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Calculus 2 : Integrals

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Example Questions

Example Question #631 : Finding Integrals

Evaluate the following integral:

$\int 2t\cos(\sec(e^{3t^2}))dt$

Possible Answers:

$\frac{1}{3}\cos(\sec(e^{3t^2}))+C$

$e^{3t^2}+C$

$\frac{1}{3}e^{3t^2}+C$

$-\frac{1}{3}e^{3t^2}+C$

Correct answer:

$\frac{1}{3}e^{3t^2}+C$

Explanation:

To evaluate the integral, we can first use the fact that cosine and secant are inverses of each other, so they cancel:

$\int 2te^{3t^2}dt$

Now, we must make the following substitution:

u=3t^2, du=6tdt

Rewriting the integral in terms of u and integrating, we get

$\frac{1}{3}\int e^udu=\frac{1}{3}e^u+C$

We used the following rule to integrate:

$\int e^xdx=e^x+C$

Finally, replace u with our original x term:

$\frac{1}{3}e^{3t^2}+C$

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Example Question #48 : Solving Integrals By Substitution

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

Possible Answers:

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

$\frac{3}{2}(u^2)^{\frac{4}{3}}+C$

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

$\frac{5}{2}(x^2-2)^{\frac{4}{3}}+C$

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

Correct answer:

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

Explanation:

To integrate this problem, you have to use "u" substitution. Assign x^2-2=u . Then, find du, which is 2x. That works out since we can then replace the other x in the original problem. We will have to offset the 2 though: $du=2x; \frac{1}{2}du=x$ . Now plug in all the parts: $\frac{1}{2}\int u^{\frac{1}{3}}du$ . Now, integrate as normal, remembering to raise the exponent by 1 and then also putting that result on the bottom: $\frac{1}{2}(\frac{3}{4})(u^{\frac{4}{3}})$ . Simplify, add a C because it is an indefinite integral, and substitute your original expression back in: $\frac{3}{2}(x^2-2)^{\frac{4}{3}}+C$ .

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Example Question #41 : Solving Integrals By Substitution

$\int \frac{x^2}{x^3-1}dx$

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

To integrate this problem, use "u" substitution. Assign u=x^3-1 , $du=3x^2; \frac{1}{3}du=x^2$ . Substitute everything in so you can integrate: $\frac{1}{3}\int \frac{1}{u}du$ . Recall that when there is a single variable on the denominator, the integral is ln of that term. Therefore, after integrating, you get $\frac{1}{3}ln\left | u \right |$ . Sub back in your original expression and add C because it is an indefinite integral: $\frac{1}{3}ln\left | x^3-1 \right |+C$ .

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Example Question #50 : Solving Integrals By Substitution

Evaluate the following integral using the substitution method:

$\int^1_0 x^2(x^3+4)^3dx$

Possible Answers:

$\frac{123}{4}$

$\frac{61}{9}$

$\frac{1}{12}$

$\frac{5}{4}$

$\frac{1031}{24}$

Correct answer:

$\frac{123}{4}$

Explanation:

$\int^1_0 x^2(x^3+4)^3dx$

Make the substitution:

u=x^3+4

du=3x^2dx

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

$\int^1_0 \frac{u^3}{3}du=\frac{1}{12}u^4\left.\begin{matrix} \\ \end{matrix}\right|^1_0=\frac{1}{12}(x^3+4)^4 \left.\begin{matrix} \\ \end{matrix}\right|^1_0=\frac{123}{4}$

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Example Question #2731 : Calculus Ii

Solve: $\int10e^{10x}$ .

Possible Answers:

$\frac{e^x}{10} + C$

10e^x + C

e^x + C

$e^{10x}+C$

Correct answer:

$e^{10x}+C$

Explanation:

Substitute u=10x, du = 10 dx :

$\int e^{u}du$ ,

which is equal to

$e^{u}+ C$ .

Replace u with 10x:

$e^{10x}+C$ .

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Example Question #51 : Solving Integrals By Substitution

$\int cos^3xsinxdx$

Possible Answers:

cos^4x+C

$\frac{cos^4x}{4} + C$

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

$\frac{sin^4x}{4}$

$\frac{cos^2x}{2} +C$

Correct answer:

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

Explanation:

Substitute u=cosx, du=-sinxdx :

$\int- u^3du=-\frac{u^4}{4} + C$ .

Replace u=cos x :

$\frac{-cos^4x}{4} + C$ .

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Example Question #51 : Solving Integrals By Substitution

Solve $\int sin^3x cos x dx$ .

Possible Answers:

$\frac{sin^2x}{2} +C$

3sin^2x +C

sin^4x +C

$\frac{sin^4x}{4} +C$

cos^4x+C

Correct answer:

$\frac{sin^4x}{4} +C$

Explanation:

Substitute u=sin x, du=cosx :

$\int u^3 du=\frac{u^4}{4} +C$ .

Replace u=sin x :

$\frac{sin^4x}{4} +C$ .

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Example Question #52 : Solving Integrals By Substitution

Solve $\int 5e^{5x}dx$ .

Possible Answers:

$\frac{e^{6x}}{6} + C$

$5e^{4x} +C$

$e^{4x}+C$

$e^{5x} + C$

$5e^{x} + C$

Correct answer:

$e^{5x} + C$

Explanation:

Substitue u=5x, du=5 .

$\int e^udu=e^u+C$ .

Replace u=5x :

$e^{5x} + C$ .

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Example Question #981 : Integrals

Find $\int \frac{dx}{(x+2)^4}$ .

Possible Answers:

ln((x+2)^4)+C

${(x+2)^{-4}}+C$

4ln(x+2)+C

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

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

Correct answer:

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

Explanation:

Substitute u=x+2, du=dx :

$\int \frac{1}{u^4}du=-\frac{u^{-3}}{3}$ .

Replace u=x+2 :

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

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Example Question #53 : Solving Integrals By Substitution

Solve $\int \frac{x^2}{x^3+1}$ .

Possible Answers:

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

$ln\left | x+1\right | + C$

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

$ln\left | x^3+1\right | + C$

Correct answer:

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

Explanation:

Substitute u=x^3 + 1, du=3x^2 dx .

$\int \frac{1}{3}u^{-1}du=\frac{1}{3}ln\left | u \right |+C$ .

Replace u=x^3+1 :

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

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Calculus 2 : Integrals

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #631 : Finding Integrals

Example Question #48 : Solving Integrals By Substitution

Example Question #41 : Solving Integrals By Substitution

Example Question #50 : Solving Integrals By Substitution

Example Question #2731 : Calculus Ii

Example Question #51 : Solving Integrals By Substitution

Example Question #51 : Solving Integrals By Substitution

Example Question #52 : Solving Integrals By Substitution

Example Question #981 : Integrals

Example Question #53 : Solving Integrals By Substitution

All Calculus 2 Resources

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