Calculus Examples

f (x) = - \frac{4}{x - 2}

$f (x) = - \frac{4}{x - 2}$ ,

[0, 1]

$[0, 1]$

Step 1

To find the average value of a function, the function should be continuous on the closed interval

[a, b]

$[a, b]$ . To find whether

f (x)

$f (x)$ is continuous on

[0, 1]

$[0, 1]$ or not, find the domain of

f (x) = - \frac{4}{x - 2}

$f (x) = - \frac{4}{x - 2}$ .

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Step 1.1

Set the denominator in

\frac{4}{x - 2}

$\frac{4}{x - 2}$ equal to

0

$0$ to find where the expression is undefined.

x - 2 = 0

$x - 2 = 0$

Step 1.2

Add

2

$2$ to both sides of the equation.

x = 2

$x = 2$

Step 1.3

The domain is all values of

x

$x$ that make the expression defined.

Interval Notation:

(- \infty, 2) \cup (2, \infty)

$(- \infty, 2) \cup (2, \infty)$

Set-Builder Notation:

{x | x \neq 2}

${x | x \neq 2}$

Interval Notation:

(- \infty, 2) \cup (2, \infty)

$(- \infty, 2) \cup (2, \infty)$

Set-Builder Notation:

{x | x \neq 2}

${x | x \neq 2}$

Step 2

f (x)

$f (x)$ is continuous on

[0, 1]

$[0, 1]$ .

f (x)

$f (x)$ is continuous

Step 3

The average value of function

f

$f$ over the interval

[a, b]

$[a, b]$ is defined as

A (x) = \frac{1}{b - a} \int_{a}^{b} f (x) d x

$A (x) = \frac{1}{b - a} \int_{a}^{b} f (x) d x$ .

A (x) = \frac{1}{b - a} \int_{a}^{b} f (x) d x

$A (x) = \frac{1}{b - a} \int_{a}^{b} f (x) d x$

Step 4

Substitute the actual values into the formula for the average value of a function.

A (x) = \frac{1}{1 - 0} (\int_{0}^{1} - \frac{4}{x - 2} d x)

$A (x) = \frac{1}{1 - 0} (\int_{0}^{1} - \frac{4}{x - 2} d x)$

Step 5

Since

- 1

$- 1$ is constant with respect to

x

$x$ , move

- 1

$- 1$ out of the integral.

A (x) = \frac{1}{1 - 0} (- \int_{0}^{1} \frac{4}{x - 2} d x)

$A (x) = \frac{1}{1 - 0} (- \int_{0}^{1} \frac{4}{x - 2} d x)$

Step 6

Since

4

$4$ is constant with respect to

x

$x$ , move

4

$4$ out of the integral.

A (x) = \frac{1}{1 - 0} (- (4 \int_{0}^{1} \frac{1}{x - 2} d x))

$A (x) = \frac{1}{1 - 0} (- (4 \int_{0}^{1} \frac{1}{x - 2} d x))$

Step 7

Multiply

4

$4$ by

- 1

$- 1$ .

A (x) = \frac{1}{1 - 0} (- 4 \int_{0}^{1} \frac{1}{x - 2} d x)

$A (x) = \frac{1}{1 - 0} (- 4 \int_{0}^{1} \frac{1}{x - 2} d x)$

Step 8

Let

u = x - 2

$u = x - 2$ . Then

d u = d x

$d u = d x$ . Rewrite using

u

$u$ and

d

$d$

u

$u$ .

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Step 8.1

Let

u = x - 2

$u = x - 2$ . Find

\frac{d u}{d x}

$\frac{d u}{d x}$ .

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Step 8.1.1

Differentiate

x - 2

$x - 2$ .

\frac{d}{d x} [x - 2]

$\frac{d}{d x} [x - 2]$

Step 8.1.2

By the Sum Rule, the derivative of

x - 2

$x - 2$ with respect to

x

$x$ is

\frac{d}{d x} [x] + \frac{d}{d x} [- 2]

$\frac{d}{d x} [x] + \frac{d}{d x} [- 2]$ .

\frac{d}{d x} [x] + \frac{d}{d x} [- 2]

$\frac{d}{d x} [x] + \frac{d}{d x} [- 2]$

Step 8.1.3

Differentiate using the Power Rule which states that

\frac{d}{d x} [x^{n}]

$\frac{d}{d x} [x^{n}]$ is

n x^{n - 1}

$n x^{n - 1}$ where

n = 1

$n = 1$ .

1 + \frac{d}{d x} [- 2]

$1 + \frac{d}{d x} [- 2]$

Step 8.1.4

Since

- 2

$- 2$ is constant with respect to

x

$x$ , the derivative of

- 2

$- 2$ with respect to

x

$x$ is

0

$0$ .

1 + 0

$1 + 0$

Step 8.1.5

Add

1

$1$ and

0

$0$ .

1

$1$

1

$1$

Step 8.2

Substitute the lower limit in for

x

$x$ in

u = x - 2

$u = x - 2$ .

u_{lower} = 0 - 2

$u_{lower} = 0 - 2$

Step 8.3

Subtract

2

$2$ from

0

$0$ .

u_{lower} = - 2

$u_{lower} = - 2$

Step 8.4

Substitute the upper limit in for

x

$x$ in

u = x - 2

$u = x - 2$ .

u_{upper} = 1 - 2

$u_{upper} = 1 - 2$

Step 8.5

Subtract

2

$2$ from

1

$1$ .

u_{upper} = - 1

$u_{upper} = - 1$

Step 8.6

The values found for

u_{lower}

$u_{lower}$ and

u_{upper}

$u_{upper}$ will be used to evaluate the definite integral.

u_{lower} = - 2

$u_{lower} = - 2$

u_{upper} = - 1

$u_{upper} = - 1$

Step 8.7

Rewrite the problem using

u

$u$ ,

d u

$d u$ , and the new limits of integration.

A (x) = \frac{1}{1 - 0} (- 4 \int_{- 2}^{- 1} \frac{1}{u} d u)

$A (x) = \frac{1}{1 - 0} (- 4 \int_{- 2}^{- 1} \frac{1}{u} d u)$

Step 9

The integral of

$\frac{1}{u}$ with respect to

$u$ is

$\ln (| u |)$ .

$A (x) = \frac{1}{1 - 0} (- 4 (\ln (| u |)]_{- 2}^{- 1}))$

Step 10

Evaluate

$\ln (| u |)$ at

$- 1$ and at

$- 2$ .

$A (x) = \frac{1}{1 - 0} (- 4 (\ln (| - 1 |) - \ln (| - 2 |)))$

Step 11

Use the quotient property of logarithms,

$\log_{b} (x) - \log_{b} (y) = \log_{b} (\frac{x}{y})$ .

$A (x) = \frac{1}{1 - 0} (- 4 \ln (\frac{| - 1 |}{| - 2 |}))$

Step 12

Simplify.

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Step 12.1

The absolute value is the distance between a number and zero. The distance between

$- 1$ and

$0$ is

$1$ .

$A (x) = \frac{1}{1 - 0} (- 4 \ln (\frac{1}{| - 2 |}))$

Step 12.2

The absolute value is the distance between a number and zero. The distance between

$- 2$ and

$0$ is

$2$ .

$A (x) = \frac{1}{1 - 0} (- 4 \ln (\frac{1}{2}))$

Step 13

Simplify the denominator.

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Step 13.1

Multiply

$- 1$ by

$0$ .

$A (x) = \frac{1}{1 + 0} \cdot (- 4 \ln (\frac{1}{2}))$

Step 13.2

Add

$1$ and

$0$ .

$A (x) = \frac{1}{1} \cdot (- 4 \ln (\frac{1}{2}))$

Step 14

Reduce the expression by cancelling the common factors.

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Step 14.1

Cancel the common factor of

$1$ .

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Step 14.1.1

Cancel the common factor.

$A (x) = \frac{1}{1} \cdot (- 4 \ln (\frac{1}{2}))$

Step 14.1.2

Rewrite the expression.

$A (x) = 1 \cdot (- 4 \ln (\frac{1}{2}))$

Step 14.2

Multiply

$- 4 \ln (\frac{1}{2})$ by

$1$ .

$A (x) = - 4 \ln (\frac{1}{2})$

Step 15

Simplify

$- 4 \ln (\frac{1}{2})$ by moving

$4$ inside the logarithm.

$A (x) = - \ln ({(\frac{1}{2})}^{4})$

Step 16

Apply the product rule to

$\frac{1}{2}$ .

$A (x) = - \ln (\frac{1^{4}}{2^{4}})$

Step 17

One to any power is one.

$A (x) = - \ln (\frac{1}{2^{4}})$

Step 18

Raise

$2$ to the power of

$4$ .

$A (x) = - \ln (\frac{1}{16})$

Step 19