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AP Calculus AB : AP Calculus AB

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3 Diagnostic Tests 164 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #15 : Antiderivatives By Substitution Of Variables

$\int (2x-1)(x^2-x+8)^7dx=?$

Possible Answers:

${(x^2-x+8)^8}+C$

$\frac{(x^2-x+8)^8}{8}+C$

$\frac{(\frac{1}{3}x^3-\frac{x^2}{2}+8x)^8}{8}+C$

8(x^2-x+8)^8+C

Correct answer:

$\frac{(x^2-x+8)^8}{8}+C$

Explanation:

This is a u-substitution problem. We need to find a function and its derivative in the integral.

$u=x^2-x+8.\; du=(2x-1)dx.$

Now, replace your variables, and integrate.

$\int (2x-1)(x^2-x+8)^7dx=\int (x^2-x+8)^7(2x-1)dx$

$=\int u^7du=\frac{u^8}{8}+C=\frac{(x^2-x+8)^8}{8}+C$ .

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Example Question #16 : Antiderivatives By Substitution Of Variables

$\int x\sqrt{3+x}dx=?$

Possible Answers:

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

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

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

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

$\sqrt{3+x}+\frac{x}{2\sqrt{3+x}}+C$

Correct answer:

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

Explanation:

This problem is an application of the u-substitution method.

$u=3+x.\; x=u-3\; du=1dx.$

Now, be careful that you replace everything in the original integral in terms of our new variables. This includes the term!

$\int x\sqrt{3+x}dx=\int(u-3)u^{1/2}du=\int u^{3/2}-3u^{1/2}du$

$=\frac{u^{5/2}}{5/2}-3\frac{u^{3/2}}{3/2}=\frac{2}{5}u^{5/2}-2u^{3/2}+C$

$=\frac{2}{5}(3+x)^{5/2}-2(3+x)^{3/2}+C$ .

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Example Question #951 : Ap Calculus Ab

$\int \frac{2x-4}{2x^2-8x+3}dx=?$

Possible Answers:

2ln|4x-8|+C

$\frac{1}{2}ln|4x-8|+C$

$\frac{1}{2}ln|2x^2-8x+3|+C$

ln|2x^2-8x+3|+C

2ln|2x^2-8x+3|+C

Correct answer:

$\frac{1}{2}ln|2x^2-8x+3|+C$

Explanation:

To simplify the integral, we need to substitute new variables:

$\\u=2x^2-8x+3.\; \\du=(4x-8)dx \\du=2(2x-4)dx \\\frac{1}{2}du=(2x-4)dx$

Now, we can replace our original variables, and integrate!

$\int \frac{2x-4}{2x^2-8x+3}dx=\int \frac{du}{2u}=\frac{1}{2}\int\frac{du}{u}$

$\\=\frac{1}{2}ln|u|+C \\\\ =\frac{1}{2}ln{|2x^2-8x+3|}+C$ .

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Example Question #952 : Ap Calculus Ab

$\int \frac{2lnx}{x}dx=?$

Possible Answers:

(lnx)^2+C

2lnx^2+C

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

2ln(x)+C

$\frac{2}{lnx}+C$

Correct answer:

(lnx)^2+C

Explanation:

This is a hidden u-substitution problem! Remember, to use substitution, we need to have an integral where a function and its derivative live inside. If you look closely, you will see we have just that!

$u=lnx.\; du=\frac{1}{x}dx.$

Now, rewrite the integral, and integrate:

$\int \frac{2lnx}{x}==2\int lnx*\frac{1}{x}dx=2\int u du=2\frac{u^2}{2}+C$

=u^2+C=(lnx)^2+C.

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Example Question #61 : Techniques Of Antidifferentiation

Integrate:

$\int 5y^2\cos(y^3)dy$

Possible Answers:

$\frac{5}{3}\sin(y^3)$

$\frac{5y^3}{3}\sin(y^3)+C$

$\frac{1}{15}\sin(y^3)+C$

$\frac{5}{3}\sin(y^3)+C$

$-\frac{5}{3}\sin(y^3)+C$

Correct answer:

$\frac{5}{3}\sin(y^3)+C$

Explanation:

To integrate, we must make the following substitution:

u=y^3, du=3y^2dy

The derivative was found using the following rule:

$\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$

Now, we rewrite the integral in terms of u and solve:

$\frac{5}{3}\int\cos(u)du=\frac{5}{3}\sin(u)+C$

The integral was found using the following rule:

$\int \cos(x)dx=\sin(x)+C$

Finally, replace u with our original x term:

$\frac{5}{3}\sin(y^3)+C$

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Example Question #953 : Ap Calculus Ab

Integrate:

$\int 3xe^{x^2}dx$

Possible Answers:

$\frac{3}{2}e^{x^2}$

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

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

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

$\frac{1}{6}e^{x^2}+C$

Correct answer:

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

Explanation:

To integrate, we must perform the following substitution:

u=x^2, du=2xdx

The derivative was found using the following rule:

$\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$

Now, we rewrite the integral in terms of u and solve:

$\frac{3}{2}\int\ e^udu=\frac{3}{2}e^u+C$

The integral was found using the following rule:

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

Finally, replace u with our original x term:

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

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Example Question #21 : Antiderivatives By Substitution Of Variables

Integrate:

$\int 7x^4 \sin(3x^5)dx$

Possible Answers:

$-\frac{7}{15}\cos(3x^5)+C$

$-\cos(3x^5)+C$

$-\frac{7}{15}\cos(3x^5)$

$\frac{7}{15}\cos(3x^5)+C$

Correct answer:

$-\frac{7}{15}\cos(3x^5)+C$

Explanation:

To integrate, the following substitution was made:

u=3x^5, du=15x^4dx

Now, we rewrite the integral in terms of u and integrate:

$\frac{7}{15}\int\sin(u)du=-\frac{7}{15}\cos(u)+C$

The following rule was used for integration:

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

Finally, rewrite the final answer in terms of our original x term:

$-\frac{7}{15}\sin(3x^5)+C$

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Example Question #231 : Integrals

Evaluate the integral

$\int x^2e^{x^3}dx$

Possible Answers:

$\frac{1}{3}e^{x^3}+C$

$e^{u}+C$

$\frac{1}{3}e^{x^3}$

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

Correct answer:

$\frac{1}{3}e^{x^3}+C$

Explanation:

We can make a u substitution in the following way:

u=x^3 , and therefore du=3x^2dx

Simplifying the integral, we get

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

Rewriting in terms of x, we get

$\frac{1}{3}e^{x^3}+C$

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Example Question #231 : Integrals

Solve:

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

Possible Answers:

$\frac{1}{2\sqrt{3}}\arcsin(\frac{\sqrt{3}x}{2})+C$

$\frac{1}{\sqrt{3}}\arccos(\frac{\sqrt{3}x}{2})+C$

$\frac{1}{\sqrt{3}}\arcsin(\frac{\sqrt{3}x}{4})+C$

$\frac{1}{2}\arcsin(\frac{\sqrt{3}x}{2})+C$

$\frac{1}{\sqrt{3}}\arcsin(\frac{\sqrt{3}x}{2})+C$

Correct answer:

$\frac{1}{\sqrt{3}}\arcsin(\frac{\sqrt{3}x}{2})+C$

Explanation:

To integrate, we must make the following substitution:

$u=\sqrt{3}x, du=\sqrt{3}dx$

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

$\frac{1}{\sqrt{3}}\int \frac{du}{\sqrt{4-u^2}}=\frac{1}{\sqrt{3}}\arcsin(\frac{u}{2})+C$

The following rule was used for integration:

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

Replacing u with our original x term, we get

$\frac{1}{\sqrt{3}}\arcsin(\frac{\sqrt{3}x}{2})+C$

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Example Question #21 : Antiderivatives By Substitution Of Variables

Solve:

$\int \frac{dx}{25+5x^2}$

Possible Answers:

$\frac{1}{\sqrt{5}}\arcsin{\frac{\sqrt{5}x}{5}}+C$

$\frac{1}{5\sqrt{5}}\arctan{\frac{\sqrt{5}x}{5}}+C$

$\frac{1}{\sqrt{5}}\arctan{\frac{\sqrt{5}x}{5}}+C$

$-\frac{1}{5\sqrt{5}}\arctan{\frac{\sqrt{5}x}{5}}+C$

$\frac{1}{5\sqrt{5}}\arctan{\frac{\sqrt{5}x}{25}}+C$

Correct answer:

$\frac{1}{5\sqrt{5}}\arctan{\frac{\sqrt{5}x}{5}}+C$

Explanation:

To integrate, we must make the following substitution:

$u=\sqrt{5}x, du=\sqrt{5}dx$

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

$\frac{1}{\sqrt{5}}\int \frac{du}{25+u^2}=\frac{1}{5\sqrt{5}}\arctan(\frac{u}{5})+C$

The following rule was used for integration:

$\int \frac{dx}{a^2+u^2}=\frac{1}{a}\arctan(\frac{u}{a})+C$

Replacing u with our original x term, we get

$\frac{1}{5\sqrt{5}}\arctan{\frac{\sqrt{5}x}{5}}+C$

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AP Calculus AB : AP Calculus AB

Study concepts, example questions & explanations for AP Calculus AB

All AP Calculus AB Resources

Example Questions

Example Question #15 : Antiderivatives By Substitution Of Variables

Example Question #16 : Antiderivatives By Substitution Of Variables

Example Question #951 : Ap Calculus Ab

Example Question #952 : Ap Calculus Ab

Example Question #61 : Techniques Of Antidifferentiation

Example Question #953 : Ap Calculus Ab

Example Question #21 : Antiderivatives By Substitution Of Variables

Example Question #231 : Integrals

Example Question #231 : Integrals

Example Question #21 : Antiderivatives By Substitution Of Variables

All AP Calculus AB Resources

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