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AP Calculus AB : AP Calculus AB

Study concepts, example questions & explanations for 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 #5 : Antiderivatives By Substitution Of Variables

Solve the following integral using substitution:

$\int{x^2\sqrt{x^3+1}}dx$

Possible Answers:

$\frac{2}{3}(x^3+1)^9+c$

$x^{\frac{7}{2}}+x^2+c$

$\frac{1}{3}(x^3+1)^{\frac{5}{2}}+c$

$\frac{2}{9}(x^3+1)^{\frac{3}{2}}+c$

$\sqrt[3]{(x^3+1)}(\frac{1}{x^2})+c$

Correct answer:

$\frac{2}{9}(x^3+1)^{\frac{3}{2}}+c$

Explanation:

To solve the integral, we have to simplify it by using a variable to substitute for a variable of . For this problem, we will let u replace the expression x^3+1 . Next, we must take the derivative of u. Its derivative is 3x^2dx . Next, solve this equation for so that we may replace it in the integral. Plug in place of x^3+1 and $\frac{du}{3x^2}$ in place of into the original integral and simplify. The x^2 in the denominator cancels out the remaining x^2 in the integral, leaving behind a $\frac{1}{3}$ . We can pull the $\frac{1}{3}$ out front of the integral. Next, take the anti-derivative of the integrand and replace with the original expression, adding the constant to the answer. The specific steps are as follows:

1. $\int{x^2\sqrt{x^3+1}}dx$

2. = x^3+1

3. du=3x^2dx

4. $\int{x^2\sqrt{u}\frac{du}{3x^2}}$

5. $\frac{1}{3}\int{\sqrt{u}du}$

6. $\frac{1}{3}(\frac{2}{3}u^{\frac{3}{2}})+c$

7. $\frac{2}{9}u^{\frac{3}{2}}+c$

8. $\frac{2}{9}(x^3+1)^{\frac{3}{2}}+c$

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Example Question #6 : Antiderivatives By Substitution Of Variables

Solve the following integral using substitution:

$\int{cos^3xsinxdx}$

Possible Answers:

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

$\frac{-1}{3}(cosx)^3+c$

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

cotxcos^2x+c

$\frac{1}{3}(cosx)^3+c$

Correct answer:

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

Explanation:

To solve the integral, we have to simplify it by using a variable to substitute for a variable of . For this problem, we will let u replace the expression cosx . Next, we must take the derivative of . Its derivative is -sinxdx . Next, solve this equation for so that we may replace it in the integral. Plug in place of cosx and $\frac{du}{-sinx}$ in place of into the original integral and simplify. The sinx in the denominator cancels out the remaining sinx in the integral, leaving behind a . Next, take the anti-derivative of the integrand and replace with the original expression, adding the constant to the answer. The specific steps are as follows:

1. $\int{cos^3xsinxdx}$

2. u=cosx

3. du=-sinxdx

4. $\int{u^3sinx(\frac{du}{-sinx})}$

5. $\int-u^3du$

6. $\frac{-1}{4}u^4+c$

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

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Example Question #7 : Antiderivatives By Substitution Of Variables

Solve the following integral using substitution:

$\int{xsin(x^2)dx}$

Possible Answers:

2cos(x^2)+c

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

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

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

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

Correct answer:

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

Explanation:

To solve the integral, we have to simplify it by using a variable to substitute for a variable of . For this problem, we will let u replace the expression x^2 . Next, we must take the derivative of . Its derivative is . Next, solve this equation for so that we may replace it in the integral. Plug in place of x^2 and $\frac{du}{2x}$ in place of into the original integral and simplify. The in the denominator cancels out the remaining in the integral, leaving behind a $\frac{1}{2}$ . We can pull the $\frac{1}{2}$ out front of the integral. Next, take the anti-derivative of the integrand and replace with the original expression, adding the constant to the answer. The specific steps are as follows:

1. $\int{xsin(x^2)dx}$

2. u=x^2

3. du=2xdx

4. $\int{x(sinu)\frac{du}{2x}}$

5. $\frac{1}{2}\int{sinudu}$

6. $\frac{-1}{2}cosu+c$

7. $\frac{-1}{2}cos(x^2)+c$

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Example Question #2 : Antiderivatives By Substitution Of Variables

Solve the following integral using substitution:

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

Possible Answers:

$\frac{1}{2}e^{x2}+c$

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

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

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

$2xe^{x^3}+c$

Correct answer:

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

Explanation:

To solve the integral, we have to simplify it by using a variable to substitute for a variable of . For this problem, we will let u replace the expression x^3 . Next, we must take the derivative of . Its derivative is 3x^2 . Next, solve this equation for so that we may replace it in the integral. Plug in place of x^3 and $\frac{du}{3x^2}$ in place of into the original integral and simplify. The x^2 in the denominator cancels out the remaining x^2 in the integral, leaving behind a $\frac{1}{3}$ . We can pull the $\frac{1}{3}$ out front of the integral. Next, take the anti-derivative of the integrand and replace with the original expression, adding the constant to the answer. The specific steps are as follows:

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

2. u=x^3

3. du=3x^2

4. $\int{x^2e^u(\frac{du}{3x^2})}$

5. $\frac{1}{3}\int{e^udu}$

6. $\frac{1}{3}e^u+c$

7. $\frac{1}{3}e^{x^3}+c$

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Example Question #9 : Antiderivatives By Substitution Of Variables

Solve the following integral using substitution:

$\int{e^xcos(e^x)dx}$

Possible Answers:

-cos(e^x)+c

-sin(e^x)+c

sin(e^x)+c

$e^{cosx}+c$

$e^{sinx}+c$

Correct answer:

sin(e^x)+c

Explanation:

To solve the integral, we have to simplify it by using a variable to substitute for a variable of . For this problem, we will let u replace the expression e^x . Next, we must take the derivative of . Its derivative is e^x . Next, solve this equation for so that we may replace it in the integral. Plug in place of e^x and $\frac{du}{e^x}$ in place of into the original integral and simplify. The e^x in the denominator cancels out the remaining e^x in the integral. Next, take the anti-derivative of the integrand and replace with the original expression, adding the constant to the answer. The specific steps are as follows:

1. $\int{e^xcos(e^x)dx}$

2. u=e^x

3. du=e^xdx

4. $\int{e^xcosu(\frac{du}{e^x})}$

5. $\int{cosudu}$

6. sinu+c

7. sin(e^x)+c

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Example Question #9 : Antiderivatives By Substitution Of Variables

Solve the following integral using substitution:

$\int{\frac{sin(lnx)}{x}dx}$

Possible Answers:

$sinx(\frac{1}{x})+c$

-cos(lnx)+c

cos(lnx)+c

$\frac{1}{\sqrt{1-lnx}}+c$

$-sinx(\frac{1}{x})+c$

Correct answer:

-cos(lnx)+c

Explanation:

To solve the integral, we have to simplify it by using a variable to substitute for a variable of . For this problem, we will let u replace the expression lnx . Next, we must take the derivative of . Its derivative is $\frac{1}{x}$ . Next, solve this equation for so that we may replace it in the integral. Plug in place of lnx and dux in place of into the original integral and simplify. The in the denominator cancels out the remaining in the integral. Next, take the anti-derivative of the integrand and replace with the original expression, adding the constant to the answer. The specific steps are as follows:

1. $\int{\frac{sin(lnx)}{x}dx}$

2. u=lnx

3. $du=(\frac{1}{x}dx)$

4. dux=dx

5. $\int{\frac{sinu}{x}dux}$

6. $\int{sinudu}$

7. -cosu+c

8. -cos(lnx)+c

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Example Question #11 : Antiderivatives By Substitution Of Variables

Solve the following integral using substitution:

$\int{(x^2+e^{3x})(x^3+e^{3x})^{\frac{6}{7}}dx}$

Possible Answers:

$\frac{7}{39}(x^3+e^{3x})^{\frac{13}{7}}+c$

$\frac{12}{7}(x^3+e^{3x})^{\frac{7}{12}}+c$

$-21(x^3+e^{3x})^{\frac{1}{3}}+c$

$\frac{39}{7}(x^3+e^{3x})^{\frac{7}{6}}+c$

$3x^2(e^{3x})^{\frac{-1}{7}}+c$

Correct answer:

$\frac{7}{39}(x^3+e^{3x})^{\frac{13}{7}}+c$

Explanation:

To solve the integral, we have to simplify it by using a variable to substitute for a variable of . For this problem, we will let u replace the expression $x^3+e^{3x}$ . Next, we must take the derivative of . Its derivative is $3x^2+e^{3x}$ . Next, solve this equation for so that we may replace it in the integral. Plug in place of $x^3+e^{3x}$ and $\frac{du}{3x^2+e^{3x}}$ in place of into the original integral and simplify. The $x^2+e^{3x}$ in the denominator cancels out the remaining $x^2+e^{3x}$ in the integral, leaving behind a $\frac{1}{3}$ . We can pull the $\frac{1}{3}$ out front of the integral. Next, take the anti-derivative of the integrand and replace with the original expression, adding the constant to the answer. The specific steps are as follows:

1. $\int{(x^2+e^{3x})(x^3+e^{3x})^{\frac{6}{7}}dx}$

2. $u=x^3+e^{3x}$

3. $du=(3x^2+3e^{3x})dx$

4. $\int(x^2+e^{3x})u^{\frac{6}{7}(\frac{du}{x^2+e^{3x}})}$

5. $\frac{1}{3}\int{u^{\frac{6}{7}}}du$

6. $\frac{1}{3}(\frac{7}{13})u^{\frac{13}{7}}$

7. $\frac{7}{39}u^{\frac{13}{7}}+c$

8. $\frac{7}{39}(x^3+e^{3x})^{\frac{13}{7}}+c$

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Example Question #12 : Antiderivatives By Substitution Of Variables

$\int e^{4x-2}dx=?$

Possible Answers:

$4e^{4x-2}+C$

$4e^{4x-2}$

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

$e^{4x-2}+C$

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

Correct answer:

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

Explanation:

This is a u-substitution integral. We need to substitute the new function, which is modifying our base function (the exponential).

$\int e^xdx=e^x+C$ , but instead of that, our problem is e^u . We can solve this integral by completing the substitution.

$u=4x-2.\; du=4dx. \;dx=\frac{du}{4}.$

Now, we can replace everything in our integrand and rewrite in terms of our new variables:

$\int e^{4x-2}dx=\int e^u\frac{du}{4}=\frac{1}{4}\int e^udu.$

$=\frac{1}{4}e^u+C=\frac{1}{4}e^{4x-2}+C$ .

Remember to plug your variable back in and include the integration constant since we have an indefinite integral.

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Example Question #941 : Ap Calculus Ab

$\int 2cos(2x)dx=?$

Possible Answers:

4sin(2x)+C

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

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

sin(2x)+C

-sin(2x)+C

Correct answer:

sin(2x)+C

Explanation:

This is a u-substitution integral. We need to choose the following substitutions:

$u=2x.\; du=2dx.$

Now, we can replace our original problem with our new variables:

$\int2cos(2x)dx=\int cos(2x)2dx=\int cos(u)du$

=sin(u)+C=sin(2x)+C.

In the last step, we need to plug in our original function and add the integration constant.

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Example Question #14 : Antiderivatives By Substitution Of Variables

$\int \sqrt{3+4x}dx=?$

Possible Answers:

$\frac{1}{4}(3+4x)^{3/2}+C$

$\frac{1}{6}(3+4x)^{3/2}+C$

$(3+4x)^{3/2}+C$

$\frac{3}{8}(3+4x)^{3/2}+C$

$\frac{8}{3}(3+4x)^{3/2}+C$

Correct answer:

$\frac{1}{6}(3+4x)^{3/2}+C$

Explanation:

This is a hidden u-substitution problem! Because we have a function under our square root, we cannot just simply integrate it. Therefore, we need to choose the function under the square root as our substitution variable!

$u=3+4x.\; du=4dx.\; dx=\frac{du}{4}.$

Now, let us rewrite our original equation in terms of our new variable!

$\int \sqrt{3+4x}dx=\int \sqrt{u}\frac{du}{4}=\frac{1}{4}\int \sqrt udu.$

$=\frac{1}{4}\int u^{1/2}=\frac{1}{4}\frac{u^{3/2}}{3/2}+C=\frac{1}{4}*\frac{2}{3}u^{3/2}+C$

$=\frac{2}{12}u^{3/2}+C=\frac{1}{6}({3+4x)}^{3/2}+C$ .

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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 #5 : Antiderivatives By Substitution Of Variables

Example Question #6 : Antiderivatives By Substitution Of Variables

Example Question #7 : Antiderivatives By Substitution Of Variables

Example Question #2 : Antiderivatives By Substitution Of Variables

Example Question #9 : Antiderivatives By Substitution Of Variables

Example Question #9 : Antiderivatives By Substitution Of Variables

Example Question #11 : Antiderivatives By Substitution Of Variables

Example Question #12 : Antiderivatives By Substitution Of Variables

Example Question #941 : Ap Calculus Ab

Example Question #14 : Antiderivatives By Substitution Of Variables

All AP Calculus AB Resources

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