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High School Math : Calculus II — Integrals

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Example Question #71 : Calculus Ii — Integrals

What is the indefinite integral of x+1 ?

Possible Answers:

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

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

2x^2+x+c

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

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 #72 : Calculus Ii — Integrals

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

Possible Answers:

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

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

144x+169

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

24x+13

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 #41 : Finding Integrals

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

Possible Answers:

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

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

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

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

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

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 #74 : Calculus Ii — Integrals

What is the indefinite integral of 3x+12 ?

Possible Answers:

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

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

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

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 #75 : Calculus Ii — Integrals

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

Possible Answers:

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

5x+12+c

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

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

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

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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Example Question #2231 : High School Math

What is the indefinite integral of -8x^2+15 ?

Possible Answers:

-16x

$\frac{8}{3}x^3-15x+c$

$-\frac{8}{3}x^3+15x$

$-\frac{8}{3}x^3+15x+c$

Correct answer:

$-\frac{8}{3}x^3+15x+c$

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.

$\int -8x^2+15{\mathrm{d} x}=\frac{-8x^{2+1}}{2+1}+\frac{15x^{0+1}}{0+1}+c$

Remember to include a when doing integrals. This is a placeholder for any constant that might be in the new expression.

$\int -8x^2+15{\mathrm{d} x}=\frac{-8x^{3}}{3}+\frac{15x^{1}}{1}+c$

$\int -8x^2+15{\mathrm{d} x}=-\frac{8}{3}x^3+15x+c$

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Example Question #51 : Integrals

What is the indefinite integral of x^3+2x^2+5 ?

Possible Answers:

$\frac{1}{4}x^4+\frac{2}{3}x^3+5x$

4x^4+6x^3+5x

6x+c

3x^2+4x+c

$\frac{1}{4}x^4+\frac{2}{3}x^3+5x+c$

Correct answer:

$\frac{1}{4}x^4+\frac{2}{3}x^3+5x+c$

Explanation:

To find the indefinite integral, we can use the reverse power rule. Raise the exponent of the variable by one and then divide by that new exponent.

We're going to treat as 5x^0 .

$\int x^3+2x^2+5{\mathrm{d} x}=\frac{x^{3+1}}{3+1}+\frac{2x^{2+1}}{2+1}+\frac{5x^{0+1}}{0+1}+c$

Remember to include the when taking the integral to compensate for any constant.

$\int x^3+2x^2+5{\mathrm{d} x}=\frac{x^{4}}{4}+\frac{2x^{3}}{3}+\frac{5x^{1}}{1}+c$

Simplify.

$\int x^3+2x^2+5{\mathrm{d} x}=\frac{1}{4}x^4+\frac{2}{3}x^3+5x+c$

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Example Question #78 : Calculus Ii — Integrals

What is the indefinite integral of $4x^2+\frac{2}{3}x$ ?

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

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

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

Remember to include a to cover any potential constant that might be in our new equation.

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

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

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Example Question #101 : Asymptotic And Unbounded Behavior

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

Possible Answers:

$\cos(x)+c$

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

$\tan(x)+c$

$-\cos(x)+c$

$-\sin(x)+c$

Correct answer:

$-\cos(x)+c$

Explanation:

Just like with the derivatives, the indefinite integrals or anti-derivatives of trig functions must be memorized.

$\int \sin(x){\mathrm{d} x}=-\cos(x)+c$

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Example Question #80 : Calculus Ii — Integrals

$\int 4x^2+5x-3\: dx=?$

Possible Answers:

$\frac{4}{3}x^3+\frac{5}{x^2}+3x+c$

$\frac{4}{3}x^3+\frac{5}{2}x^2-3x$

$\frac{3}{4}x^2+\frac{5}{2}x-3+c$

$\frac{4}{3}x^3+\frac{5}{2}x^2-3x+c$

8x+5

Correct answer:

$\frac{4}{3}x^3+\frac{5}{2}x^2-3x+c$

Explanation:

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

$\int 4x^2+5x-3\: dx=\frac{4x^{2+1}}{2+1}+\frac{5x^{1+1}}{1+1}-\frac{3x^{0+1}}{0+1}+c$

Don't forget to include a to compensate for any constant!

$\int 4x^2+5x-3\: dx=\frac{4x^{3}}{3}+\frac{5x^{2}}{2}-\frac{3x^{1}}{1}+c$

$\int 4x^2+5x-3\: dx=\frac{4}{3}x^3+\frac{5}{2}x^2-3x+c$

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High School Math : Calculus II — Integrals

Study concepts, example questions & explanations for High School Math

All High School Math Resources

Example Questions

Example Question #71 : Calculus Ii — Integrals

Example Question #72 : Calculus Ii — Integrals

Example Question #41 : Finding Integrals

Example Question #74 : Calculus Ii — Integrals

Example Question #75 : Calculus Ii — Integrals

Example Question #2231 : High School Math

Example Question #51 : Integrals

Example Question #78 : Calculus Ii — Integrals

Example Question #101 : Asymptotic And Unbounded Behavior

Example Question #80 : Calculus Ii — Integrals

All High School Math Resources

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