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High School Math : Finding Indefinite Integrals

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Example Questions

Example Question #1 : Finding Indefinite Integrals

$\int \frac{3}{x}=$

Possible Answers:

$3x^{-2}+C$

$-3e^{-x}$

$-3x^{-2}+C$

$3\ln \left | x \right |+C$

$3x^{-1}+C$

Correct answer:

$3\ln \left | x \right |+C$

Explanation:

The integral of $\frac{1}{x}$ is $\ln \left | x \right |+C$ . The constant 3 is simply multiplied by the integral.

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Example Question #2 : Finding Indefinite Integrals

$\int cos(x)\sin(x){\mathrm{d} x}=?$

Possible Answers:

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

$\sin(x)\cos(x)+c$

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

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

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

Correct answer:

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

Explanation:

To integrate $\cos(x)\sin(x)$ , we need to get the two equations in terms of each other. We are going to use "u-substitution" to create a new variable, , which will equal $\cos(x)$ .

Now, if $u=\cos(x)$ , then

$\frac{\mathrm{d}u }{\mathrm{d} x}=\frac{\mathrm{d} }{\mathrm{d} x}\cos(x)=-sin(x)$

Multiply both sides by $\mathrm d {x}$ to get the more familiar:

${\mathrm{d} u}=-\sin(x)\mathrm{d}x$

Note that our $\mathrm d{u}=-\sin(x)$ , and our original equation was asking for a positive $\sin(x)$ .

That means if we want $\int\cos(x)\sin(x)$ in terms of , it looks like this:

$\int u\cdot (-\mathrm d{u}})$

Bring the negative sign to the outside:

$-\int u\mathrm d{u}$ .

We can use the power rule to find the integral of :

$-(\frac{1}{2}u^2+c)$

Since we said that $u=\cos(x)$ , we can plug that back into the equation to get our answer:

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

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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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High School Math : Finding Indefinite Integrals

Study concepts, example questions & explanations for High School Math

All High School Math Resources

Example Questions

Example Question #1 : Finding Indefinite Integrals

Example Question #2 : Finding Indefinite Integrals

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

All High School Math Resources

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