Call Now to Set Up Tutoring:

(888) 888-0446

Calculus 2 : Integrals

Study concepts, example questions & explanations for Calculus 2

Create An Account

All Calculus 2 Resources

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

Example Questions

Example Question #254 : Indefinite Integrals

$\int2x^2+x-3 dx$

Possible Answers:

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

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

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

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

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

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

Report an Error

Example Question #255 : Indefinite Integrals

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

Possible Answers:

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

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

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

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

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

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 #531 : Finding 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

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

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

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

Report an Error

Example Question #1 : Improper Integrals

Evaluate:

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

Possible Answers:

$\frac{1}{2}$

$\frac{1}{e}$

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

$\frac{1}{2e}$

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 #1 : Improper Integrals

Evaluate:

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

Possible Answers:

$\ln 4$

$-\frac{1}{4}$

The integral does not converge

$-\ln 4$

$\frac{1}{4}$

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 #881 : Integrals

Evaluate:

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

Possible Answers:

$\frac{1}{4e }$

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

$\frac{3}{4e }$

The integral does not converge.

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

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:

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

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

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

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

Report an Error

Example Question #531 : Finding Integrals

Evaluate:

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

Possible Answers:

$\frac{1}{2}$

$\frac{2}{e}$

$\frac{e}{2}$

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

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

Report an Error

Example Question #883 : Integrals

Evaluate:

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

Possible Answers:

$\frac{1}{2}$

The integral does not converge.

$\frac{1}{e}$

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.

Report an Error

Example Question #884 : Integrals

Evaluate the improper integral:

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

Possible Answers:

e^2

The integral does not converge.

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,

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

Report an Error

Example Question #881 : Integrals

Evaluate the improper integral:

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

Possible Answers:

$\frac{1}{2}$

$\frac{e}{2}$

$\frac{e}{4}$

$\frac{1}{4}$

The integral does not converge.

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.

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

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

Report an Error

View Calculus 2 Tutors

Dossevi
Certified Tutor

cole Ste Genevive Versailles, Bachelor of Science, Mathematics. Institut National Polytechnique de Grenoble, Master of Scienc...

View Calculus 2 Tutors

Scott
Certified Tutor

Florida State University, Bachelor of Science, Applied Mathematics. Florida State University, Bachelor of Science, Astrophysi...

View Calculus 2 Tutors

Sharif
Certified Tutor

University of South Florida-Main Campus, Bachelor of Science, Cellular and Molecular Biology.

All Calculus 2 Resources

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

Popular Subjects

SAT Tutors in Washington DC, Statistics Tutors in Philadelphia, Math Tutors in Denver, Calculus Tutors in Boston, GMAT Tutors in Boston, Algebra Tutors in San Diego, Spanish Tutors in Dallas Fort Worth, SSAT Tutors in Washington DC, Math Tutors in Boston, Math Tutors in New York City

Popular Courses & Classes

GMAT Courses & Classes in Washington DC, SAT Courses & Classes in Atlanta, MCAT Courses & Classes in Philadelphia, Spanish Courses & Classes in Phoenix, ISEE Courses & Classes in Los Angeles, SSAT Courses & Classes in Atlanta, ISEE Courses & Classes in Miami, MCAT Courses & Classes in Seattle, MCAT Courses & Classes in Boston, Spanish Courses & Classes in San Francisco-Bay Area

Popular Test Prep

GRE Test Prep in San Diego, ACT Test Prep in New York City, LSAT Test Prep in New York City, ISEE Test Prep in Los Angeles, MCAT Test Prep in Houston, ISEE Test Prep in Seattle, ISEE Test Prep in Dallas Fort Worth, SSAT Test Prep in Washington DC, LSAT Test Prep in San Francisco-Bay Area, ACT Test Prep in Dallas Fort Worth

DMCA Complaint

If you believe that content available by means of the Website (as defined in our Terms of Service) infringes one or more of your copyrights, please notify us by providing a written notice (“Infringement Notice”) containing the information described below to the designated agent listed below. If Varsity Tutors takes action in response to an Infringement Notice, it will make a good faith attempt to contact the party that made such content available by means of the most recent email address, if any, provided by such party to Varsity Tutors.

Your Infringement Notice may be forwarded to the party that made the content available or to third parties such as ChillingEffects.org.

Please be advised that you will be liable for damages (including costs and attorneys’ fees) if you materially misrepresent that a product or activity is infringing your copyrights. Thus, if you are not sure content located on or linked-to by the Website infringes your copyright, you should consider first contacting an attorney.

Please follow these steps to file a notice:

You must include the following:

A physical or electronic signature of the copyright owner or a person authorized to act on their behalf; An identification of the copyright claimed to have been infringed; A description of the nature and exact location of the content that you claim to infringe your copyright, in \ sufficient detail to permit Varsity Tutors to find and positively identify that content; for example we require a link to the specific question (not just the name of the question) that contains the content and a description of which specific portion of the question – an image, a link, the text, etc – your complaint refers to; Your name, address, telephone number and email address; and A statement by you: (a) that you believe in good faith that the use of the content that you claim to infringe your copyright is not authorized by law, or by the copyright owner or such owner’s agent; (b) that all of the information contained in your Infringement Notice is accurate, and (c) under penalty of perjury, that you are either the copyright owner or a person authorized to act on their behalf.

Send your complaint to our designated agent at:

Charles Cohn Varsity Tutors LLC
101 S. Hanley Rd, Suite 300
St. Louis, MO 63105

Or fill out the form below:

Do not fill in this field

Contact Information

* Full legal name

Company name

* Phone number

* Email address

Address

* Address1

Address2

* City

* State

* Zipcode

* Country

Complaint Details

* URL of copyrighted work (where your original material is located)

* Please describe the copyrighted work so that it may be easily identified

I am the owner, or an agent authorized to act on behalf of the owner of an exclusive right that is allegedly infringed.

I have a good faith belief that the use of the material in the manner complained of is not authorized by the copyright owner, its agent, or the law.

This notification is accurate.

I acknowledge that there may be adverse legal consequences for making false or bad faith allegations of copyright infringement by using this process.

* Signature (your digital signature is legally binding)

HIGH SCHOOL

GRADUATE SCHOOL

K-8

Search 50+ Tests

math tutoring

science tutoring

foreign languages

elementary tutoring

other

Search 350+ Subjects

Calculus 2 : Integrals

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #254 : Indefinite Integrals

Example Question #255 : Indefinite Integrals

Example Question #531 : Finding Integrals

Example Question #1 : Improper Integrals

Example Question #1 : Improper Integrals

Example Question #881 : Integrals

Example Question #531 : Finding Integrals

Example Question #883 : Integrals

Example Question #884 : Integrals

Example Question #881 : Integrals

All Calculus 2 Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: