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GRE Subject Test: Math : Substitution

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

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2 Diagnostic Tests 148 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #1 : Integrals

Integrate:

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

Possible Answers:

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

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

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

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

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

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 #1 : Integrals

Evaluate the following integral:

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

Possible Answers:

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

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

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

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

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

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.

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

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Example Question #1 : Integrals

Integrate the following using substitution.

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

Possible Answers:

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

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

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

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

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

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GRE Subject Test: Math : Substitution

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

All GRE Subject Test: Math Resources

Example Questions

Example Question #1 : Integrals

Example Question #1 : Integrals

Example Question #1 : Integrals

Example Question #4 : Substitution

All GRE Subject Test: Math Resources

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