answer:

$= -2 + \ln \frac{729}{16}$

Explanation:

Rewrite this as follows:

$\int_{0}^{1} \ln (x^2 + 4x + 4) \; \mathrm{d}x$

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

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

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

Substitute u = x + 2 . Then $\frac{\mathrm{d} u}{\mathrm{d} x} = 1$ and $\mathrm{d}x = \mathrm{d}x$ , and the bounds of integration become 2 and 3, making the integral equal to

$= 2 \int_{2}^{3} \ln u \; \mathrm{d}u$

$= \left.\begin{matrix} 2 \left (u \ln u - u \right ) \\ \\ \end{matrix}\right| \begin{matrix} 3\\ \\ 2 \end{matrix}$

$= 2 [(3 \ln 3 - 3 ) - (2 \ln 2 - 2)]$

$= 2 [ 3 \ln 3 - 2 \ln 2 - 1 ]$

$= 6 \ln 3 - 4 \ln 2 - 2$

$= \ln 3^6 - \ln 2^4 - 2$

$= \ln 729 - \ln 16 - 2$

$= -2 + \ln \frac{729}{16}$

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Example Question #2 : Finding Integrals

Evaluate:

$\int_{4}^{16} \log_{4} x\; \mathrm{d}x$

Possible Answers:

$36- 6 \ln 2$

$36- \frac{ 6 }{\ln 2}$

$28- 6 \ln 2$

$28- \frac{ 6 }{\ln 2}$

The integral is undefined.

Correct answer:

$28- \frac{ 6 }{\ln 2}$

Explanation:

$\int_{4}^{16} \log_{4} x\; \mathrm{d}x$

$= \int_{4}^{16} \frac{\ln x}{\ln 4}\; \mathrm{d}x$

$= \frac{1}{\ln 4} \int_{4}^{16}\ln x \; \mathrm{d}x$

$= \frac{1}{\ln 4} \cdot \left.\begin{matrix} \left (x \ln x - x \right ) \\ \\ \end{matrix}\right| \begin{matrix} 16\\ \\ 4 \end{matrix}$

$= \frac{1}{\ln 4} \cdot \left [ \left (16 \ln 16 - 16 \right ) - \left (4 \ln 4 - 4 \right ) \right ]$

$= \frac{1}{\ln 4} \cdot \left (16 \ln 16 - 4 \ln 4 - 12 \right )$

$= \frac{ 16 \ln 16 }{\ln 4} - \frac{ 4 \ln 4 }{\ln 4} - \frac{ 12 }{\ln 4}$

$= 16 \log_{4} 16 - 4 - \frac{ 12 }{\ln 4} = 16 \cdot 2 - 4 - \frac{ 12 }{2 \ln 2} = 32 - 4 - \frac{ 6 }{\ln 2} = 28- \frac{ 6 }{\ln 2}$

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Example Question #1 : Finding Integrals

Evaluate:

$\int_{1}^{ e/2} \ln 2x \; \mathrm{d}x$

Possible Answers:

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

$1 + \ln 2$

$1 - \ln 2$

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

$\ln 2$

Correct answer:

$1 - \ln 2$

Explanation:

$\int_{1}^{ e/2} \ln 2x \; \mathrm{d}x$

Substitute u = 2x ; so $\frac{\mathrm{d} u}{\mathrm{d} x} = 2$ and $\; \mathrm{d} x = \frac{1}{2}\mathrm{d} u$ , and the bounds of integration become 2 and ; the above becomes

$\frac{1}{2}\int_{2}^{ e} \ln u \; \mathrm{d}u$

$= \left.\begin{matrix} \frac{1}{2} \cdot u \ln u - u\\ \\ \end{matrix}\right| \begin{matrix} e\\ \\ 2 \end{matrix}$

$= \frac{1}{2} \cdot \left [ \left ( e \ln e - e \right) - \left ( 2 \ln 2 - 2 \right) \right]$

$= \frac{1}{2} \cdot \left [ \left ( e \cdot 1 - e \right) - \left ( 2 \ln 2 - 2 \right) \right]$

$= \frac{1}{2} \cdot \left [ \left ( e - e \right) - \left ( 2 \ln 2 - 2 \right) \right] = \frac{1}{2} \cdot \left [2 - 2 \ln 2 \right] = 1 - \ln 2$

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Example Question #2 : Finding Integrals

Which of the following functions makes the statement $\int_{0}^{3} f(x) \; \mathrm{d}x = 0$ true?

Possible Answers:

$f(x) = \sin \frac{3 \pi x}{4}$

$f(x) = \sin \frac{ \pi x}{3}$

$f(x) = \sin \frac{2 \pi x}{3}$

$f(x) = \sin \frac{3 \pi x}{2}$

$f(x) = \sin 3 \pi x$

Correct answer:

$f(x) = \sin \frac{2 \pi x}{3}$

Explanation:

$\int_{0}^{3} \sin nx \; \mathrm{d}x = \left.\begin{matrix} -\frac{\cos nx}{n}\\ \\ \end{matrix}\right|\begin{matrix} 3\\ \\ 0 \end{matrix} = -\frac{\cos 3n }{n} + \frac{\cos 0}{n} = \frac{1-\cos 3n }{n}$

Therefore, we are looking for a value of for which

$\int_{0}^{3} \sin nx \; \mathrm{d}x = \frac{1-\cos 3n }{n} = 0$ , or, equivalently,

$1-\cos 3n = 0$ , or

$\cos 3n = 1$

$3n = 0, 2 \pi, 4 \pi, 6\pi ...$

$n = 0, \frac{2 \pi}{3}, \frac{4 \pi}{3}, 2\pi ...$

The only choice that makes an element of this set is

$f(x) = \sin \frac{2 \pi x}{3}$ .

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Example Question #2 : Finding Integrals

Evaluate:

$\int_{0}^{4 \pi} | \cos x | \; \mathrm{d}x$

Possible Answers:

$8 \sqrt{2}$

$4 \sqrt{2}$

$4 \sqrt{3}$

Correct answer:

Explanation:

An easy way to look at this is to note that on the interval $[0, 4 \pi]$ , the integrand

$f(x)=| \cos x |$

can be rewritten as

$f(x)= \begin{Bmatrix} \cos x & 0 \leq x \leq\frac{\pi}{2} \\ \\ - \cos x & \frac{\pi}{2} \leq x \leq \frac{3\pi}{2} \\ \\ \cos x & \frac{3\pi}{2} \leq x \leq \frac{5\pi}{2} \\ \\ - \cos x & \frac{5\pi}{2} \leq x \leq \frac{7\pi}{2} \\ \\ \cos x & \frac{7\pi}{2} \leq x \leq 4\pi \end{matrix}$

Therefore,

$\int_{0}^{4 \pi} | \cos x | \; \mathrm{d}x$

$= \int_{0}^{4 \pi} f(x) \; \mathrm{d}x$

$= \int_{0}^{\frac{\pi}{2}} f(x) \; \mathrm{d}x + \int_{\frac{\pi}{2}}^{\frac{3\pi}{2}} f(x) \; \mathrm{d}x + \int_{\frac{3\pi}{2}}^{\frac{5\pi}{2}} f(x) \; \mathrm{d}x + \int_{\frac{5\pi}{2}}^{\frac{7\pi}{2}} f(x) \; \mathrm{d}x + \int_{\frac{7\pi}{2}}^{4 \pi} f(x) \; \mathrm{d}x$

$= \int_{0}^{\frac{\pi}{2}} \cos x \; \mathrm{d}x - \int_{\frac{\pi}{2}}^{\frac{3\pi}{2}} \cos x \; \mathrm{d}x + \int_{\frac{3\pi}{2}}^{\frac{5\pi}{2}} \cos x \; \mathrm{d}x - \int_{\frac{5\pi}{2}}^{\frac{7\pi}{2}} \cos x \; \mathrm{d}x + \int_{\frac{7\pi}{2}}^{4 \pi} \cos x \; \mathrm{d}x$

The antiderivative of $g(x)= \cos x$ is $G(x) = \sin x$ . We can evaluate at each boundary of integration:

$G (0) = \sin 0 = 0$

$G\left ( \frac{\pi }{2}\right ) = \sin \frac{\pi }{2} = 1$

$G\left ( \frac{3\pi }{2}\right ) = \sin \frac{3\pi }{2} = -1$

$G\left ( \frac{5\pi }{2}\right ) = \sin \frac{5\pi }{2} = 1$

$G\left ( \frac{7\pi }{2}\right ) = \sin \frac{7\pi }{2} = -1$

$G (4\pi) = \sin 4\pi = 0$

Then

$\int_{0}^{\frac{\pi}{2}} \cos x \; \mathrm{d}x = \left ( G \left ( \frac{\pi }{2}\right ) - G (0)\right ) = 1 - 0= 1$

$\int_{\frac{\pi}{2}}^{ \frac{3\pi}{2}} \cos x \; \mathrm{d}x = \left ( G \left ( \frac{3\pi }{2}\right ) - G \left ( \frac{\pi }{2}\right )\right ) = -1 -1= -2$

$\int_{\frac{3\pi}{2}}^{ \frac{5\pi}{2}} \cos x \; \mathrm{d}x = \left ( G \left ( \frac{5\pi }{2}\right ) - G \left ( \frac{3\pi }{2}\right )\right ) = 1 - (-1)= 2$

$\int_{\frac{5\pi}{2}}^{ \frac{7\pi}{2}} \cos x \; \mathrm{d}x = \left ( G \left ( \frac{7\pi }{2}\right ) - G \left ( \frac{5\pi }{2}\right )\right ) = -1 -1= -2$

$\int_{\frac{7\pi}{2}}^{4 \pi} \cos x \; \mathrm{d}x = \left ( G (4\pi) -G \left ( \frac{7\pi }{2}\right ) \right ) = 0 - (-1) = 1$

The original integral can be evaluated as

$\int_{0}^{4 \pi} | \cos x | \; \mathrm{d}x = 1 - (-2) + 2 - (-2) + 1 = 1 + 2 + 2 + 2 + 1 = 8$

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Example Question #3 : Finding Integrals

Evaluate:

$\int_{0}^{1}\int_{0}^{\pi} y^{2}\cos x \; \mathrm{d} x \; \mathrm{d} y$

Possible Answers:

$\frac{\pi \sqrt{2}}{2}$

$\frac{\pi}{3}$

$\frac{\pi}{2}$

$\frac{\pi \sqrt{3}}{2}$

Correct answer:

Explanation:

$\int_{0}^{1}\int_{0}^{\pi} y^{2}\cos x \; \mathrm{d} x \; \mathrm{d} y$

$=\int_{0}^{1}y^{2}\left (\int_{0}^{\pi}\cos x \; \mathrm{d} x \right )\; \mathrm{d} y$

We evaluate

$\int_{0}^{\pi}\cos x \; \mathrm{d} x$

$= \left.\begin{matrix} \sin x\\ \\ \end{matrix}\right|\begin{matrix} \pi \\ \\ 0 \end{matrix}$

$= \sin \pi - \sin 0 = 0 - 0 = 0$

The original double integral is now

$=\int_{0}^{1}y^{2} \cdot 0 \; \mathrm{d} y = \int_{0}^{1} 0 \; \mathrm{d} y= 0$

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Example Question #1 : Finding Integrals

Evaluate:

$\int_{0}^{1}\int_{0}^{\pi} y^{2}\sin x \; \mathrm{d} x \; \mathrm{d} y$

Possible Answers:

$\frac{1}{3}$

$\frac{2}{3}$

$\frac{2\sqrt{3}}{3}$

$\frac{ \sqrt{3}}{3}$

Correct answer:

$\frac{2}{3}$

Explanation:

$\int_{0}^{1}\int_{0}^{\pi} y^{2}\sin x \; \mathrm{d} x \; \mathrm{d} y$

$=\int_{0}^{1}y^{2}\left (\int_{0}^{\pi}\sin x \; \mathrm{d} x \right )\; \mathrm{d} y$

We evaluate

$\int_{0}^{\pi}\sin x \; \mathrm{d} x$

$= \left.\begin{matrix} -\cos x\\ \\ \end{matrix}\right|\begin{matrix} \pi \\ \\ 0 \end{matrix}$

$=- \cos \pi - (- \cos 0) = - (-1) - (-1) = 1 +1 = 2$

The original double integral is now

$=\int_{0}^{1}y^{2} \cdot 2 \; \mathrm{d} y$

$= 2 \int_{0}^{1}y^{2} \; \mathrm{d} y$

$= 2 \cdot \left.\begin{matrix} \frac{y^{3}}{3}\\ \\ \end{matrix}\right|\begin{matrix} 1 \\ \\ 0 \end{matrix}$

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

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Example Question #5 : Finding Integrals

Evaluate:

$\int_{0}^{1}\int_{0}^{\ln 2} e^{x}y^{2} \; \mathrm{d} x \; \mathrm{d} y$

Possible Answers:

$\frac{4}{3}$

$\frac{2}{3}$

$\frac{1}{3}$

Correct answer:

$\frac{1}{3}$

Explanation:

$\int_{0}^{1}\int_{0}^{\ln 2} e^{x}y^{2} \; \mathrm{d} x \; \mathrm{d} y$

$= \int_{0}^{1}y^{2} \left (\int_{0}^{\ln 2} e^{x} \; \mathrm{d}x \right ) \; \mathrm{d} y$

We evaluate

$\int_{0}^{\ln 2} e^{x} \; \mathrm{d}x$

$= \left.\begin{matrix} e^{x}\\ \\ \end{matrix}\right|\begin{matrix} \ln 2 \\ \\ 0 \end{matrix}$

$= e^{ln 2 } - e^{0} = 2 - 1 = 1$

The original double integral is now

$\int_{0}^{1}y^{2} \cdot 1 \; \mathrm{d} y$

$=\int_{0}^{1}y^{2} \; \mathrm{d} y$

$= \left.\begin{matrix} \frac{y^{3}}{3} \\ \\ \end{matrix}\right|\begin{matrix} 1 \\ \\ 0 \end{matrix}$

$= \frac{1^{3}}{3} - \frac{0^{3}}{3} = \frac{1}{3} - \frac{0}{3} = \frac{1}{3}$

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Example Question #6 : Finding Integrals

Evaluate:

$\int_{0}^{\ln 2}\int_{0}^{\ln 2} e^{x+y} \; \mathrm{d} x \; \mathrm{d} y$

Possible Answers:

e^2

Correct answer:

Explanation:

The problem is easier if it is written as follows:

$\int_{0}^{\ln 2}\int_{0}^{\ln 2} e^{x+y} \; \mathrm{d} x \; \mathrm{d} y$

$= \int_{0}^{\ln 2}\int_{0}^{\ln 2} e^{x} e^{y} \; \mathrm{d} x \; \mathrm{d} y$

$= \int_{0}^{\ln 2}e^{y} \left (\int_{0}^{\ln 2} e^{x} \; \mathrm{d} x \right ) \; \mathrm{d} y$

We evaluate

$\int_{0}^{\ln 2} e^{x} \; \mathrm{d}x$

$= \left.\begin{matrix} e^{x}\\ \\ \end{matrix}\right|\begin{matrix} \ln 2 \\ \\ 0 \end{matrix}$

$= e^{ln 2 } - e^{0} = 2 - 1 = 1$

The original double integral is now

$= \int_{0}^{\ln 2}e^{y} \cdot 1 \; \mathrm{d} y$

$= \int_{0}^{\ln 2}e^{y} \; \mathrm{d} y$

which, similarly to $\int_{0}^{\ln 2} e^{x} \; \mathrm{d}x$ , is equal to 1.

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Example Question #351 : Integrals

Evaluate:

$\int_{- \sqrt[3]{\pi}}^{\sqrt[3]{\pi}} \sin\left ( x^{3} \right ) \; \mathrm{d} x$

Possible Answers:

$- \frac{ \sqrt[3]{\pi ^{2} } }{2}$

$-2 \sqrt[3]{\pi ^{2} }$

$2 \sqrt[3]{\pi ^{2} }$

$\frac{ \sqrt[3]{\pi ^{2} } }{2}$

Correct answer:

Explanation:

No calculation is necessary here.

If $f (x) = \sin\left ( x^{3} \right )$ , then

$f (-x) = \sin\left [ \left (- x \right ) ^{3} \right ] = \sin \left (- x ^{3} \right )$ .

Since a sine is an odd function,

$f (-x) = \sin \left (- x ^{3} \right ) = - \sin \left ( x ^{3} \right ) = - f (x)$ .

This makes $f (x) = \sin\left ( x^{3} \right )$ an odd function, and for any ,

$\int_{-A}^{A} \sin\left ( x^{3} \right ) \; \mathrm{d} x = 0$ .

This integral meets the criterion.

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Calculus 2 : Calculus II

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #1 : Finding Integrals

Example Question #2 : Finding Integrals

Example Question #1 : Finding Integrals

Example Question #2 : Finding Integrals

Example Question #2 : Finding Integrals

Example Question #3 : Finding Integrals

Example Question #1 : Finding Integrals

Example Question #5 : Finding Integrals

Example Question #6 : Finding Integrals

Example Question #351 : Integrals

All Calculus 2 Resources

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