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Calculus 2 : Calculus II

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

Example Questions

Example Question #24 : Solving Integrals By Substitution

Evaluate the following indefinite integral using the substitution method.

$\int 3(x\sec(x^3))^2 dx$

Possible Answers:

$\tan(\frac{x^4}{4})+C$

$x^3\tan(x^3)+C$

$\tan(x^3) + C$

$\sec^3(x^3)+C$

$3\sec^2(x^3)+C$

Correct answer:

$\tan(x^3) + C$

Explanation:

The integral can be expanded by distributing the exponent.

$\int3(x\sec(x^3))^2dx=\int3x^2\sec^2(x^3)dx$

We will make the following substitution:

u=x^3 .

Differentiating both sides yields

du=3x^2dx .

We can then substitute the left hand side of each equation into our integral and evaluate it now.

$\int3x^2\sec^2(x^3)dx=\int\sec^2(u)du=\tan(u)+C.$

Finally, we substitute the original variable back into the expression:

$\tan(u)+C=\tan(x^3)+C$ .

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Example Question #28 : Solving Integrals By Substitution

Solve:

$\int_{0}^{\pi} sin(x)cos(x)dx$

Possible Answers:

$\frac{1}{4}$

$\frac{1}{2}$

$\frac{1}{8}$

Correct answer:

Explanation:

Use substitution:

u = sin(x) , du = cos(x)dx

Plug the and into the regular equation, but no need to worry about the bounds yet:

$\int udu$ $= \frac{1}{2}u^{2}$

Plug back into the integrated equation from above and evaluate from to $\pi$ .

$= \frac{1}{2} (sin(x))^2$

$\frac{1}{2} (sin(\pi)^2 - sin(0)^2) = 0$

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Example Question #2711 : Calculus Ii

Solve:

$\int \frac{(ln(x))^2}{x}dx$

Possible Answers:

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

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

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

None of the chocies.

$\frac{1}{4} (ln(x))^4+C$

Correct answer:

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

Explanation:

Use substitution integration:

u = ln(x)

$du=\frac{1}{x}dx$

$\int u^2 du=\frac{1}{3}u^3+C=\frac{1}{3}(ln(x))^3+C$

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Example Question #22 : Solving Integrals By Substitution

What is the integral of $xe^{x^2}$ ?

Possible Answers:

$\frac{e^{x^2}}{2x}$

$e^{x^2}$

$\frac{e^{x^2}}{x}$

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

Correct answer:

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

Explanation:

Use substitution:

u = x^2

du = 2xdx

$\frac{du}{2} = xdx$

$\int \frac{e^{u}}{2}du$ $= \frac{e^{u}}{2}$ $= \frac{e^{x^2}}{2}$

Substitute back in.

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Example Question #961 : Integrals

$\int \frac{dx}{\sqrt{16-4x^2}}$

Possible Answers:

$\arcsin(\frac{x}{2})+C$

$\frac{1}{2}\arcsin(x)+C$

$\frac{1}{4}\arcsin(\frac{x}{2})+C$

$\frac{1}{2}\arcsin(\frac{x}{2})+C$

$\frac{1}{2}\arcsin(\frac{x}{4})+C$

Correct answer:

$\frac{1}{2}\arcsin(\frac{x}{2})+C$

Explanation:

To evaluate the integral, we must first perform the following substitution:

u=2x, du=2dx

Now, rewrite the integral and integrate:

$\frac{1}{2}\int \frac{du}{\sqrt{16-u^2}}=\frac{1}{2}\arcsin(\frac{u}{4})+C$

The integration was performed using the following rule:

$\int \frac{dx}{\sqrt{a^2-x^2}}=\arcsin(\frac{x}{a})+C$

Finally, replace u with the original term:

$\frac{1}{2}\arcsin(\frac{x}{2})+C$

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Example Question #32 : Solving Integrals By Substitution

Evaluate the following integral:

$\int \frac{dx}{\sqrt{x+14}}$

Possible Answers:

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

$\frac{\sqrt{x+14}}{2}+C$

$2\sqrt{x+14}+C$

$\sqrt{x+14}+C$

Correct answer:

$2\sqrt{x+14}+C$

Explanation:

To evaluate the integral, we must make the following substitution:

$u=\sqrt{x+14}, du=\frac{1}{2}(x+14)^{-\frac{1}{2}}dx$

Now, rewrite the integral and integrate:

$\int2du=2u+C$

The integral was performed using the following rule:
$\int dx=x+C$

Finally, replace u with our original term:

$2\sqrt{x+14}+C$

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Example Question #31 : Solving Integrals By Substitution

Evaluate the indefinite integral $\int 10e^{-x}dx$ .

Possible Answers:

$-5e^{-x}+C$

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

None of the other answers

$10e^{-x}+C$

$-10e^{-x}+C$

Correct answer:

$-10e^{-x}+C$

Explanation:

We proceed as follows,

$\int 10e^{-x}dx$ . Start

$10\int e^{-x}dx$ . Factor out the 10.

Use u-substitution with u = -x , then taking derivates of both sides gives du = -1 dx .

$10\int -1e^u du$ . Substitute values

$-10\int e^u du$ . Factor out the negative.

-10e^u+C . The antiderivative of e^u is e^u . Don't forget .

$-10e^{-x}+C$ . Substitute back.

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Example Question #31 : Solving Integrals By Substitution

Evaluate the indefinite integral $\int xe^{x^2} dx$ .

Possible Answers:

$2xe^{x^2}+C$

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

None of the other answers

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

Not possible to integrate

Correct answer:

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

Explanation:

We evaluate the integral as follows,

$\int xe^{x^2} dx$ . Start

Use u-substitution, let u=x^2 , then taking derivatives of both sides gives du = 2xdx . Divide both sides of this equation by , giving $\frac{1}{2}du = xdx$ . Now we can substitute out xdx, x^2 , and get

$\int e^{u} \frac{1}{2}du$

$\frac{1}{2}\int e^{u} du$ . Factor out the 1/2 .

$\frac{1}{2}e^u+C$ . Integrate e^u and add .

$\frac{1}{2}e^{x^2}+C$ . Substitute back

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Example Question #31 : Solving Integrals By Substitution

Evaluate $\int^{15}_{8}x\sqrt{1+x}dx$ .

Possible Answers:

$\frac{246}{17}$

$\frac{4316}{15}$

None of the other answers.

$\frac{234}{521}$

Correct answer:

$\frac{4316}{15}$

Explanation:

We use u-substitution to evaluate this integral.

Let u = 1+x . Subtracting gives u-1=x , and taking derivatives gives du =dx (We subtract from both sides in order to make the expression under the square root as simple as possible). Then we have

$\int^{15}_{8}x\sqrt{1+x}dx$ . Start

$\int^{16}_{9}(u-1)\sqrt{u}du$ . Make our substitutions. (Make sure you change the bounds of integration too, by plugging and into u=1+x for ).

$\int^{16}_{9}(u-1)u^{\frac{1}{2}}du$ .

$\int^{16}_{9} u^{\frac{3}{2}}-u^{\frac{1}{2}}du$

$[\frac{2}{5}u^{\frac{5}{2}}-\frac{2}{3}u^{\frac{3}{2}}]^{16}_9$

$(\frac{2}{5}(16)^{\frac{5}{2}}-\frac{2}{3}(16)^{\frac{3}{2}})-(\frac{2}{5}(9)^{\frac{5}{2}}-\frac{2}{3}(9)^{\frac{3}{2}})$

$\frac{2}{5}(1024)-\frac{2}{3}(64)-\frac{2}{5}(243)+\frac{2}{3}(27)$

$\frac{2}{5}(781)-\frac{2}{3}(37)$

$\frac{1562}{5}-\frac{74}{3}$

$\frac{4316}{15}$

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Example Question #36 : Solving Integrals By Substitution

Evaluate $\int^{\pi/2}_{0}\frac{2x\tan^{-1}(x^2)}{1+x^4}dx$

Possible Answers:

$\frac{\pi}{2}$

$\frac{\pi}{3}$

None of the other answers.

Correct answer:

None of the other answers.

Explanation:

The correct answer is $\frac{1}{2}(\tan^{-1}(\frac{\pi^2}{4}))^2$ .

We proceed as follows-

$\int^{\pi/2}_{0}\frac{2x\tan^{-1}(x^2)}{1+x^4}dx$ . Start

Evaluating this integral relies on the fact $\frac{d}{dx}\tan^{-1}(x) = \frac{1}{1+x^2}$ , and the Chain Rule for derivatives.

Use u-substitution $u = \tan^{-1}(x^2)$ , then we obtain $du = \frac{1}{1+(x^2)^2} \times 2x dx$

Our integral then becomes

$\int^{\tan^{-1}(\frac{\pi^2}{4})}_{0} u du$ after substitution. (The new upper bound on the integral cannot be simplified well, so we should leave it as is).

We then integrate to get

$[\frac{1}{2}u^2]^{\tan^{-1}(\frac{\pi^2}{4})}_0$

$\frac{1}{2}(\tan^{-1}(\frac{\pi^2}{4}))^2$

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Calculus 2 : Calculus II

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #24 : Solving Integrals By Substitution

Example Question #28 : Solving Integrals By Substitution

Example Question #2711 : Calculus Ii

Example Question #22 : Solving Integrals By Substitution

Example Question #961 : Integrals

Example Question #32 : Solving Integrals By Substitution

Example Question #31 : Solving Integrals By Substitution

Example Question #31 : Solving Integrals By Substitution

Example Question #31 : Solving Integrals By Substitution

Example Question #36 : Solving Integrals By Substitution

All Calculus 2 Resources

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