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Calculus 1 : Writing Equations

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Example Questions

Example Question #91 : Equations

Evaluate the following indefinite integral:

$\int (sin(x))^{3}cos(x)dx$ $\displaystyle \int (sin(x))^{3}cos(x)dx$

Possible Answers:

$\frac{1}{3}(sin(x))^{4}+C$ $\displaystyle \frac{1}{3}(sin(x))^{4}+C$

$\frac{1}{4}(sin(x))^{3}+C$ $\displaystyle \frac{1}{4}(sin(x))^{3}+C$

$4(sin(x))^{4}+C$ $\displaystyle 4(sin(x))^{4}+C$

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

Correct answer:

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

Explanation:

To solve this problem we have to use a u substitution. A "u-sub" is done by using the following steps:

1. Set u equal to the x equation in parentheses, take the derivative, and solve:

$\displaystyle u=sin(x)$

$\displaystyle du=cos(x)dx$

2. Replace x values with u values and integrate accordingly:

$\int u^{3}du=\frac{u^{3+1}}{3+1}=\frac{u^{4}}{4}$ $\displaystyle \int u^{3}du=\frac{u^{3+1}}{3+1}=\frac{u^{4}}{4}$

3. Put the original x equation back in for u and add "C":

$\frac{u^{4}}{4}=\frac{1}{4}(sin(x))^{4}+C$ $\displaystyle \frac{u^{4}}{4}=\frac{1}{4}(sin(x))^{4}+C$

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Example Question #92 : Equations

Evaluate the following indefinite integral:

$\int (sec(2x))^{2}dx$ $\displaystyle \int (sec(2x))^{2}dx$

Possible Answers:

$\displaystyle tan(2x)+C$

$\displaystyle 2tan(2x)+C$

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

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

Correct answer:

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

Explanation:

To solve this problem we have to use a u substitution. We also need to remember our properties of trig functions. A "u-sub" is done by using the following steps:

1. Set u equal to the x equation in parentheses, take the derivative, and solve:

u=2x $\displaystyle u=2x$

du=2dx $\displaystyle du=2dx$

$dx=\frac{1}{2}du$ $\displaystyle dx=\frac{1}{2}du$

2. Replace x values with u values and integrate accordingly:

$\int (sec(u))^{2}*\frac{1}{2}du=\frac{1}{2}tan(u)$ $\displaystyle \int (sec(u))^{2}*\frac{1}{2}du=\frac{1}{2}tan(u)$

3. Put the original x equation back in for u and add "C":

$\frac{1}{2}tan(u)=\frac{1}{2}tan(2x)+C$ $\displaystyle \frac{1}{2}tan(u)=\frac{1}{2}tan(2x)+C$

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Example Question #91 : Writing Equations

Evaluate the following indefinite integral:

$\int e^{x}dx$ $\displaystyle \int e^{x}dx$

Possible Answers:

$e^{x}+C$ $\displaystyle e^{x}+C$

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

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

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

Correct answer:

$e^{x}+C$ $\displaystyle e^{x}+C$

Explanation:

To solve this problem, all we have to remember is that the integral of e doesn't change. That gives us the following:

$\int e^{x}dx=e^{x}+C$ $\displaystyle \int e^{x}dx=e^{x}+C$

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Example Question #94 : Equations

Evaluate the following indefinite integral:

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

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

To solve this problem we have to use a u substitution. A "u-sub" is done by using the following steps:

1. Set u equal to the x equation in parentheses, take the derivative, and solve:

u=2x $\displaystyle u=2x$

du=2dx $\displaystyle du=2dx$

$dx=\frac{1}{2}du$ $\displaystyle dx=\frac{1}{2}du$

2. Replace x values with u values and integrate accordingly:

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

3. Put the original x equation back in for u and add "C":

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

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Example Question #95 : Equations

Evaluate the following indefinite integral:

$\int 3e^{6x}dx$ $\displaystyle \int 3e^{6x}dx$

Possible Answers:

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

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

$\frac{1}{3}e^{6x}+C$ $\displaystyle \frac{1}{3}e^{6x}+C$

$e^{6x}+C$ $\displaystyle e^{6x}+C$

Correct answer:

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

Explanation:

To solve this problem we have to use a u substitution. A "u-sub" is done by using the following steps:

1. Set u equal to the x equation in parentheses, take the derivative, and solve:

u=6x $\displaystyle u=6x$

du=6dx $\displaystyle du=6dx$

$dx=\frac{1}{6}du$ $\displaystyle dx=\frac{1}{6}du$

2. Replace x values with u values and integrate accordingly:

$\int 3e^{u}*\frac{1}{6}du= \int \frac{1}{2}e^{u}du=\frac{1}{2}e^{u}$ $\displaystyle \int 3e^{u}*\frac{1}{6}du= \int \frac{1}{2}e^{u}du=\frac{1}{2}e^{u}$

3. Put the original x equation back in for u and add "C":

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

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Example Question #96 : Equations

Evaluate the following indefinite integral:

$\int (x + sin(x))dx$ $\displaystyle \int (x + sin(x))dx$

Possible Answers:

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

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

$x^{2}-cos(x)+C$ $\displaystyle x^{2}-cos(x)+C$

$2x^{2}+cos(x)+C$ $\displaystyle 2x^{2}+cos(x)+C$

Correct answer:

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

Explanation:

To solve this integral, we use the power rule for the first term:

$\frac{x^{1+1}}{1+1}=\frac{x^{2}}{2}$ $\displaystyle \frac{x^{1+1}}{1+1}=\frac{x^{2}}{2}$

And use the laws of trig functions for the second term:

$\int sin(x)dx= -cos(x)$ $\displaystyle \int sin(x)dx= -cos(x)$

We can combine these terms and add our "C" to get the final answer:

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

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Example Question #97 : Equations

Evaluate the following indefinite integral:

$\int (3x+2)^{2}dx$ $\displaystyle \int (3x+2)^{2}dx$

Possible Answers:

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

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

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

$9(3x+2)^{3}+C$ $\displaystyle 9(3x+2)^{3}+C$

Correct answer:

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

Explanation:

To solve this problem we have to use a u substitution. A "u-sub" is done by using the following steps:

1. Set u equal to the x equation in parentheses, take the derivative, and solve:

u=3x+2 $\displaystyle u=3x+2$

du=3dx $\displaystyle du=3dx$

$dx=\frac{1}{3}du$ $\displaystyle dx=\frac{1}{3}du$

2. Replace x values with u values and integrate accordingly:

$\int u^{2}*\frac{1}{3}du=\frac{u^{2+1}}{3*(2+1)}=\frac{u^{3}}{9}$ $\displaystyle \int u^{2}*\frac{1}{3}du=\frac{u^{2+1}}{3*(2+1)}=\frac{u^{3}}{9}$

3. Put the original x equation back in for u and add "C":

$\frac{u^{3}}{9}=\frac{1}{9}(3x+2)+C$ $\displaystyle \frac{u^{3}}{9}=\frac{1}{9}(3x+2)+C$

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Example Question #98 : Equations

Evaluate the following indefinite integral:

$\int (2x+6)^{3}dx$ $\displaystyle \int (2x+6)^{3}dx$

Possible Answers:

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

$\frac{1}{8}(2x+6)^{4}+C$ $\displaystyle \frac{1}{8}(2x+6)^{4}+C$

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

$8(2x+6)^{4}+C$ $\displaystyle 8(2x+6)^{4}+C$

Correct answer:

$\frac{1}{8}(2x+6)^{4}+C$ $\displaystyle \frac{1}{8}(2x+6)^{4}+C$

Explanation:

To solve this problem we have to use a u substitution. A "u-sub" is done by using the following steps:

1. Set u equal to the x equation in parentheses, take the derivative, and solve:

u=2x+6 $\displaystyle u=2x+6$

du=2dx $\displaystyle du=2dx$

$dx=\frac{1}{2}du$ $\displaystyle dx=\frac{1}{2}du$

2. Replace x values with u values and integrate accordingly:

$\int u^{3}*\frac{1}{2}du=\frac{u^{3+1}}{2*(3+1)}=\frac{u^{4}}{8}$ $\displaystyle \int u^{3}*\frac{1}{2}du=\frac{u^{3+1}}{2*(3+1)}=\frac{u^{4}}{8}$

3. Put the original x equation back in for u and add "C":

$\frac{u^{4}}{8}=\frac{1}{8}(2x+6)^{4}+C$ $\displaystyle \frac{u^{4}}{8}=\frac{1}{8}(2x+6)^{4}+C$

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Example Question #99 : Equations

Evaluate the following indefinite integral:

$\int sin(3x)dx$ $\displaystyle \int sin(3x)dx$

Possible Answers:

$\displaystyle -3cos(3x)+C$

$\displaystyle 3cos(3x)+C$

$-\frac{1}{3}cos(3x)+C$ $\displaystyle -\frac{1}{3}cos(3x)+C$

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

Correct answer:

$-\frac{1}{3}cos(3x)+C$ $\displaystyle -\frac{1}{3}cos(3x)+C$

Explanation:

To solve this problem we have to use a u substitution. A "u-sub" is done by using the following steps:

1. Set u equal to the x equation in parentheses, take the derivative, and solve:

u=3x $\displaystyle u=3x$

du=3dx $\displaystyle du=3dx$

$dx=\frac{1}{3}du$ $\displaystyle dx=\frac{1}{3}du$

2. Replace x values with u values and integrate accordingly:

$\int sin(u)*\frac{1}{3}du=-\frac{1}{3}cos(u)$ $\displaystyle \int sin(u)*\frac{1}{3}du=-\frac{1}{3}cos(u)$

3. Put the original x equation back in for u and add "C":

$-\frac{1}{3}cos(u)=-\frac{1}{3}cos(3x)+C$ $\displaystyle -\frac{1}{3}cos(u)=-\frac{1}{3}cos(3x)+C$

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Example Question #100 : Equations

Evaluate the following indefinite integral:

$\int (x^{2}+3x)dx$ $\displaystyle \int (x^{2}+3x)dx$

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

To solve this integral, use the power rule. Applying it to this problem gives us the following for the first term:

$\frac{x^{2+1}}{2+1}=\frac{1}{3}x^{3}$ $\displaystyle \frac{x^{2+1}}{2+1}=\frac{1}{3}x^{3}$

And the following for the second term:

$\frac{x^{1+1}}{1+1}=\frac{3}{2}x^{2}$ $\displaystyle \frac{x^{1+1}}{1+1}=\frac{3}{2}x^{2}$

We can combine these terms and add our "C" to get the final answer:

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

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Calculus 1 : Writing Equations

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #91 : Equations

Example Question #92 : Equations

Example Question #91 : Writing Equations

Example Question #94 : Equations

Example Question #95 : Equations

Example Question #96 : Equations

Example Question #97 : Equations

Example Question #98 : Equations

Example Question #99 : Equations

Example Question #100 : Equations

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