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High School Math : Finding Integrals

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Example Questions

Example Question #3 : Finding Indefinite Integrals

What is the indefinite integral of x^3+2 ?

Possible Answers:

6x+6+c

$\frac{1}{3}x^4+2+c$

4x^4+2x

3x^2

$\frac{1}{4}x^4+2x+c$

Correct answer:

$\frac{1}{4}x^4+2x+c$

Explanation:

To find the indefinite integral of our equation, we can use the reverse power rule.

To use the reverse power rule, we raise the exponent of the by one and then divide by that new exponent.

Remember that, when taking the integral, we treat constants as that number times $x^{0}$ since anything to the zero power is . For example, treat as 2x^0 .

$\int x^3+2{\mathrm{d} x}=\frac{x^{3+1}}{3+1}+\frac{2x^{0+1}}{0+1}+c$

When taking an integral, be sure to include a at the end of everything. stands for "constant". Since taking the derivative of a constant whole number will always equal , we include the to anticipate the possiblity of the equation actually being x^2+5 or x^2-100 instead of just x^2 .

$\int x^3+2{\mathrm{d} x}=\frac{x^{4}}{4}+\frac{2x^{1}}{1}+c$

$\int x^3+2{\mathrm{d} x}=\frac{1}{4}x^4+2x+c$

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Example Question #4 : Finding Indefinite Integrals

What is the indefinite integral of $\frac{1}{3}x^2$ ?

Possible Answers:

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

x^2

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

2x+3

x^3+c

Correct answer:

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

Explanation:

To find the indefinite integral of our equation, we can use the reverse power rule.

To use the reverse power rule, we raise the exponent of the by one and then divide by that new exponent.

$\int \frac{1}{3}x^2{\mathrm{d} x}=\frac{1}{3}*\frac{x^{2+1}}{2+1}+c$

When taking an integral, be sure to include a . stands for "constant". Since taking the derivative of a constant whole number will always equal , we include the to anticipate the possiblity of the equation actually being x^2+5 or x^2-100 instead of just x^2 .

$\int \frac{1}{3}x^2{\mathrm{d} x}=\frac{1}{3}*\frac{x^{3}}{3}+c$

$\int \frac{1}{3}x^2{\mathrm{d} x}=\frac{1}{9}x^3+c$

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Example Question #81 : Functions, Graphs, And Limits

What is the indefinite integral of x+18 ?

Possible Answers:

$\frac{1}{2}x^2+18x+c$

2x^2+18x+c

$\frac{1}{2}x^3+18x^2+cx$

Undefined

Correct answer:

$\frac{1}{2}x^2+18x+c$

Explanation:

To find the indefinite integral of our equation, we can use the reverse power rule.

To use the reverse power rule, we raise the exponent of the by one and then divide by that new exponent.

Remember that, when taking the integral, we treat constants as that number times $x^{0}$ , since anything to the zero power is . Treat as 18x^0 .

$\int x+18{\mathrm{d} x}=\frac{x^{1+1}}{1+1}+\frac{18x^{0+1}}{0+1}+c$

$\int x+18{\mathrm{d} x}=\frac{x^{2}}{2}+\frac{18x^{1}}{1}+c$

$\int x+18{\mathrm{d} x}=\frac{1}{2}x^2+18x+c$

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Example Question #2223 : High School Math

What is the indefinite integral of $\sin(x)$ ?

Possible Answers:

$-\sin(x)+c$

$-\cos(x)$

$-\cos(x)+c$

$2\sin(x^2)+c$

$\cos(x)$

Correct answer:

$-\cos(x)+c$

Explanation:

$\sin(x)$ is a special function.

The indefinite integral is $\int \sin(x) {\mathrm{d} x}=-\cos(x)+c$ .

Even though it is a special function, we still need to include a . stands for "constant". Since taking the derivative of a constant whole number will always equal , we include the to anticipate the possiblity of the equation actually being x^2+5 or x^2-100 instead of just x^2 .

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Example Question #2224 : High School Math

What is the indefinite integral of 6x^2+8 ?

Possible Answers:

$\frac{2}{3}x^3+8x+c$

2x^3-8x-c

12x

2x^3+8x+c

Correct answer:

2x^3+8x+c

Explanation:

To solve this problem, we can use the anti-power rule or reverse power rule. We raise the exponent on the variables by one and divide by the new exponent.

For this problem, we'll treat as 8x^0 since anything to the zero power is one.

$\int 6x^2+8{\mathrm{d} x}=\frac{6x^{2+1}}{2+1}+\frac{8x^{0+1}}{0+1}+c$

Since the derivative of any constant is , when we take the indefinite integral, we add a to compensate for any constant that might be there.

From here we can simplify.

$\int 6x^2+8{\mathrm{d} x}=\frac{6x^{3}}{3}+\frac{8x^{1}}{1}+c$

$\int 6x^2+8{\mathrm{d} x}=2x^3+8x+c$

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Example Question #71 : Calculus Ii — Integrals

What is the indefinite integral of x+1 ?

Possible Answers:

2x^2+x+c

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

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

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

Correct answer:

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

Explanation:

To solve this problem, we can use the anti-power rule or reverse power rule. We raise the exponent on the variables by one and divide by the new exponent.

For this problem, we'll treat as 1x^0 since anything to the zero power is one.

$\int x+1{\mathrm{d} x}=\frac{x^{1+1}}{1+1}+\frac{1x^{0+1}}{0+1}+c$

Since the derivative of any constant is , when we take the indefinite integral, we add a to compensate for any constant that might be there.

From here we can simplify.

$\int x+1{\mathrm{d} x}=\frac{x^{2}}{2}+\frac{x^{1}}{1}+c$

$\int x+1{\mathrm{d} x}=\frac{1}{2}x^2+x+c$

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Example Question #91 : Functions, Graphs, And Limits

What is the indefinite integral of 12x^2+13x+4 ?

Possible Answers:

$4x^3+\frac{13}{2}x^2+4x+c$

$12x^3+\frac{13}{2}x^2+4x+c$

$4x^3-\frac{13}{2}x^2+4x-c$

24x+13

144x+169

Correct answer:

$4x^3+\frac{13}{2}x^2+4x+c$

Explanation:

To solve this problem, we can use the anti-power rule or reverse power rule. We raise the exponent on the variables by one and divide by the new exponent.

For this problem, we'll treat as 4x^0 since anything to the zero power is one.

$\int 12x^2+13x+4{\mathrm{d} x}=\frac{12x^{2+1}}{2+1}+\frac{13x^{1+1}}{1+1}+\frac{4x^{0+1}}{0+1}+c$

Since the derivative of any constant is , when we take the indefinite integral, we add a to compensate for any constant that might be there.

From here we can simplify.

$\int 12x^2+13x+4{\mathrm{d} x}=\frac{12x^{3}}{3}+\frac{13x^{2}}{2}+\frac{4x^{1}}{1}+c$

$\int 12x^2+13x+4{\mathrm{d} x}=4x^3+\frac{13}{2}x^2+4x+c$

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Example Question #72 : Calculus Ii — Integrals

What is the indefinite integral of $2x^4+\frac{1}{2}x$ ?

Possible Answers:

$8x^2+\frac{1}{2}+c$

$\frac{1}{2}x^4+x^{2}+c$

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

$\frac{2}{5}x^5+\frac{1}{4}x^{2}$

$\frac{2}{5}x^5+\frac{1}{4}x^{2}+c$

Correct answer:

$\frac{2}{5}x^5+\frac{1}{4}x^{2}+c$

Explanation:

To find the indefinite integral, we can use the reverse power rule: we raise the exponent by one and then divide by our new exponent.

$\int 2x^4+\frac{1}{2}x{\mathrm{d} x}=\frac{2x^{4+1}}{4+1}+\frac{\frac{1}{2}x^{1+1}}{1+1}+c$

Remember when taking the indefinite integral to include a to cover any potential constants.

Simplify.

$\int 2x^4+\frac{1}{2}x{\mathrm{d} x}=\frac{2x^{5}}{5}+\frac{\frac{1}{2}x^{2}}{2}+c$

$\int 2x^4+\frac{1}{2}x{\mathrm{d} x}=\frac{2}{5}x^5+\frac{1}{4}x^{2}+c$

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Example Question #71 : Calculus Ii — Integrals

What is the indefinite integral of 3x+12 ?

Possible Answers:

$\frac{2}{3}x^2-12+c$

$\frac{3}{2}x^2+12x+c$

$\frac{3}{2}x^2+12x$

Correct answer:

$\frac{3}{2}x^2+12x+c$

Explanation:

To find the indefinite integral, we can use the reverse power rule: we raise the exponent by one and then divide by our new exponent.

We are going to treat as 12x^0 since anything to the zero power is one.

$\int 3x+12{\mathrm{d} x}=\frac{3x^{1+1}}{1+1}+\frac{12x^{0+1}}{0+1}+c$

Remember when taking the indefinite integral to include a to cover any potential constants.

Simplify.

$\int 3x+12{\mathrm{d} x}=\frac{3x^{2}}{2}+\frac{12x^{1}}{1}+c$

$\int 3x+12{\mathrm{d} x}=\frac{3}{2}x^2+12x+c$

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Example Question #73 : Calculus Ii — Integrals

What is the indefinite integral of 5x^2+12x ?

Possible Answers:

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

5x+12+c

$\frac{5}{3}x^3+6x^2+cx$

$\frac{3}{5}x^3+\frac{1}{6}x^2+c$

$\frac{5}{3}x^3+6x^2+c$

Correct answer:

$\frac{5}{3}x^3+6x^2+c$

Explanation:

To find the indefinite integral, we use the reverse power rule. That means we raise the exponent on the variables by one and then divide by the new exponent.

$\int 5x^2+12x{\mathrm{d} x}=\frac{5x^{2+1}}{2+1}+\frac{12x^{1+1}}{1+1}+c$

Remember to include a when computing integrals. This is a place holder for any constant that might be in the new expression.

$\int 5x^2+12x{\mathrm{d} x}=\frac{5x^{3}}{3}+\frac{12x^{2}}{2}+c$

$\int 5x^2+12x{\mathrm{d} x}=\frac{5}{3}x^3+\frac{12}{2}x^2+c$

$\int 5x^2+12x{\mathrm{d} x}=\frac{5}{3}x^3+6x^2+c$

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High School Math : Finding Integrals

Study concepts, example questions & explanations for High School Math

All High School Math Resources

Example Questions

Example Question #3 : Finding Indefinite Integrals

Example Question #4 : Finding Indefinite Integrals

Example Question #81 : Functions, Graphs, And Limits

Example Question #2223 : High School Math

Example Question #2224 : High School Math

Example Question #71 : Calculus Ii — Integrals

Example Question #91 : Functions, Graphs, And Limits

Example Question #72 : Calculus Ii — Integrals

Example Question #71 : Calculus Ii — Integrals

Example Question #73 : Calculus Ii — Integrals

All High School Math Resources

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