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AP Calculus AB : Antiderivatives by substitution of variables

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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:

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

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

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

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

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

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:

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

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

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

$4e^{4x-2}$

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

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 #13 : Antiderivatives By Substitution Of Variables

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

Possible Answers:

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

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

-sin(2x)+C

sin(2x)+C

4sin(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 #61 : Techniques Of Antidifferentiation

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

Possible Answers:

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

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

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

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

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

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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Example Question #11 : Antiderivatives By Substitution Of Variables

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

Possible Answers:

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

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

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

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

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 #61 : Techniques Of Antidifferentiation

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

Possible Answers:

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

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

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

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

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

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 #17 : Antiderivatives By Substitution Of Variables

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

Possible Answers:

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

2ln|4x-8|+C

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

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

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

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 #18 : Antiderivatives By Substitution Of Variables

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

Possible Answers:

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

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

2ln(x)+C

2lnx^2+C

(lnx)^2+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 #222 : Integrals

Integrate:

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

Possible Answers:

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

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

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

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

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

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 #11 : Antiderivatives By Substitution Of Variables

Integrate:

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

Possible Answers:

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

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

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

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

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

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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AP Calculus AB : Antiderivatives by substitution of variables

Study concepts, example questions & explanations for AP Calculus AB

All AP Calculus AB Resources

Example Questions

Example Question #11 : Antiderivatives By Substitution Of Variables

Example Question #12 : Antiderivatives By Substitution Of Variables

Example Question #13 : Antiderivatives By Substitution Of Variables

Example Question #61 : Techniques Of Antidifferentiation

Example Question #11 : Antiderivatives By Substitution Of Variables

Example Question #61 : Techniques Of Antidifferentiation

Example Question #17 : Antiderivatives By Substitution Of Variables

Example Question #18 : Antiderivatives By Substitution Of Variables

Example Question #222 : Integrals

Example Question #11 : Antiderivatives By Substitution Of Variables

All AP Calculus AB Resources

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