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High School Math : Calculus II — Integrals

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Example Questions

Example Question #41 : Comparing Relative Magnitudes Of Functions And Their Rates Of Change

Evaluate the integral below:

$f(x)=\int \frac{dx}{x^2-16}$

Possible Answers:

$\frac{1}{8} ln\left | \frac{x-4}{x+4} \right |+c$

$\frac{1}{4} ln\left | \frac{x-2}{x+2} \right |+c$

$\frac{1}{4} ln\left | \frac{x+2}{x-2} \right |+c$

$\frac{1}{8} ln\left | \frac{x+4}{x-4} \right |+c$

Correct answer:

$\frac{1}{8} ln\left | \frac{x-4}{x+4} \right |+c$

Explanation:

In this case we have a rational function as $\frac{N(x)}{D(x)}$ , where

N(x)=1

and

$D(x)=\frac{1}{x^2-16}$

D(x) can be written as a product of linear factors:

$\frac{1}{x^2-16}=\frac{1}{(x-4)(x+4)}\equiv \frac{A}{x-4}+\frac{B}{x+4}$

It is assumed that A and B are certain constants to be evaluated. Denominators can be cleared by multiplying both sides by (x - 4)(x + 4). So we get:

1=A(x+4)+B(x-4)

First we substitute x = -4 into the produced equation:

$1=-8B\Rightarrow B=-\frac{1}{8}$

Then we substitute x = 4 into the equation:

$1=8A\Rightarrow A=\frac{1}{8}$

Thus:

$\frac{1}{(x-4)(x+4)}=\frac{1}{8}\frac{1}{x-4}-\frac{1}{8}\frac{1}{x+4}$

Hence:

$\int \frac{dx}{x^2-16}=\int(\frac{1}{8}\frac{1}{x-4}-\frac{1}{8}\frac{1}{x+4})dx$

$=\frac{1}{8}ln\left | x-4 \right |-\frac{1}{8}ln\left |x+4 \right |+c$

$=\frac{1}{8}(ln\left | x-4 \right |-ln\left |x+4 \right |)+c$

$=\frac{1}{8}ln\left | \frac{x-4}{x+4}\right |+c$

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Example Question #42 : Comparing Relative Magnitudes Of Functions And Their Rates Of Change

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

Possible Answers:

10x^2+c

x^4+5x^2+c

3x+5

2x+5

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

Correct answer:

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

Explanation:

To solve for the indefinite integral, we can use the reverse power rule. We raise the power of the exponents by one and divide by that new exponent. For this problem, that would look like:

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

Remember, when taking an integral, definite or indefinite, we always add , as there could be a constant involved.

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

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

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Example Question #43 : Comparing Relative Magnitudes Of Functions And Their Rates Of Change

What is the indefinite integral of 5x+8 ?

Possible Answers:

16x^3+c

$\frac{5}{6}x^3+4x^2+cx+b$

$\frac{5}{2}x^2+8x +c$

Correct answer:

$\frac{5}{2}x^2+8x +c$

Explanation:

To solve for the indefinite integral, we can use the reverse power rule. We raise the power of the exponents by one and divide by that new exponent. For this problem, that would look like:

$\int 5x+8{\mathrm{d} x}=\frac{5x^{1+1}}{1+1}+\frac{8x^{0+1}}{0+1}+c$

Remember, when taking an integral, definite or indefinite, we always add , as there could be a constant involved.

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

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

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Example Question #44 : Comparing Relative Magnitudes Of Functions And Their Rates Of Change

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

Possible Answers:

4x^4+x^2+5x+c

3x^2+2

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

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

Correct answer:

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

Explanation:

To solve for the indefinite integral, we can use the reverse power rule. We raise the power of the exponents by one and divide by that new exponent.

We're going to treat as 5x^0 , as anything to the zero power is one.

For this problem, that would look like:

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

Remember, when taking an integral, definite or indefinite, we always add , as there could be a constant involved.

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

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

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Example Question #71 : Asymptotic And Unbounded Behavior

What is the anti-derivative of ?

Possible Answers:

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

2+c

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

x^2+c

Correct answer:

x^2+c

Explanation:

To find the indefinite integral of our expression, 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.

First we need to realize that 2x=2x^1 . From there we can solve:

$\int2x{\mathrm{d} x}=\frac{2x^{1+1}}{1+1}+c$

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 .

$\int2x{\mathrm{d} x}=\frac{2x^{2}}{2}+c$

$\int2x{\mathrm{d} x}=x^2+c$

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Example Question #72 : Asymptotic And Unbounded Behavior

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

Possible Answers:

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

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

4x^4+2x

3x^2

6x+6+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 .

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

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

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

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Example Question #73 : Asymptotic And Unbounded Behavior

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

Possible Answers:

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

x^2

x^3+c

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

2x+3

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 #41 : Finding Integrals

What is the indefinite integral of x+18 ?

Possible Answers:

2x^2+18x+c

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

Undefined

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

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 #42 : Finding Integrals

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

Possible Answers:

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

$-\cos(x)$

$-\cos(x)+c$

$-\sin(x)+c$

$\cos(x)$

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 #43 : Finding Integrals

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

Possible Answers:

12x

2x^3+8x+c

2x^3-8x-c

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

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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High School Math : Calculus II — Integrals

Study concepts, example questions & explanations for High School Math

All High School Math Resources

Example Questions

Example Question #41 : Comparing Relative Magnitudes Of Functions And Their Rates Of Change

Example Question #42 : Comparing Relative Magnitudes Of Functions And Their Rates Of Change

Example Question #43 : Comparing Relative Magnitudes Of Functions And Their Rates Of Change

Example Question #44 : Comparing Relative Magnitudes Of Functions And Their Rates Of Change

Example Question #71 : Asymptotic And Unbounded Behavior

Example Question #72 : Asymptotic And Unbounded Behavior

Example Question #73 : Asymptotic And Unbounded Behavior

Example Question #41 : Finding Integrals

Example Question #42 : Finding Integrals

Example Question #43 : Finding Integrals

All High School Math Resources

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