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GRE Subject Test: Math : Integrals

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Example Questions

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Example Question #1 : Substitution

Integrate:

$\int \frac{18x-6}{3x^2-2x}dx$

Possible Answers:

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

$\frac{1}{6}\: ln\left | 3x^2-2x\right |+C$

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

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

$6\: ln\left | 3x^2-2x\right |+C$

Correct answer:

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

Explanation:

This problem requires U-Substitution. Let u=3x^2-2x and find .

$du=6x-2\:dx$

Notice that the numerator in $\int \frac{18x-6}{3x^2-2x}dx$ has common factor of 2, 3, or 6. The numerator can be factored so that the term can be a substitute. Factor the numerator using 3 as the common factor.

$\int \frac{18x-6}{3x^2-2x}dx= \int\frac{3(6x-2)}{3x^2-2x}dx=3\int \frac{(6x-2)}{3x^2-2x}dx$

Substitute and terms, integrate, and resubstitute the term.

$3\int \frac{du}{u} = 3\: ln\left | u\right |+C=3\: ln\left | 3x^2-2x\right |+C$

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Example Question #2 : Substitution

Evaluate the following integral:

$\int5x^4(x^5-9)^6dx$

Possible Answers:

$\frac{(x^4-9)^7}{7}+c$

$\frac{(x^5+9)^7}{7}+c$

$\frac{(x^5-9)^7}{7}$

$\frac{(x^5+9)^6}{6}+c$

$\frac{(x^5-9)^7}{7}+c$

Correct answer:

$\frac{(x^5-9)^7}{7}+c$

Explanation:

To calculate this integral, we could expand that whole binomial, but it would be very time consuming and a bit of a pain. Instead, let's use u substitution:

Given this:

$\int5x^4(x^5-9)^6dx$

We can say that

u=x^5-9

du=5x^4dx

Then, plug it back into our original expression

$\int u^6du$

Evaluate this integral to get

$\frac{u^7}{7}+c$

Then, replace u with what we substituted it for to get our final answer. Note because this is an indefinite integral, we need a plus c in it.

$\frac{(x^5-9)^7}{7}+c$

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Example Question #3 : Substitution

Integrate the following using substitution.

$\int \frac{1}{4-5x}dx$

Possible Answers:

$-\frac{1}{5}\ln\left | 4-5x\right |+c$

$\ln\left | 4-5x\right |+c$

$\frac{-5}{4-5x}+c$

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

Correct answer:

$-\frac{1}{5}\ln\left | 4-5x\right |+c$

Explanation:

$\int \frac{1}{4-5x}dx$

Using substitution, we set u=4-5x which means its derivative is du=-5dx .

Substituting for 4-5x , and $- \frac{1}{5}du$ for we have:

$\int \frac{1}{4-5x}dx=-\frac{1}{5} \int \frac{1}{u} du$

Now we just integrate:

$-\frac{1}{5} \int \frac{1}{u} du=-\frac{1}{5}\ln\left | u\right |+c$

Finally, we remove our substitution u=4-5x to arrive at an expression with our original variable:

$-\frac{1}{5}\ln\left | 4-5x\right |+c$

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Example Question #4 : Substitution

Evaluate the following integral:

$\int5x^4(x^5-9)^6dx$

Possible Answers:

$\frac{(x^5-9)^7}{7}+c$

$\frac{(x^5-9)^7}{7}$

$\frac{(x^5+9)^6}{6}+c$

$\frac{(x^5+9)^7}{7}+c$

$\frac{(x^4-9)^7}{7}+c$

Correct answer:

$\frac{(x^5-9)^7}{7}+c$

Explanation:

To calculate this integral, we could expand that whole binomial, but it would be very time consuming and a bit of a pain. Instead, let's use u substitution:

Given this:

$\int5x^4(x^5-9)^6dx$

We can say that

u=x^5-9

du=5x^4dx

Then, plug it back into our original expression

$\int u^6du$

Evaluate this integral to get

$\frac{u^7}{7}+c$

Then, replace u with what we substituted it for to get our final answer. Note because this is an indefinite integral, we need a plus c in it.

$\frac{(x^5-9)^7}{7}+c$

Report an Error

Example Question #1 : Integration By Parts

Integrate the following.

$\int x \cos x \ dx$

Possible Answers:

$x\sin x+ \sin x + c$

$x \sin x + \cos x +c$

$x\sin x-\cos x + c$

$x\cos x + \sin x + c$

Correct answer:

$x \sin x + \cos x +c$

Explanation:

Integration by parts follows the formula:

$\int u \ dv=uv- \int v \ du$

So, our substitutions will be u=x and $dv=\cos x \ dx$

which means du=dx and $v=\sin x$

Plugging our substitutions into the formula gives us:

$x\sin x-\int \sin x \ dx$

Since $\int \sin x \ dx=-\cos x + c$ , we have:

$x\sin x - (- \cos x + c)$ , or

$x\sin x + \cos x + c$

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Example Question #2 : Integration By Parts

Evaluate the following integral.

$\int x e^{4x}dx$

Possible Answers:

$xe^{4x}+\frac{1}{4}e^{4x}+c$

$\frac{x}{4}e^{4x}-\frac{1}{16}e^{4x}+c$

$\frac{x}{4}e^{4x} +\frac{1}{16}e^{4x}+c$

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

Correct answer:

$\frac{x}{4}e^{4x}-\frac{1}{16}e^{4x}+c$

Explanation:

Integration by parts follows the formula:

$\int u \ dv=uv-\int v\ du$

In this problem we have $\int x e^{4x}dx$ so we'll assign our substitutions:

u=x and $dv=e^{4x} dx$

which means du=dx and $v=\frac{1}{4}e^{4x}$

Including our substitutions into the formula gives us:

$x\frac{1}{4}e^{4x}-\int \frac{1}{4}e^{4x}dx$

We can pull out the fraction from the integral in the second part:

$\frac{x}{4}e^{4x}-\frac{1}{4} \int e^{4x}dx$

Completing the integration gives us:

$\frac{x}{4}e^{4x}-\frac{1}{4} \left(\frac{1}{4}e^{4x}\right) +c$

$\frac{x}{4}e^{4x}-\frac{1}{16} e^{4x} +c$

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Example Question #1 : Integration By Parts

Evaluate the following integral.

$\int x^2 \ln x \ dx$

Possible Answers:

$\frac{x^{3}}{3}\ln x-\frac{1}{9}x^3+c$

$\frac{x^{3}}{3}\ln x+\frac{1}{9}x^3+c$

$\frac{x^3}{3}\ln x + c$

$\frac{x^{3}}{3}\ln x-\frac{1}{3}x^3+c$

Correct answer:

$\frac{x^{3}}{3}\ln x-\frac{1}{9}x^3+c$

Explanation:

Integration by parts follows the formula:

$\int u \ dv=uv-\int v \ du$

Our substitutions will be $u=\ln x$ and dv= x^2 dx

which means $du= \frac{1}{x}dx$ and $v=\frac{1}{3}x^3$ .

Plugging our substitutions into the formula gives us:

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

Look at the integral: we can pull out the $\frac{1}{3}$ and simplify the remaining $\frac{x^3}{x}$ as

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

We now solve the integral: $\int x^2 dx = \frac{1}{3} x^3 + c$ , so:

$\frac{x^3}{3} \ln x -\frac{1}{3} \left(\frac{1}{3}x^3 + c\right)$

$\frac{x^3}{3} \ln x - \frac{1}{9} x^3+c$

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Example Question #4 : Integration By Parts

Evaluate the following integral.

$\int x \cos (2x+1)dx$

Possible Answers:

$\frac{1}{2}x \sin (2x+1)- \frac{1}{2}\cos (2x+1)+c$

$\frac{1}{2}x \sin (2x+1)+ \frac{1}{2}\cos (2x+1)+c$

$\frac{1}{2}x \sin (2x+1)+ \frac{1}{4}\cos (2x+1)+c$

$\frac{1}{2}x \sin (2x+1)$

Correct answer:

$\frac{1}{2}x \sin (2x+1)+ \frac{1}{4}\cos (2x+1)+c$

Explanation:

Integration by parts follows the formula:

$\int u \ dv=uv - \int v \ du$ .

Our substitutions are u=x and $dv= \cos (2x+1) dx$

which means du=dx and $v=\frac{1}{2} \sin (2x+1)$ .

Plugging in our substitutions into the formula gives us

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

We can pull $\frac{1}{2}$ outside of the integral.

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

Since $\int \sin (2x+1) dx =-\frac{1}{2} \cos (2x+1)+ c$ , we have

$\frac{x}{2} \sin (2x+1)- \frac{1}{2} \left(-\frac{1}{2} \cos (2x+1) + c\right)$

$\frac{x}{2} \sin (2x+1)+\frac{1}{4} \cos (2x+1) + c$

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Example Question #1 : Trigonometric Integrals

Integrate the following.

$\int \sin (7x+3) \ dx$

Possible Answers:

$-\frac{1}{7} \cos \ (7x+3) +c$

${7} \cos \ (7x+3) +c$

$\frac{1}{7} \cos \ (7x+3) +c$

$-{7} \cos \ (7x+3) +c$

Correct answer:

$-\frac{1}{7} \cos \ (7x+3) +c$

Explanation:

We can integrate the function by using substitution where 7x+3=u so $du=7 \ dx$ .

$\int \sin (7x+3) \ dx= \frac{1}{7}\int \sin u \ du$

Just focus on integrating sine now:

$\frac{1}{7}\int \sin u \ du = \frac{1}{7} (- \cos u +c)$

The last step is to reinsert the substitution:

$-\frac{1}{7} \cos \ (7x+3) +c$

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Example Question #2 : Trigonometric Integrals

Integrate the following.

$\int \ \cos (2x+6) \ dx$

Possible Answers:

$2\ \sin (2x+6) +c$

$\frac{1}{2} \sin (2x+6) +c$

$-\frac{1}{2} \sin (2x+6) +c$

$2\ \cos (2x+6) +c$

Correct answer:

$\frac{1}{2} \sin (2x+6) +c$

Explanation:

We can integrate using substitution:

2x+6=u and $du=2\ dx$ so $dx=\frac{1}{2} \ du$

$\int \ \cos (2x+6) \ dx= \frac{1}{2}\int \cos u \ du$

Now we can just focus on integrating cosine:

$\frac{1}{2}\int \cos u \ du = \frac{1}{2} \sin u +c$

Once the integration is complete, we can reinsert our substitution:

$\frac{1}{2} \sin (2x+6) +c$

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GRE Subject Test: Math : Integrals

Study concepts, example questions & explanations for GRE Subject Test: Math

All GRE Subject Test: Math Resources

Example Questions

Example Question #1 : Substitution

Example Question #2 : Substitution

Example Question #3 : Substitution

Example Question #4 : Substitution

Example Question #1 : Integration By Parts

Example Question #2 : Integration By Parts

Example Question #1 : Integration By Parts

Example Question #4 : Integration By Parts

Example Question #1 : Trigonometric Integrals

Example Question #2 : Trigonometric Integrals

All GRE Subject Test: Math Resources

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