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High School Math : Finding Indefinite Integrals

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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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High School Math : Finding Indefinite Integrals

Study concepts, example questions & explanations for High School Math

All High School Math Resources

Example Questions

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

All High School Math Resources

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