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High School Math : Integrals

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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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Example Question #81 : Comparing Relative Magnitudes Of Functions And Their Rates Of Change

What is the indefinite integral of 4x+3 with respect to ?

Possible Answers:

2x^3-c

2x^2+3x+c

2x^2+3x

4+c

Correct answer:

2x^2+3x+c

Explanation:

To find the indefinite integral, we're going to use the reverse power rule: raise the exponent of the variable by one and then divide by that new exponent.

$\int 4x+3 dx=(\frac{4x^{1+1}}{1+1})+(\frac{3x^{0+1}}{0+1})+c$

Be sure to include + c to compensate for any constant!

$\int 4x+3dx=(\frac{4x^{1+1}}{1+1})+(\frac{3x^{0+1}}{0+1})+c$

$\int 4x+3dx=(\frac{4x^{2}}{2})+(\frac{3x^{1}}{1})+c$

$\int 4x+3dx=2x^2+3x+c$

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Example Question #82 : Calculus Ii — Integrals

$\int x^4-3x^2+8\ {\mathrm{d} x}=?$

Possible Answers:

$\frac{1}{5}x^5-x^3+8x+c$

5x^5-3x^3+8x+c

x^7

$\frac{1}{5}x^5-x^3+8x$

12x^2-6x

Correct answer:

$\frac{1}{5}x^5-x^3+8x+c$

Explanation:

To find the indefinite integral, or anti-derivative, we can use the reverse power rule. We raise the exponent of each variable by one and divide by that new exponent.

$\int x^4-3x^2+8\ {\mathrm{d} x}=(\frac{x^{4+1}}{4+1})-(\frac{3x^{2+1}}{2+1})+(\frac{8x^{0+1}}{0+1})+c$

Don't forget to include a to cover any constant!

Simplify.

$\int x^4-3x^2+8\ {\mathrm{d} x}=(\frac{x^{5}}{5})-(\frac{3x^{3}}{3})+(\frac{8x^{1}}{1})+c$

$\int x^4-3x^2+8\ {\mathrm{d} x}=\frac{1}{5}x^5-x^3+8x+c$

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Example Question #2241 : High School Math

$\int \frac{2}{3}x^3+7x^2-12x \ {\mathrm{d} x}=?$

Possible Answers:

$\frac{1}{6}x^4+\frac{7}{3}x^3-12x+c$

$\frac{1}{6}x^4+\frac{7}{3}x^3-6x^2+c$

2x^2+14x

$6x^4+\frac{7}{3}x^3-6x^2$

Correct answer:

$\frac{1}{6}x^4+\frac{7}{3}x^3-6x^2+c$

Explanation:

To find the indefinite integral of $\frac{2}{3}x^3+7x^2-12x$ , we can use the reverse power rule. To do this, we raise our exponent by one and then divide the variable by that new exponent.

$\int \frac{2}{3}x^3+7x^2-12x\ {\mathrm{d} x}=\frac{\frac{2}{3}x^{3+1}}{3+1}+\frac{7x^{2+1}}{2+1}-\frac{12x^{1+1}}{1+1}+c$

Don't forget to include a to cover any constant!

$\int \frac{2}{3}x^3+7x^2-12x\ {\mathrm{d} x}=\frac{\frac{2}{3}x^{4}}{4}+\frac{7x^{3}}{3}-\frac{12x^{2}}{2}+c$

$\int \frac{2}{3}x^3+7x^2-12x\ {\mathrm{d} x}=\frac{2}{12}x^4+\frac{7}{3}x^3-6x^2+c$

$\int \frac{2}{3}x^3+7x^2-12x\ {\mathrm{d} x}=\frac{1}{6}x^4+\frac{7}{3}x^3-6x^2+c$

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Example Question #1 : Finding Integrals By Substitution

Determine the indefinite integral:

$\dpi{120}\int\frac{\sqrt{\log_{10} x}}{x}\; \; dx$

Possible Answers:

$\frac{ 2 \sqrt{ \log_{10} x} \cdot \log_{10} x} {3 } + C$

$\frac{ 2 \sqrt{ \log_{10} x} \cdot \ln x} {3 } + C$

$\frac{ 2 \log_{10} x} {3 } + C$

$\frac{ 2 \sqrt{ \log_{10} x} } {3 \ln x} + C$

$\frac{ 2 \sqrt{ \log_{10} x} } {3 } + C$

Correct answer:

$\frac{ 2 \sqrt{ \log_{10} x} \cdot \ln x} {3 } + C$

Explanation:

$\log_{10} x = \frac{\ln x}{\ln 10}$ , so this can be rewritten as

$\dpi{120}\int\frac{\sqrt{\log_{10} x}}{x}\; \; dx$

$\dpi{120}=\int\frac{\sqrt{\ln x}}{x \sqrt{\ln 10} }\; \; dx$

$\dpi{120}= \frac{1}{\sqrt{\ln 10}}\int\frac{\sqrt{\ln x}}{x }\; \; dx$

Set $u = \ln x$ . Then

$\frac{\mathrm{d} u}{\mathrm{d} x} = \frac{\mathrm{d} }{\mathrm{d} x} \ln x = \frac{1}{x}$

and

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

Substitute:

$\dpi{120} \frac{1}{\sqrt{\ln 10}}\int\frac{\sqrt{\ln x}}{x }\; \; dx$

$\dpi{120} =\frac{1}{\sqrt{\ln 10}} \int\sqrt{u} \; du$

$\dpi{120} =\frac{1}{\sqrt{\ln 10}} \int u ^{\frac{1}{2}} \; du$

$\dpi{120} = \frac{1}{\sqrt{\ln 10}} \left (\frac{u^{\frac{3}{2}}}{{\frac{3}{2}}} + C \right )$

$\dpi{120} = \frac{1}{\sqrt{\ln 10}} \left (\frac{ 2u \sqrt{u}} {3} + C \right )$

The outer factor can be absorbed into the constant, and we can substitute back:

$\dpi{120} = \frac{1}{\sqrt{\ln 10}} \left (\frac{ 2 \ln x \cdot \sqrt{ \ln x}} {3} \right )+ C$

$= \frac{ 2 \sqrt{ \ln x} \cdot \ln x} {3 \cdot \sqrt{\ln 10}} + C$

$= \frac{ 2 \sqrt{ \log_{10} x} \cdot \ln x} {3 } + C$

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Example Question #71 : Functions, Graphs, And Limits

Evaluate:

$\dpi{120}\int_{1}^{e^{ 3}}\frac{\sqrt{\ln x}}{x}\; \; dx$

Possible Answers:

$\frac{2}{3}$

$\frac{2\sqrt{3}}{3}$

$2 \ln 3$

$2\sqrt{3}$

$2 \sqrt{\ln 3}$

Correct answer:

$2\sqrt{3}$

Explanation:

Set $u = \ln x$ . Then

$\frac{\mathrm{d} u}{\mathrm{d} x} = \frac{\mathrm{d} }{\mathrm{d} x} \ln x = \frac{1}{x}$

and

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

Also, since $u = \ln x$ , the limits of integration change to $\ln e^{3} = 3$ and $\ln 1= 0$ .

Substitute:

$\dpi{120}\int_{1}^{e^{ 3}}\frac{\sqrt{\ln x}}{x}\; \; dx$

$\dpi{120} = \int_{0}^{3}\sqrt{u} \; du$

$= \int_{0}^{3} u ^{\frac{1}{2}} \; du$

$\dpi{150} = \left.\begin{matrix} \frac{u^{\frac{3}{2}}}{{\frac{3}{2}}} & \\ & \end{matrix}\right|$ $\begin{matrix} 3\\ \\ 0 \end{matrix}$

$\dpi{150} = \left.\begin{matrix} \\ \frac{ 2u \sqrt{u}} {3} & \\ & \end{matrix}\right|$ $\begin{matrix} 3\\ \\ 0 \end{matrix}$

$=\frac{ 2\cdot 3\cdot \sqrt{3}} {3} - \frac{ 2\cdot 0\cdot \sqrt{0}} {3} =2\sqrt{3}$

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High School Math : Integrals

Study concepts, example questions & explanations for High School Math

All High School Math Resources

Example Questions

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

Example Question #81 : Comparing Relative Magnitudes Of Functions And Their Rates Of Change

Example Question #82 : Calculus Ii — Integrals

Example Question #2241 : High School Math

Example Question #1 : Finding Integrals By Substitution

Example Question #71 : Functions, Graphs, And Limits

All High School Math Resources

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