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Calculus 1 : Equations

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Example Questions

Example Question #81 : How To Find Integral Expressions

Evaluate the following indefinite integral:

$\int (2x+3)dx$

Possible Answers:

$x^{2}+3+C$

$2x^{2}+3x+C$

$x^{2}+3x+C$

$x^{2}+3x$

Correct answer:

$x^{2}+3x+C$

Explanation:

To evaluate the integral, use the inverse power rule:

$\int x^{n}dx=\frac{x^{n+1}}{n+1} n\neq1$

Applying that rule to this problem gives us the following for the first term:

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

And the following for the second term:

$\frac{3x^{0+1}}{0+1}=\frac{3x^{1}}{1}=3x$

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

$x^{2}+3x+C$

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Example Question #81 : Integral Expressions

Evaluate the following indefinite integral:

$\int (56x^{6}+9x^{2})dx$

Possible Answers:

$7x^{7}+9x^{3}+C$

$8x^{7}+3x^{3}$

$7x^{7}+9x^{3}$

$8x^{7}+3x^{3}+C$

Correct answer:

$8x^{7}+3x^{3}+C$

Explanation:

To evaluate the integral, use the inverse power rule:

$\int x^{n}dx=\frac{x^{n+1}}{n+1} n\neq1$

Applying that rule to this problem gives us the following for the first term:

$\frac{56x^{6+1}}{6+1}=\frac{56x^{7}}{7}=8x^{7}$

And the following for the second term:

$\frac{9x^{2+1}}{2+1}=\frac{9x^{3}}{3}=3x^{3}$

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

$8x^{7}+3x^{3}+C$

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Example Question #81 : How To Find Integral Expressions

Evaluate the following indefinite integral:

$\int (12x^{4}-6x^{2}-9)dx$

Possible Answers:

$\frac{12x^{5}}{5}+2x^{3}-9x+C$

$5x^{5}+2x^{3}-9x+C$

$12x^{5}-2x^{3}-9x+C$

$\frac{12x^{5}}{5}-2x^{3}-9x+C$

Correct answer:

$\frac{12x^{5}}{5}-2x^{3}-9x+C$

Explanation:

To evaluate the integral, use the inverse power rule:

$\int x^{n}dx=\frac{x^{n+1}}{n+1} n\neq1$

Applying that rule to this problem gives us the following for the first term:

$\frac{12x^{4+1}}{4+1}=\frac{12x^{5}}{5}$

The following for the second term:

$\frac{-6x^{2+1}}{2+1}=\frac{-6x^{3}}{3}=-2x^{3}$

And the following for our thrid term:

$\frac{-9x^{0+1}}{0+1}=\frac{-9x^{1}}{1}=-9x$

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

$\frac{12x^{5}}{5}-2x^{3}-9x+C$

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Example Question #81 : Equations

Evaluate the following indefinite integral:

$\int 15x^{2}dx$

Possible Answers:

$15x^{3}+C$

$5x^{2}+C$

$5x^{4}+C$

$5x^{3}+C$

Correct answer:

$5x^{3}+C$

Explanation:

To evaluate the integral, use the inverse power rule:

$\int x^{n}dx=\frac{x^{n+1}}{n+1} n\neq1$

Applying that rule to this problem gives us the following for the first term:

$\frac{15x^{2+1}}{2+1}=\frac{15x^{3}}{3}=5x^{3}$

Add our "C" to get the final answer:

$5x^{3}+C$

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Example Question #84 : How To Find Integral Expressions

Evaluate the following indefinite integral:

$\int cos(x+3)dx$

Possible Answers:

sin(x+3)+C

3cos(x+3)+C

3sin(x+3)+C

sin(x)+3+C

Correct answer:

sin(x+3)+C

Explanation:

To evaluate this integral we have to remember the laws of trig functions. In this problem we also use a "u substitution" to account for the function inside of the cosine. The steps for a "u-sub" are as follows:

1. Set the function equal to u and take the derivative of both sides.

u=x+3

du=dx

2. Substitute each of the x values for u values in the integral then solve the integral:

$\int cos(x+3)dx=\int cos(u)du= sin(u)+C$

3. Put the x values back into the equation to get the final answer:

sin(u)+C = sin(x+3)+C

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Example Question #81 : Equations

Evaluate the following indefinite integral:

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

Possible Answers:

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

$3(2x+6)^{3}+C$

$(2x+6)^{2}+C$

$(2x+6)^{3}+C$

Correct answer:

$(2x+6)^{3}+C$

Explanation:

In this problem we use a "u substitution" to account for the function inside of the parentheses. The steps for a "u-sub" are as follows:

1. Set the function equal to u, take the derivative of both sides, and solve for dx:

u=2x+6

du=2dx

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

2. Substitute each of the x values for u values in the integral then solve accordingly:

$\int 6(u)^{2}*\frac{du}{2}= \int3u^{2}du$

$\int3u^{2}du=\frac{3u^{2+1}}{2+1}=\frac{3u^{3}}{3}=u^{3}$

3. Put the x values back into the equation and add "C" to get the final answer:

$u^{3}=(2x+6)^{3}+C$

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Example Question #82 : How To Find Integral Expressions

Evaluate the following indefinite integral:

$\int (cos(x))^{2}sin(x)dx$

Possible Answers:

$(cos(x))^{3}+C$

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

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

$-(cos(x))^{3}+C$

Correct answer:

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

Explanation:

In this problem we use a "u substitution" to account for the function inside of the parentheses. The steps for a "u-sub" are as follows:

1. Set the function equal to u, take the derivative of both sides, and solve for dx:

u=cos(x)dx

du=-sin(x)dx

-du=sin(x)dx

2. Substitute each of the x values for u values in the integral then solve accordingly:

$\int u^{2}*-du= \int -u^{2}du$

$\int -u^{2}du=\frac{-u^{2+1}}{2+1}=\frac{-u^{3}}{3}$

3. Put the x values back into the equation and add "C" to get the final answer:

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

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Example Question #86 : How To Find Integral Expressions

Evaluate the following indefinite integral:

$\int (2x+5)^{2}dx$

Possible Answers:

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

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

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

$\frac{(2x+5)^{3}}{6}+C$

Correct answer:

$\frac{(2x+5)^{3}}{6}+C$

Explanation:

In this problem we use a "u substitution" to account for the function inside of the parentheses. The steps for a "u-sub" are as follows:

1. Set the function equal to u, take the derivative of both sides, and solve for dx:

u=2x+5

du=2dx

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

2. Substitute each of the x values for u values in the integral then solve accordingly:

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

$\int \frac{u^{2}}{2}du=\frac{u^{2+1}}{2*(2+1)}=\frac{u^{3}}{2*3}=\frac{u^{3}}{6}$

3. Put the x values back into the equation and add "C" to get the final answer:

$\frac{u^{3}}{6}=\frac{(2x+5)^{3}}{6}+C$

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Example Question #82 : Equations

Evaluate the following indefinite integral:

$\int 12(4x+2)^{3}dx$

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

In this problem we use a "u substitution" to account for the function inside of the parentheses. The steps for a "u-sub" are as follows:

1. Set the function equal to u, take the derivative of both sides, and solve for dx:

u=4x+2

du=4dx

$dx=\frac{du}{4}$

2. Substitute each of the x values for u values in the integral then solve accordingly:

$\int 12u^{3}*\frac{du}{4}= \int 3u^{3}du$

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

3. Put the x values back into the equation and add "C" to get the final answer:

$\frac{3u^{4}}{4}=\frac{3}{4}(4x+2)^{4}+C$

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Example Question #83 : Equations

Evaluate the following indefinite integral:

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

Possible Answers:

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

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

$(x+6)^{3}+C$

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

Correct answer:

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

Explanation:

In this problem we use a "u substitution" to account for the function inside of the parentheses. The steps for a "u-sub" are as follows:

1. Set the function equal to u, take the derivative of both sides, and solve for dx:

u=x+6

du=dx

2. Substitute each of the x values for u values in the integral then solve accordingly:

$\int u^{2}du = \frac{u^{2+1}}{2+1}=\frac{u^{3}}{3}$

3. Put the x values back into the equation and add "C" to get the final answer:

$\frac{u^{3}}{3}=\frac{1}{3}(x+6)^{3}+C$

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Calculus 1 : Equations

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #81 : How To Find Integral Expressions

Example Question #81 : Integral Expressions

Example Question #81 : How To Find Integral Expressions

Example Question #81 : Equations

Example Question #84 : How To Find Integral Expressions

Example Question #81 : Equations

Example Question #82 : How To Find Integral Expressions

Example Question #86 : How To Find Integral Expressions

Example Question #82 : Equations

Example Question #83 : Equations

All Calculus 1 Resources

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