Call Now to Set Up Tutoring:

(888) 888-0446

Calculus 2 : Calculus II

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 #21 : Integrals

Find the indefinite integral.

$\int x^2\sin(x)dx$

Possible Answers:

$-x^2\cos(x)+2x\cos(x)+C$

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

$-x^2\sin(x)+2x\sin(x)+2\sin(x)+2+C$

$-x^2\cos(x)+2x\sin(x)+2\cos(x)+C$

Correct answer:

$-x^2\cos(x)+2x\sin(x)+2\cos(x)+C$

Explanation:

Use Integration By Parts (IBP) to simplify the integral, using the IBP formula

$\int udv = uv - \int vdu$

Assign the variables and to appropriate expressions from the original integral,

$u = x^{2} \rightarrow du = 2xdx$

$dv = \sin(x)dx \rightarrow v = -\cos(x)$

Substitute these new expressions into the IBP formula and our integral becomes as follows,

$\int x^2\sin(x)dx = -x^{2}\cos(x)- \int -\cos(x)2xdx$

$= -x^{2}\cos(x) + \int \cos(x)2xdx$

Use IBP again on the integral $\int \cos(x)2xdx$ and assign the variables and to appropriate expressions from the integral,

$u = 2x \rightarrow du= 2dx$

$dv = \cos(x)dx \rightarrow v= \sin(x)$

Substitute these new expressions into the IBP formula again and our full expression becomes as follows,

$-x^{2}\cos(x) + \left [ 2x\sin(x)-\int 2\sin(x)dx \right ]$

Integrate using the trig integration rules as follows,

$\int \sin(x)dx = - \cos(x)$

So the final indefinite integral expression becomes

$-x^{2}\cos(x) + 2x\sin(x)- (-2\cos(x)) +C$

And the final answer is as follows,

$-x^{2}\cos(x) + 2x\sin(x)+ 2\cos(x) +C$

Report an Error

Example Question #21 : Integrals

Find the indefinite integral.

$\int \frac{1}{x^2+x\sqrt{x}}dx$

Possible Answers:

$ln\left | x^2+x\sqrt{x}\right | + C$

$-2ln\left | \sqrt{x} \right |-\frac{2}{\sqrt{x}}+2ln\left | \sqrt{x}+1 \right | + C$

$2ln\left | x^\frac{3}{2}+x\right | + C$

$-2ln\left | x^\frac{3}{2}\right | + 2ln\left | x\right | + C$

$ln\left | \sqrt{x} \right |+\sqrt{x}+ln\left | \frac{3}{\sqrt{x}} \right | + C$

Correct answer:

$-2ln\left | \sqrt{x} \right |-\frac{2}{\sqrt{x}}+2ln\left | \sqrt{x}+1 \right | + C$

Explanation:

Make an appropriate u-substitution to simplify the integral.

$u = \sqrt{x} \rightarrow u^{2}=x \rightarrow u^{4}= x^{2}$

$du = \frac{1}{2}x^{-\frac{1}{2}}dx = \frac{1}{2\sqrt{x}}dx$

$2\sqrt{x}du=dx$

Since $u = \sqrt{x}$ ,

2udu=dx

Substitute the new expressions containing and back into the original integral as follows,

$\int \frac{1}{u^{4}+u^{3}}2udu \Leftrightarrow 2\int \frac{u}{u(u^{3}+u^{2})}du \Leftrightarrow 2\int \frac{1}{u^{3}+u^{2}}du$

Use partial fraction decomposition on the integral to simplify the expression as follows,

$\frac{1}{u^{3}+u^{2}} = \frac{1}{u^{2}(u+1)} = \frac{A}{u}+ \frac{B}{u^{2}}+ \frac{C}{u+1}$

$1=A(u)(u+1)+B(u+1)+C(u^{2})$

Set u = 0 then,

$1 = A(0)(0+1)+B(0+1)+C(0^{2}) \rightarrow B=1$

Set u = -1 then,

$1 = A(-1)(-1+1)+B(-1+1)+C(-1)^{2} \rightarrow C=1$

Set u = 1 then,

$1 = A(1)(1+1)+B(1+1)+C(1)^{2}$

1 = A(2)+ (1)(2)+ (1)(1)

$1 = 2A + 2 + 1 \rightarrow A = -1$

So,

$\frac{1}{u^{3}+u^{2}} = \frac{-1}{u}+ \frac{1}{u^{2}}+ \frac{1}{u+1}$

And our integral expression becomes

$2\int (\frac{-1}{u}+ \frac{1}{u^{2}}+ \frac{1}{u+1})du$

Integrate each term. We will need to use the power rule and natural log integration as follows,

$\int \frac{1}{x}dx = ln\left | x\right | \rightarrow \int \frac{1}{u+1}du = ln\left | u + 1\right |$

Where $n\neq 1$ , $\int \frac{1}{x^{n}}dx = \int x^{-n}dx = \frac{x^{-n+1}}{-n+1} \rightarrow \int \frac{1}{u^{2}}du = -\frac{1}{u}$

This results in our final indefinite integral,

$2(-ln\left | u\right |-\frac{1}{u}+ln\left | u + 1\right |) + C$

Substitute back in our original u-substitution $u = \sqrt{x}$ and simplify to give us our final answer

$-2ln\left | \sqrt{x}\right |-\frac{2}{\sqrt{x}}+2ln\left | \sqrt{x}+ 1\right | + C$

Report an Error

Example Question #23 : Integrals

Find the definite integral.

$\int_{0}^{8\pi ^2}\cos(\sqrt{x+\pi ^2})dx$

Possible Answers:

$-3\pi$

Correct answer:

Explanation:

Simplify the integral using an appropriate u-substitution,

$u = \sqrt{x+\pi^{2}}$

$du = \frac{1}{2}(x+\pi^{2})^{-\frac{1}{2}}dx = \frac{1}{2\sqrt{x+\pi^{2}}}dx$

$2\sqrt{x+\pi^{2}}du=dx$

Since $u = \sqrt{x+\pi^{2}}$ ,

2udu=dx

Substitute the new expressions containing and into the original integral as follows,

$\int \cos(u)2udu$

$2\int u \cos(u)du$

Use Integration By Parts (IBP) to simplify the integral further, using the IBP formula

$\int wdv = wv - \int vdw$

Assign the variables and to appropriate expressions from the integral

$w = u \rightarrow dw = du$

$dv = \cos(u)du \rightarrow v = \sin(u)$

Substitute these expressions back into the IBP formula and our integral becomes as follows,

$2\left ( u\sin(u)-\int \sin(u)du\right )$

Integrate using the trig integration rule

$\int \sin(x)dx = -\cos(x)$

Our indefinite integral becomes

$2(u\sin(u)-(-\cos(u))) +C \Leftrightarrow 2u\sin(u)+2\cos(u)+C$

Substitute back in our original u-substitution $u = \sqrt{x+\pi^{2}}$ to get our final indefinite integral,

$2\sqrt{x+\pi^{2}}\sin(\sqrt{x+\pi^{2}})+2\cos(\sqrt{x+\pi^{2}})+C$

Apply the given boundaries x = 0 to $x = 8\pi^2$ to get our final answer.

$\left [ 2\sqrt{8\pi^2+\pi^{2}}\sin(\sqrt{8\pi^2+\pi^{2}})+2\cos(\sqrt{8\pi^2+\pi^{2}})\right ]-\left [ 2\sqrt{0+\pi^{2}}\sin(\sqrt{0+\pi^{2}})+2\cos(\sqrt{0+\pi^{2}})\right ]$

$\left [ 2(3\pi)(0)+2(-1)\right ] - \left [ 2(\pi)(0)+2(-1)\right ]$

$\left [ -2\right ]- \left [ -2\right ] = 0$

Report an Error

Example Question #21 : Integrals

Find the indefinite integral.

$\int x(x+7)^{100} dx$

Possible Answers:

$\frac{1}{2}x^2+\frac{1}{101}(x+7)^{101}+C$

$\frac{1}{102}x^{102}-\frac{7}{102}x^{101}+C$

$\frac{(x^2+7x)^{101}}{101}*(\frac{1}{3}x^3+\frac{7}{2}x^2)+C$

$\frac{1}{102}(x+7)^{102}-\frac{7}{101}(x+7)^{101}+C$

Correct answer:

$\frac{1}{102}(x+7)^{102}-\frac{7}{101}(x+7)^{101}+C$

Explanation:

Make an appropriate u-substitution to transform the integral into something we are familiar with.

$u = x+7 \rightarrow u-7=x$

du = dx

Substitute the new expressions containing and into the original equation as follows,

$\int (u-7)(u)^{100}du$

$\int \left ( u^{101}-7u^{100}\right )du$

Integrate each term using the power rule.

$\int cx^{n}dx = \frac{cx^{n+1}}{n+1} \rightarrow \int -7u^{100}du = -\frac{7u^{101}}{101}$

After integration we get the expression as follows,

$\frac{u^{102}}{102}-\frac{7u^{101}}{101}+C$

Substitute back in our original u-substitution u = x+7 to get the final answer.

$\frac{(x+7)^{102}}{102}-\frac{7(x+7)^{101}}{101}+C$

Report an Error

Example Question #25 : Integrals

Find the indefinite integral.

$\int x^2ln\left | x+8\right |dx$

Possible Answers:

$\frac{1}{3}x^3ln\left | x+8\right |-\frac{1}{12}x^4ln\left | x+8\right |+C$

$\frac{1}{3}x^3ln\left | x+8\right |-\frac{1}{9}x^3+\frac{4}{3}x^2-\frac{64}{3}x+\frac{512}{3}ln\left | x+8\right |+C$

$\frac{1}{3}x^3ln\left | x+8\right |+C$

$\frac{1}{3(x+8)}x^3+C$

Correct answer:

$\frac{1}{3}x^3ln\left | x+8\right |-\frac{1}{9}x^3+\frac{4}{3}x^2-\frac{64}{3}x+\frac{512}{3}ln\left | x+8\right |+C$

Explanation:

First use Integration By Parts (IBP) to make the integral doable, using the IBP formula,

$\int udv = uv-\int vdu$

Assign appropriate expressions from the original integral to the variables and to use in the IBP formula.

$u = ln\left | x+8\right |\rightarrow du = \frac{1}{x+8}dx$

$dv = x^{2}dx \rightarrow v = \frac{1}{3}x^{3}$

Substitute these expressions into the IBP formula and our integral becomes as follows,

$\int x^2ln\left | x+8\right |dx = \frac{1}{3}x^{3}ln\left | x+8\right |- \int \frac{1}{3}x^3*\frac{1}{x+8}dx$

Use long division to simplify the expression $\frac{x^{3}}{x+8}$ ,

$x^{2} -8x+64-\frac{512}{x+8}$

$x+8|\overline{x^3 \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; }$

$\\ -(x^{3}+8x^{2})\\ \overline{\ \ \ \ \ \ \ \ -8x^{2}\ \ \ \ }\\ {\ \ \ \ \ -(-8x^{2}-64x)}\\ \overline{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 64x}\\ {\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ -(64x+512)}\\ \overline{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 512}$

So,

$\frac{x^{3}}{x+8} = x^{2} -8x+64-\frac{512}{x+8}$

Substitute this back into its appropriate place in our expression and it becomes as follows,

$\frac{1}{3}x^{3}ln\left | x+8\right |- \frac{1}{3} \int (x^{2} -8x+64-\frac{512}{x+8})dx$

Integrate each term using the power rule and natural log integration as follows,

$\int x^{n}dx = \frac{x^{n+1}}{n+1} \rightarrow \int x^{2}dx= \frac{1}{3}x^{3}$

$\int \frac{c}{x}dx = cln\left | x\right | \rightarrow \int \frac{-512}{x+8}dx = -512ln\left | x+8\right |$

Our indefinite integral becomes as follows,

$\frac{1}{3}x^{3}ln\left | x+8\right |- \frac{1}{3}(\frac{1}{3}x^{3}-4x^{2}+64x-512ln\left | x+8\right |)+C$

Simplifying, our answer is

$\frac{1}{3}x^{3}ln\left | x+8\right |- \frac{1}{9}x^{3}-\frac{4}{3}x^{2}+\frac{64}{3}x-\frac{512}{3}ln\left | x+8\right |+C$

Report an Error

Example Question #21 : Introduction To Integrals

Find the indefinite integral.

$\int \frac{x^3-4x+7}{x^2+2x-3}dx$

Possible Answers:

$\frac{\frac{1}{4}x^4-2x^2+7x}{\frac{1}{3}x^3+x^2-3x}+C$

$\frac{1}{2}x^2-2x+ln\left | x-1\right |+2ln\left | x+3\right | + C$

$2ln\left | x-1\right |+ln\left | x+3\right | + C$

$\frac{1}{2}x^2-2x-\frac{7}{3}x + C$

Correct answer:

$\frac{1}{2}x^2-2x+ln\left | x-1\right |+2ln\left | x+3\right | + C$

Explanation:

First divide the numerator by the denominator, using long division.

$x-2 +\frac{3x+1}{x^2+2x-3}{}$

$x^2+2x-3|\overline{x^3-4x+7}$

$\\ -x^3-2x^2-3x\\ \overline{-2x^2-x+7\ \ \ \ }\\+2x^2+4x-6\\ \overline{\ \ \ \ \ \ \ \ \ \ \ 3x-1}$

Now we can write our integral as follows.

$\int x-2 +\frac{3x+1}{(x+3)(x-1)}dx$

From here we will use partial fraction decomposition.

$\frac{3x+1}{(x+3)(x-1)}=\frac{A}{(x+3)}+\frac{B}{x-1}$

3x+1=A(x-1)+B(x+3)

Set x=1 then,

$4=A(0)+b(4)\rightarrow B=1$ .

Set x=-3 then,

$-8=A(-4)+B(0)\rightarrow A=2$

Therefore we get,

$\frac{3x+1}{(x+3)(x-1)}=\frac{2}{(x+3)}+\frac{1}{(x-1)}$ .

Now we can rewrite our integral as follows,

$\int x-2 +\frac{2}{x+2}+\frac{1}{x-1}dx$ .

From here integrate each term. To integrate we will need to use the power rule, integration of a constant, and nautral log integration we are as follows.

$\\ \int x dx =\frac{1}{2}x^2 \rightarrow \int x^n=\frac{1}{n+1}x^{n+1}\\ \int 2dx=2x \rightarrow \int c\ dx=cx \\ \int \frac{2}{x+3}dx=2ln|x+3| \rightarrow \int\frac{dx}{x}=ln|x|$

Resulting in our final answer,

$\frac{1}{2}x^2-2x+ln\left | x-1\right |+2ln\left | x+3\right | + C$ .

Report an Error

Example Question #27 : Integrals

Find the indefinite integral.

$\int \frac{x^4+7}{x^3+x^2}dx$

Possible Answers:

$\frac{1}{2}x^2-x-7ln\left | x\right |-\frac{7}{x}+8ln\left | x+1\right |+C$

$ln\left | x\right |-\frac{28}{x^4}+C$

$(\frac{1}{5}x^5+7x)*ln\left | x^3+x^2\right |+C$

$\frac{1}{3}x^3ln\left | x+1\right |+7ln\left | x^3+x^2\right |+C$

Correct answer:

$\frac{1}{2}x^2-x-7ln\left | x\right |-\frac{7}{x}+8ln\left | x+1\right |+C$

Explanation:

Simplify the integrand using long division.

$x - 1 + \frac{x^2 +7}{x^3 + x^2}$

$x^3+x^2|\overline{x^4 + 7 \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \;}$

$\\ -(x^4+x^3)\\ \overline{\ \ \ \ \ \ \ \ \ -x^3\ \ \ \ }\\ {\ \ \ \ \ -(-x^3-x^2)}\\ \overline{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x^2+7}$

The integral becomes

$\int \left ( x - 1 + \frac{x^2 +7}{x^3 + x^2} \right ) dx$

Use partial fraction decomposition to simplify the expression $\frac{x^2 +7}{x^3 + x^2}$ ,

$\frac{x^2 +7}{x^3 + x^2} = \frac{x^2 + 7}{x^2(x+1)} = \frac{A}{x}+ \frac{B}{x^2}+\frac{C}{x+1}$

x^2 + 7 = A(x)(x+1)+B(x+1)+C(x^2)

Set x = 0 then,

$(0)^2+7 = A(0)(0+1)+B(0+1)+C(0^2) \rightarrow B = 7$

Set x = -1 then,

$(-1)^2 +7 = A(-1)(-1+1)+B(-1+1)+C(-1)^2 \rightarrow C = 8$

Set x = 1 then,

(1)^2 + 7 = A(1)(1+1)+B(1+1)+C(1^2)

8 = A(2)+ (7)(2)+ (8)(1)

$8 = 2A + 14 + 8 \rightarrow A = -7$

Therefore we get

$\frac{x^2 +7}{x^3 + x^2} = \frac{-7}{x}+ \frac{7}{x^2}+\frac{8}{x+1}$

Now we can rewrite our integral as follows,

$\int \left ( x - 1 + \frac{-7}{x}+ \frac{7}{x^2}+\frac{8}{x+1}\right ) dx$

Integrate each term using the power rule, integration of a constant, and natural log integration as follows,

$\int x^n=\frac{1}{n+1}x^{n+1} \rightarrow \int xdx = \frac{x^2}{2}$

$\int c\ dx=cx \rightarrow \int 1dx = x$

$\int\frac{c}{x}dx=cln|x| \rightarrow \int -\frac{7}{x}dx = -7ln\left | x\right |$

Resulting in our final answer,

$\frac{1}{2}x^2 - x -7ln\left | x\right | - \frac{7}{x} + 8ln\left | x+1\right |+C$

Report an Error

Example Question #28 : Integrals

Find the indefinite integral.

$\int \frac{ln\left | x\right |}{(x+7)^2}dx$

Possible Answers:

$-\frac{ln\left | x\right |}{x+7}+\frac{1}{7}ln\left | x\right |-\frac{1}{7}ln\left | x+7\right |+C$

$ln\left | x\right |+2ln\left | x+7\right |+C$

$\frac{1}{2x^2+14x}+C$

$xln\left | x\right |-\frac{1}{x+7}+C$

Correct answer:

$-\frac{ln\left | x\right |}{x+7}+\frac{1}{7}ln\left | x\right |-\frac{1}{7}ln\left | x+7\right |+C$

Explanation:

Use Integration By Parts (IBP) to make the integral doable, using the IBP formula.

$\int udv = uv - \int vdu$

Assign appropriate expressions from the original integral to the variables and to use in the IBP formula.

$u = ln\left | x\right | \rightarrow du = \frac{1}{x}dx$

$dv = (x+7)^{-2}dx \rightarrow v = -(x+7)^{-1}$

Substitute these expressions into the IBP formula and our integral becomes as follows,

$-(x+7)^{-1}ln\left | x\right | - \int -(x+7)^{-1}\frac{1}{x}dx$

$-\frac{1}{x+7}ln\left | x\right |+ \int \frac{1}{x(x+7)}dx$

Use partial fraction decomposition to further simplify

$\frac{1}{x(x+7)} = \frac{A}{x} + \frac{B}{x+7}$

1 = A(x+7)+B(x)

Set x = 0 then,

$1 = A(0+7)+B(0) \rightarrow 1 = 7A \rightarrow A = \frac{1}{7}$

Set x = -7 then,

$1 = A(-7+7)+B(-7) \rightarrow 1 = -7B \rightarrow B = -\frac{1}{7}$

So,

$\frac{1}{x(x+7)} = \frac{\frac{1}{7}}{x} + \frac{-\frac{1}{7}}{x+7}$

Therefore we get

$-\frac{1}{x+7}ln\left | x\right |+ \int \left ( \frac{\frac{1}{7}}{x} + \frac{-\frac{1}{7}}{x+7}\right ) dx$

Finish the integration using the natural log integration rule

$\int \frac{c}{x}dx = cln\left | x\right | \rightarrow \int \frac{\frac{1}{7}}{x}dx = \frac{1}{7}ln\left | x\right |$

Resulting in the final answer,

$-\frac{1}{x+7}ln\left | x\right |+ \frac{1}{7}ln\left | x\right | -\frac{1}{7}ln\left | x + 7\right |+ C$

Report an Error

Example Question #21 : Integrals

Find the indefinite integral.

$\int x^2\sqrt{x^3+5} dx$

Possible Answers:

$\frac{3}{2}(x^5+5)^\frac{2}{3} + C$

$\frac{1}{3}x^3+ (x^3+5)^\frac{3}{2} + C$

$\frac{2}{9}(x^3+5)^\frac{3}{2} + C$

$\frac{2}{9}x^\frac{3}{2} + C$

Correct answer:

$\frac{2}{9}(x^3+5)^\frac{3}{2} + C$

Explanation:

Simplify the integral choosing an appropriate u-substitution

$u = x^{3}+5$

$du = 3x^{2}dx\rightarrow \frac{1}{3}du=x^{2}dx$

Substitute the new expressions containing back into the original integral as follows,

$\int \frac{1}{3}du\sqrt{u}$

Since $\sqrt{u} = u^{\frac{1}{2}}$ ,

$\frac{1}{3}\int \sqrt{u}du = \frac{1}{3}\int u^{\frac{1}{2}}du$

Integrate using the power rule of integration,

$\int x^{n}dx = \frac{x^{n+1}}{n+1} \rightarrow \int u^{\frac{1}{2}}du = \frac{2u^{\frac{3}{2}}}{3}$

Resulting in the answer after integrating,

$\frac{1}{3}\left [ \frac{2}{3}u^{\frac{3}{2}} \right ] +C$

$\frac{2}{9}u^{\frac{3}{2}}+C$

Substitute back in our original $u = x^{3}+5$ for our final answer,

$\frac{2}{9}\left ( x^{3}+5\right )^{\frac{3}{2}}+C$

Report an Error

Example Question #21 : Integrals

Find the indefinite integral.

$\int e^{\sqrt{x}}dx$

Possible Answers:

$xe^{\sqrt{x}} + C$

$2e^{\sqrt{x}}(\sqrt{x}-1)+C$

$2e^{\sqrt{x}} + C$

$\bigca\sqrt{x}e^{\sqrt{x}}+ C$

Correct answer:

$2e^{\sqrt{x}}(\sqrt{x}-1)+C$

Explanation:

Make an appropriate u-substitution to turn the integral into something we are familiar with,

$u=\sqrt{x}$

$du = \frac{1}{2}x^{-\frac{1}{2}}dx =\frac{1}{2} *\frac{1}{\sqrt{x}}dx$

$2\sqrt{x}du = dx$ .

And since $u = \sqrt{x}$ ,

2udu =dx

Substitute the new expressions containing back into the original integral as follows,

$\int e^{u}2udu$

$2\int ue^{u}du$ .

From here we will use Integration By Parts (IBP) to simplify the integral further, using the IBP formula

$\int wdv = wv - \int vdw$ .

Assign the variables and to appropriate expressions from the original integral,

$w = u \rightarrow dw = du$

$dv = e^{u}du \rightarrow v = e^{u}$ .

Substitute these new expressions back into the IBP formula and our integral becomes as follows,

$2\int ue^{u}du = 2(ue^{u}-\int e^{u}du)$ .

Integrate using the following integration rule:

$\int e^{x}dx = e^{x} \rightarrow \int e^{u}du = e^{u}$ .

This results in the expression

$2(ue^{u}-e^{u}) +C$ .

Substitute back in our original u-substitution $u = \sqrt{x}$ ,

$2(\sqrt{x}e^{\sqrt{x}}-e^{\sqrt{x}}) +C$ .

Simplify to get the final answer,

$2e^{\sqrt{x}}(\sqrt{x}-1) +C$ .

Report an Error

View Calculus 2 Tutors

Satweka
Certified Tutor

The University of Texas at Dallas, Master of Science, Biomedical Engineering.

View Calculus 2 Tutors

Amelia
Certified Tutor

Austin College, Bachelor in Arts, Applied Mathematics. University of North Carolina at Charlotte, Master of Science, Applied ...

View Calculus 2 Tutors

Pankaj
Certified Tutor

University of Delhi, Electrical Engineer, Electrical Engineering. Columbia University in the City of New York, Master of Arts...

All Calculus 2 Resources

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

Popular Subjects

SSAT Tutors in Houston, ACT Tutors in Washington DC, Calculus Tutors in San Diego, Computer Science Tutors in Boston, Physics Tutors in Los Angeles, Algebra Tutors in Seattle, Statistics Tutors in Denver, Biology Tutors in Miami, ACT Tutors in Houston, SAT Tutors in Dallas Fort Worth

Popular Courses & Classes

SAT Courses & Classes in Los Angeles, GRE Courses & Classes in Phoenix, SAT Courses & Classes in New York City, LSAT Courses & Classes in New York City, LSAT Courses & Classes in Denver, MCAT Courses & Classes in San Francisco-Bay Area, SSAT Courses & Classes in Phoenix, GMAT Courses & Classes in Phoenix, MCAT Courses & Classes in Denver, MCAT Courses & Classes in Miami

Popular Test Prep

SAT Test Prep in Miami, SSAT Test Prep in Chicago, GMAT Test Prep in San Francisco-Bay Area, MCAT Test Prep in Washington DC, LSAT Test Prep in Dallas Fort Worth, SAT Test Prep in Houston, LSAT Test Prep in New York City, SAT Test Prep in Los Angeles, MCAT Test Prep in Dallas Fort Worth, GMAT Test Prep in Philadelphia

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 : Calculus II

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #21 : Integrals

Example Question #21 : Integrals

Example Question #23 : Integrals

Example Question #21 : Integrals

Example Question #25 : Integrals

Example Question #21 : Introduction To Integrals

Example Question #27 : Integrals

Example Question #28 : Integrals

Example Question #21 : Integrals

Example Question #21 : Integrals

All Calculus 2 Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: