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Calculus 2 : Calculus II

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

Example Questions

Example Question #881 : Integrals

$\int2x^2+x-3 dx$

Possible Answers:

$\frac{2x^3}{3}+\frac{x^2}{3}-3x+C$

$\frac{2x^3}{3}+\frac{x^2}{2}-3x+C$

$\frac{2x^3}{3}+\frac{x^2}{2}-3x^3+C$

$\frac{2x^3}{3}+\frac{x^2}{2}-3x$

$\frac{2x^3}{4}+\frac{x^2}{2}-3x+C$

Correct answer:

$\frac{2x^3}{3}+\frac{x^2}{2}-3x+C$

Explanation:

To find this integral, look at each term separately.

For the first term, raise the exponent by 1 and also put that result on the denominator: $2(\frac{x^3}{3})=\frac{2x^3}{3}$ .

For the next term, do the same: $\frac{x^2}{2}$ .

Same for the third term (since it's a constant, multiply the coefficient by x): -3x .

Put those all together to get: $\frac{2x^3}{3}+\frac{x^2}{2}-3x$ . Since this is an indefinite integral, make sure to add C at the end: $\frac{2x^3}{3}+\frac{x^2}{2}-3x+C$ .

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Example Question #882 : Integrals

$\int \frac{3x^2-4x}{x}dx$

Possible Answers:

$\frac{3x^2}{2}-4x^3+C$

$\frac{3x^2}{2}-4x+C$

$\frac{3x^2}{2}+4x+C$

$\frac{3x^2}{2}-4x$

$\frac{x^2}{2}-4x+C$

Correct answer:

$\frac{3x^2}{2}-4x+C$

Explanation:

First chop up the fraction into two separate terms and simplify: $\frac{3x^2-4x}{x}=\frac{3x^2}{x}-\frac{4x}{x}=3x-4$ .

Now, integrate that expression. Remember to raise the exponent by 1 and also put that result on the denominator:

$\frac{3x^2}{2}-4x$

Since it's an indefinite integral, remember to add C at the end: $\frac{3x^2}{2}-4x+C$ .

Report an Error

Example Question #883 : Integrals

Evaluate:

$\int_{1}^{\infty } \frac{4\; \mathrm{d}x}{x^{2}}$

Possible Answers:

The integral does not converge

Correct answer:

Explanation:

$\int_{1}^{\infty } \frac{4\; \mathrm{d}x}{x^{2}} = 4 \lim_{b\rightarrow \infty } \int_{1}^{b}\frac{\; \mathrm{d}x}{x^{2}}$

$\int \frac{ \; \mathrm{d}x}{x^{2}} = \int x^{-2} \mathrm{d}x} = \frac{ x^{-1}}{-1}= \left ( - \frac{1}{x} \right )= - \frac{1}{x}$

$\lim_{x\rightarrow \infty } \frac{1}{x} = 0$ , so

$4 \lim_{b\rightarrow \infty } \int_{1}^{b}\frac{\; \mathrm{d}x}{x^{2}}$

$= 4 \lim_{b\rightarrow \infty } \left.\begin{matrix} - \frac{1}{x}\\ \textrm{ } \\ \textrm{ } \end{matrix}\right|\begin{matrix} b\\ \\1 \end{matrix}$

$=4 \left [ -0 -\left ( - \frac{1}{1}\right ) \right ]= 4 \cdot 1 = 4$

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Example Question #1 : Improper Integrals

Evaluate:

$\int_{0}^{\infty } \frac{\cos x\; \mathrm{d}x}{e^{x}}$

Possible Answers:

$\frac{1}{2e}$

$\frac{1}{e}$

$\frac{\sqrt{2}}{2e}$

$\frac{1}{2}$

Correct answer:

$\frac{1}{2}$

Explanation:

First, we will find the indefinite integral $\int \frac{\cos x\; \mathrm{d}x}{e^{x}} = \int e^{-x} \cos x\; \mathrm{d}x$ using integration by parts.

We will let $u = \cos x$ and $\mathrm{d}v= e^{-x}\; \mathrm{d}x$ .

Then $\frac{\mathrm{d} u}{\mathrm{d} x} = - \sin x$ and $\mathrm{d} u = - \sin x \; \mathrm{d} x$ .

$v= \int e^{-x}\; \mathrm{d}x = - e^{-x} = - \frac{1}{e^{x} }$

$\int \frac{\cos x\; \mathrm{d}x}{e^{x}} = \int e^{-x} \cos x\; \mathrm{d}x$

$=\int u \; \mathrm{d} v$

$= uv - \int v \; \mathrm{d} u$

$= \cos x \cdot \left (- \frac{1}{e^{x} } \right ) - \int \left (- \frac{1}{e^{x} } \right ) \; \cdot \left (- \sin x \right )\; \mathrm{d} x$

$= - \frac{\cos x}{e^{x} } - \int e^{-x} \sin x \; \mathrm{d} x$

To find $\int e^{-x} \sin x \; \mathrm{d} x$ , we use another integration by parts:

$u = \sin x$ , which means that $\mathrm{d} u = \cos x \; \mathrm{d} x$ , and

$\mathrm{d}v= e^{-x}\; \mathrm{d}x$ , which means that, again, $v= - \frac{1}{e^{x} }$ .

$\int e^{-x} \sin x \; \mathrm{d} x$

$=\int u \; \mathrm{d} v$

$= uv - \int v \; \mathrm{d} u$

$= \sin x \cdot \left (- \frac{1}{e^{x} } \right ) - \int \left (- \frac{1}{e^{x} } \right ) \; \cdot \cos x \; \mathrm{d} x$

$= - \frac{ \sin x }{e^{x} } + \int \frac{\cos x \; \mathrm{d} x}{e^{x} }$

$\int \frac{\cos x\; \mathrm{d}x}{e^{x}} = - \frac{\cos x}{e^{x} } - \int e^{-x} \sin x \; \mathrm{d} x$

$\int \frac{\cos x\; \mathrm{d}x}{e^{x}} = - \frac{\cos x}{e^{x} } - \left ( - \frac{ \sin x }{e^{x} } + \int \frac{\cos x \; \mathrm{d} x}{e^{x} } \right )$

$\int \frac{\cos x\; \mathrm{d}x}{e^{x}} = - \frac{\cos x}{e^{x} } + \frac{ \sin x }{e^{x} } - \int \frac{\cos x \; \mathrm{d} x}{e^{x} }$

$2 \int \frac{\cos x\; \mathrm{d}x}{e^{x}} = - \frac{\cos x}{e^{x} } + \frac{ \sin x }{e^{x} }$

$2 \int \frac{\cos x\; \mathrm{d}x}{e^{x}} = \frac{\sin x -\cos x}{e^{x} }$

$\int \frac{\cos x\; \mathrm{d}x}{e^{x}} = \frac{\sin x -\cos x}{2e^{x} }$

Since

$\frac{-2}{2e^{x} } \leq \frac{\sin x -\cos x}{2e^{x} } \leq \frac{2}{2e^{x} }$ , or,

$- \frac{1}{e^{x} } \leq \frac{\sin x -\cos x}{2e^{x} } \leq \frac{1}{e^{x} }$

for all real , and

$- \lim_{x\rightarrow \infty } \frac{1}{e^{x} } = \lim_{x\rightarrow \infty } \frac{1}{e^{x} } = 0$ ,

by the Squeeze Theorem,

$\lim_{x\rightarrow \infty } \frac{\sin x -\cos x}{2e^{x} } = 0$ .

$\int_{0}^{\infty } \frac{\cos x\; \mathrm{d}x}{e^{x}}$

$= \lim_{b\rightarrow \infty }\int_{0}^{b } \frac{\cos x\; \mathrm{d}x}{e^{x}}$

$= \lim_{ b\rightarrow \infty }$ $\left.\begin{matrix} \frac{\sin x -\cos x}{2e^{x} } \\ \textrm{ } \end{matrix}\right|$ $\begin{matrix} b\\ \\ 0 \end{matrix}$

$= 0 - \frac{\sin 0 -\cos 0}{2e^{0} }$

$= - \frac{ 0 -1}{2 \cdot 1 } = \frac{1}{2}$

Report an Error

Example Question #2 : Improper Integrals

Evaluate:

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

Possible Answers:

$-\ln 4$

The integral does not converge

$\ln 4$

$\frac{1}{4}$

$-\frac{1}{4}$

Correct answer:

$-\frac{1}{4}$

Explanation:

First, we will find the indefinite integral, $\int x \ln x \; \mathrm{d}x$ .

We will let $u = \ln x$ and $\mathrm{d}v= x \; \mathrm{d}x$ .

Then,

$\frac{\mathrm{d} u }{\mathrm{d} x}= \frac{1}{x}$ and $\mathrm{d} u = \frac{\mathrm{d} x}{x}$ .

$v= \int x \; \mathrm{d}x =\frac{ x ^{2}}{2}$

and

$\int x \ln x \; \mathrm{d}x$

$=\int u \; \mathrm{d} v$

$= uv - \int v \; \mathrm{d} u$

$= \ln{x} \cdot \frac{ x ^{2}}{2} - \int \left (\frac{ x ^{2}}{2} \right ) \frac{\mathrm{d} x}{x}$

$= \frac{ x ^{2}\ln{x}}{2} - \frac{1}{2}\int x\; \mathrm{d} x$

$= \frac{ x ^{2}\ln{x}}{2} - \frac{1}{2} \cdot \frac{x^{2}}{2}$

$= \frac{ x ^{2}\ln{x}}{2} - \frac{x^{2}}{4}$

Now, this expression evaluated at x = 1 is equal to

$\frac{ 1 ^{2}\ln{1}}{2} - \frac{1^{2}}{4} = 0 - \frac{1}{4} = - \frac{1}{4}$ .

At x = 0 it is undefined, because $\ln 0$ does not exist.

We can use L'Hospital's rule to find its limit as $x\rightarrow 0$ , as follows:

$\lim_{x\rightarrow 0}\frac{2 x ^{2}\ln{x} - x^{2}}{4} =\lim_{x\rightarrow 0} \frac{2 \ln{x} - 1}{\frac{4}{x ^{2}}}$

$\lim_{x\rightarrow 0} \left (2 \ln{x} - 1 \right ) = - \infty$ and $\lim_{x\rightarrow 0} \frac{4}{x ^{2}} = \infty$ , so by L'Hospital's rule,

$\lim_{x\rightarrow 0} \frac{2 \ln{x} - 1}{\frac{4}{x ^{2}}}=\lim_{x\rightarrow 0} \frac{\frac {2}{x} }{\frac{-8}{x ^{3}}} = \lim_{x\rightarrow 0}\left ( -\frac{x^{2}}{4} \right ) = 0$

Therefore,

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

$= \left.\begin{matrix} \frac{ x ^{2}\ln{x}}{2} - \frac{x^{2}}{4} \\ \\ \end{matrix}\right| \begin{matrix} 1\\ \\ 0 \end{matrix}$

$=\lim _{b\rightarrow 0}\left [ \left.\begin{matrix} \frac{ x ^{2}\ln{x}}{2} - \frac{x^{2}}{4} \\ \\ \end{matrix}\right| \begin{matrix} 1\\ \\ b \end{matrix} \right ]$

$= -\frac{1}{4} - 0$

$= -\frac{1}{4}$

Report an Error

Example Question #884 : Integrals

Evaluate:

$\int_{1}^{\infty } \frac{x\; \mathrm{d}x}{e^{2x}}$

Possible Answers:

The integral does not converge.

$\frac{1}{4e }$

$\frac{3}{4e^{2 }}$

$\frac{3}{4e }$

$\frac{1}{4e^{2 }}$

Correct answer:

$\frac{3}{4e^{2 }}$

Explanation:

First, we will find the indefinite integral

$\int \frac{x\; \mathrm{d}x}{e^{2x}} = \int xe^{-2x}\; \mathrm{d}x$

using integration by parts.

We will let u = x and $\mathrm{d}v= e^{-2x}\; \mathrm{d}x$ .

Then $\frac{\mathrm{d} u }{\mathrm{d} x}= 1$ and $\mathrm{d} u =\mathrm{d} x$ .

Also,

$v= \int e^{-2x}\; \mathrm{d}x$ .

To find , we can substitute for -2x . Then $\frac{\mathrm{d} t}{\mathrm{d} x} = -2$ or $\mathrm{d} x = -\frac{1}{2} \mathrm{d} t$ , so the antiderivative is

$v= - \frac{1}{2} \int e^{t}\; \mathrm{d}t = - \frac{1}{2} e^{t} = - \frac{1}{2} e^{-2x} = - \frac{1}{2e^{2x}}$ .

Now we can integrate by parts:

$\int \frac{x\; \mathrm{d}x}{e^{2x}} = \int xe^{-2x}\; \mathrm{d}x$

$=\int u \; \mathrm{d} v$

$= uv - \int v \; \mathrm{d} u$

$= x \left ( - \frac{1}{2e^{2x}} \right ) - \int\left ( - \frac{1}{2e^{2x}} \right )\; \mathrm{d} x$

$= - \frac{x}{2e^{2x}} + \frac{1}{2} \int e^{-2x} } \; \mathrm{d} x$

$= - \frac{x}{2e^{2x}} + \frac{1}{2}\left ( - \frac{1}{2e^{2x}} \right )$

$= - \frac{x}{2e^{2x}} - \frac{1}{4e^{2x}}$

$= - \frac{2x}{4e^{2x}} - \frac{1}{4e^{2x}}$

$= - \frac{2x+1}{4e^{2x}}$

By L'Hospital's rule, since both the numerator and the denominator approach $\infty$ as $x\rightarrow \infty$ ,

$\lim_{x\rightarrow \infty } - \frac{2x+1}{4e^{2x}} = \lim_{x\rightarrow \infty } - \frac{2}{8e^{2x}} = \lim_{x\rightarrow \infty } - \frac{1}{4e^{2x}} = 0$ .

So:

$\int_{1}^{\infty } \frac{x\; \mathrm{d}x}{e^{2x}}$

$= \lim_{b\rightarrow \infty } \int_{1}^{b } \frac{x\; \mathrm{d}x}{e^{2x}}$

$= \lim_{b\rightarrow \infty } \left.\begin{matrix} - \frac{2x+1}{4e^{2x}} \\ \textrm{ } \end{matrix}\right| \begin{matrix} b \\ 1 \end{matrix}$

$= 0 -\left ( - \frac{2 \cdot 1 +1}{4e^{2\cdot 1 }} \\ \textrm{ } \right ) = \frac{3}{4e^{2 }}$

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Example Question #1 : Improper Integrals

Evaluate:

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

Possible Answers:

$\frac{e}{2}$

$\frac{1}{2}$

$\frac{2}{e}$

Correct answer:

$\frac{1}{2}$

Explanation:

Rewrite the integral as

$\int_{0 }^{\infty} \frac{x\; \mathrm{d}x}{e^{x^{2}}} = \int_{0 }^{\infty}e^{-x^{2}}\cdot x \; \mathrm{d}x} }$ .

Substitute $u = - x^{2}$ . Then

$\frac{\mathrm{d} u}{\mathrm{d} x} = - 2x$ and $-\frac{1}{2} \mathrm{d} u = x \; \mathrm{d} x$ . The lower bound of integration stays $u = - 0^{2} = 0$ , and the upper bound becomes $\lim_{x\rightarrow \infty } - x^{2} = - \infty$ , so

$\int_{0 }^{\infty} \frac{x\; \mathrm{d}x}{e^{x^{2}}} = \int_{0 }^{\infty}e^{-x^{2}}\cdot x \; \mathrm{d}x} }=-\frac{1}{2} \int_{0 }^{-\infty}e^{u} \mathrm{d}u}= \frac{1}{2} \int_{-\infty }^{0}e^{u} \mathrm{d}u}$

$=\frac{1}{2} \lim_{b\rightarrow -\infty } \left.\begin{matrix} e ^{u}\\ \textrm{ } \\ \textrm{ } \end{matrix}\right| \begin{matrix} 0\\ \\ b \end{matrix}$

Since $\lim_{b\rightarrow -\infty } e ^{u} = 0$ , the above is equal to

$=\frac{1}{2} (e^{0}-0) =\frac{1}{2} (1) = \frac{1}{2}$ .

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Example Question #885 : Integrals

Evaluate:

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

Possible Answers:

The integral does not converge.

$\frac{1}{e}$

$\frac{1}{2}$

Correct answer:

The integral does not converge.

Explanation:

Substitute $u = 1 + x^{2}$ .

$\frac{\mathrm{d} u}{\mathrm{d} x} = 2 x$

$\frac{1}{2}\mathrm{d} u = x \; \mathrm{d} x$

The lower bound of integration becomes $u = 1 + 0^{2} = 1$ ;

the upper bound becomes $\lim_{x\rightarrow \infty }\left ( x^{2}+ 1 \right )= \infty$ .

The integral therefore becomes

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

$= \int_{1}^{\infty} \frac{ \mathrm{d} u}{2u} = \frac{1}{2} \int_{1}^{\infty} \frac{ \mathrm{d} u}{u}$

$= \frac{1}{2} \lim_{b\rightarrow \infty} \int_{1}^{b} \frac{ \mathrm{d} u}{u}$

$= \lim_{b\rightarrow \infty} \left.\begin{matrix} \frac{1}{2} \ln u \\ \\ \end{matrix}\right| \begin{matrix} b \\ \\ 1 \end{matrix}$

$= \lim_{b\rightarrow \infty} \frac{1}{2} \ln b - \frac{1}{2} \ln 1$

$= \lim_{b\rightarrow \infty} \frac{1}{2} \ln b - \frac{1}{2} \cdot0$

$= \lim_{b\rightarrow \infty} \frac{1}{2} \ln b = \infty$

The integral, therefore, does not converge.

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Example Question #886 : Integrals

Evaluate the improper integral:

$\int_{0}^{\infty} x^2 e^{-x} \mathrm{d}x$

Possible Answers:

The integral does not converge.

e^2

Correct answer:

Explanation:

First, we will perform an integration by parts on the indefinite integral:

$\int x^2 e^{-x} \mathrm{d}x$

Let u = x^2 and $\mathrm {d}v = e^{-x} \; \mathrm {d}x$ .

Then,

$\frac{\mathrm{d} u}{\mathrm{d} x} = 2x$ , $\mathrm{d} u =2 x \; \mathrm{d} x$ ,

and

$\mathrm {d}v = e^{-x} \; \mathrm {d}x$ .

$\int \mathrm {d}v = \int e^{-x} \; \mathrm {d}x$

$v = -e^{-x}$

$\int x^2 e^{-x} \mathrm{d}x$

$= \int u \; \mathrm{d}v$

$= uv- \int v \; \mathrm{d}u$

$= x^2 (-e^{-x})- \int (-e^{-x}) \cdot 2x \; \mathrm{d}x$

$= -x^2 e^{-x}+ 2\int xe^{-x} \mathrm{d}x$

We do another integration by parts, setting

$\widetilde{u }= x$ and $\mathrm {d}\widetilde{v} = e^{-x} \; \mathrm {d}x$ .

Then,

$\frac{\mathrm{d} \widetilde{u}}{\mathrm{d} x} = 1$ and $\mathrm{d} \widetilde{u} = \mathrm{d} x$ .

Also,

$\mathrm {d}\widetilde{v} = e^{-x} \; \mathrm {d}x$ and, again, $\widetilde{v }= -e^{-x}$ .

$=- x^2 e^{-x}+ 2\int xe^{-x} \mathrm{d}x$

$=- x^2 e^{-x}+ 2 \int \widetilde{u} \; \mathrm{d}\widetilde{v}$

$=- x^2 e^{-x}+ 2 \left ( \widetilde{u} \widetilde{v} - \int \widetilde{v} \mathrm{d}\widetilde{u} \right )$

$= -x^2 e^{-x}+ 2 \left [ x \left ( -e^{-x} \right ) - \int \left ( -e^{-x} \right ) \mathrm{d}x\right]$

$=- x^2 e^{-x}- 2 x e^{-x} +2 \int e^{-x} \mathrm{d}x\right]$

$=- x^2 e^{-x}- 2 x e^{-x} + 2\left ( - e^{-x} \right )$

$=- x^2 e^{-x}- 2 x e^{-x} - 2e^{-x}$

$=\left (- x^2 - 2 x - 2 \right )e^{-x}$

$=\frac{ -x^2 - 2 x -2 }{e^x}$

Therefore, the antiderivative of $f(x ) = x^2 e^{-x}$ is $F(x)=\frac{ x^2 - 2 x -2 }{e^x}$ .

$\int_{0}^{\infty} x^2 e^{-x} \mathrm{d}x = \lim_{b\rightarrow \infty } \int_{0}^{b} x^2 e^{-x} \mathrm{d}x =\lim_{b\rightarrow \infty } F(b) - F(0)$

$\lim_{b\rightarrow \infty } F(b) = \lim_{x\rightarrow \infty } \frac{ x^2 - 2 x -2 }{e^x} = 0$ , which two applications of L'Hospital's rule can easily reveal.

$F(0)=\frac{ 0^2 - 2\cdot 0 -2 }{e^0} =\frac{-2}{1} = -2$

Therefore,

$\int_{0}^{\infty} x^2 e^{-x} \mathrm{d}x = 0 - (-2) = 2$ .

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Example Question #2631 : Calculus Ii

Evaluate the improper integral:

$\int_{0}^{\infty} xe^{-2x} \mathrm{d}x$

Possible Answers:

The integral does not converge.

$\frac{e}{4}$

$\frac{1}{2}$

$\frac{1}{4}$

$\frac{e}{2}$

Correct answer:

$\frac{1}{4}$

Explanation:

First, we will perform an integration by parts on the indefinite integral

$\int xe^{-2x} \mathrm{d}x$ .

Let u = x and $\mathrm {d}v = e^{-2x} \; \mathrm {d}x$ .

Then,

$\frac{\mathrm{d} u}{\mathrm{d} x} = 1$ and $\mathrm{d} u = \mathrm{d} x$ .

Also,

$\mathrm {d}v = e^{-2x} \; \mathrm {d}x$ .

$\int \mathrm {d}v = \int e^{-2x} \; \mathrm {d}x$

$v = \frac{e^{-2x}}{-2} =- \frac{1}{2}e^{-2x}$

Therefore,

$\int xe^{-2x} \mathrm{d}x$

$= \int u \; \mathrm{d}v$

$= uv- \int v \; \mathrm{d}u$

$= x \cdot - \frac{1}{2}e^{-2x}- \int \left (- \frac{1}{2}e^{-2x} \right ) \cdot \; \mathrm{d}x$

$= - \frac{x}{2}e^{-2x} + \frac{1}{2} \int e^{-2x} \; \mathrm{d}x$

$= - \frac{x}{2}e^{-2x} + \frac{1}{2}\left ( - \frac{1}{2}e^{-2x} \right )$

$= - \frac{x}{2}e^{-2x} - \frac{1}{4} e^{-2x}$

The antiderivative of $f(x) = xe^{-2x}$ is

$F(x) = - \frac{x}{2}e^{-2x} - \frac{1}{4} e^{-2x}$

and

$\int_{0}^{ \infty} xe^{-2x} \mathrm{d}x = \lim_{b\rightarrow \infty }\int_{0}^{b} xe^{-2x} \mathrm{d}x= \lim_{b\rightarrow \infty } F(b) - F (0)$ .

$\lim_{b\rightarrow \infty } F(b) = -\lim_{b\rightarrow \infty } \frac{x}{2}e^{-2x} -\lim_{b\rightarrow \infty } - \frac{1}{4} e^{-2x}$

$\lim_{b\rightarrow \infty } F(b) = -\lim_{b\rightarrow \infty } \frac{x}{2e^{2x}} -\lim_{b\rightarrow \infty } \frac{1}{4e^{2x}} = 0$ , as can be proved by L'Hospital's rule.

$F(0) = - \frac{0}{2}\cdot e^{-2 \cdot 0} - \frac{1}{4}\cdot e^{-2 \cdot 0} = 0 - \frac{1}{4} = - \frac{1}{4}$

$\int_{0}^{ \infty} xe^{-2x} \mathrm{d}x = 0 - \left ( - \frac{1}{4}\right ) = \frac{1}{4}$

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Calculus 2 : Calculus II

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Example Question #881 : Integrals

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