Calculus Examples

y = \frac{3}{x - 2}

$y = \frac{3}{x - 2}$ ,

[4, 7]

$[4, 7]$

Step 1

Write

y = \frac{3}{x - 2}

$y = \frac{3}{x - 2}$ as a function.

f (x) = \frac{3}{x - 2}

$f (x) = \frac{3}{x - 2}$

Step 2

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

[4, 7]

$[4, 7]$ or not, find the domain of

f (x) = \frac{3}{x - 2}

$f (x) = \frac{3}{x - 2}$ .

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Step 2.1

Set the denominator in

\frac{3}{x - 2}

$\frac{3}{x - 2}$ equal to

0

$0$ to find where the expression is undefined.

x - 2 = 0

$x - 2 = 0$

Step 2.2

Add

2

$2$ to both sides of the equation.

x = 2

$x = 2$

Step 2.3

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 3

f (x)

$f (x)$ is continuous on

[4, 7]

$[4, 7]$ .

f (x)

$f (x)$ is continuous

Step 4

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 5

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

A (x) = \frac{1}{7 - 4} (\int_{4}^{7} \frac{3}{x - 2} d x)

$A (x) = \frac{1}{7 - 4} (\int_{4}^{7} \frac{3}{x - 2} d x)$

Step 6

Since

3

$3$ is constant with respect to

x

$x$ , move

3

$3$ out of the integral.

A (x) = \frac{1}{7 - 4} (3 \int_{4}^{7} \frac{1}{x - 2} d x)

$A (x) = \frac{1}{7 - 4} (3 \int_{4}^{7} \frac{1}{x - 2} d x)$

Step 7

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 7.1

Let

u = x - 2

$u = x - 2$ . Find

\frac{d u}{d x}

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

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Step 7.1.1

Differentiate

x - 2

$x - 2$ .

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

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

Step 7.1.2

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 7.1.3

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 7.1.4

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 7.1.5

Add

1

$1$ and

0

$0$ .

1

$1$

1

$1$

Step 7.2

Substitute the lower limit in for

x

$x$ in

u = x - 2

$u = x - 2$ .

u_{lower} = 4 - 2

$u_{lower} = 4 - 2$

Step 7.3

Subtract

2

$2$ from

4

$4$ .

u_{lower} = 2

$u_{lower} = 2$

Step 7.4

Substitute the upper limit in for

x

$x$ in

u = x - 2

$u = x - 2$ .

u_{upper} = 7 - 2

$u_{upper} = 7 - 2$

Step 7.5

Subtract

2

$2$ from

7

$7$ .

u_{upper} = 5

$u_{upper} = 5$

Step 7.6

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} = 5

$u_{upper} = 5$

Step 7.7

Rewrite the problem using

u

$u$ ,

d u

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

A (x) = \frac{1}{7 - 4} (3 \int_{2}^{5} \frac{1}{u} d u)

$A (x) = \frac{1}{7 - 4} (3 \int_{2}^{5} \frac{1}{u} d u)$

A (x) = \frac{1}{7 - 4} (3 \int_{2}^{5} \frac{1}{u} d u)

$A (x) = \frac{1}{7 - 4} (3 \int_{2}^{5} \frac{1}{u} d u)$

Step 8

The integral of

\frac{1}{u}

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

u

$u$ is

ln (| u |)

$\ln (| u |)$ .

A (x) = \frac{1}{7 - 4} (3 (ln (| u |)]_{2}^{5}))

$A (x) = \frac{1}{7 - 4} (3 (\ln (| u |)]_{2}^{5}))$

Step 9

Evaluate

ln (| u |)

$\ln (| u |)$ at

5

$5$ and at

2

$2$ .

A (x) = \frac{1}{7 - 4} (3 (ln (| 5 |) - ln (| 2 |)))

$A (x) = \frac{1}{7 - 4} (3 (\ln (| 5 |) - \ln (| 2 |)))$

Step 10

Use the quotient property of logarithms,

{log}_{b} (x) - {log}_{b} (y) = {log}_{b} (\frac{x}{y})

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

A (x) = \frac{1}{7 - 4} (3 ln (\frac{| 5 |}{| 2 |}))

$A (x) = \frac{1}{7 - 4} (3 \ln (\frac{| 5 |}{| 2 |}))$

Step 11

Simplify.

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Step 11.1

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

0

$0$ and

5

$5$ is

5

$5$ .

A (x) = \frac{1}{7 - 4} (3 ln (\frac{5}{| 2 |}))

$A (x) = \frac{1}{7 - 4} (3 \ln (\frac{5}{| 2 |}))$

Step 11.2

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

0

$0$ and

2

$2$ is

2

$2$ .

A (x) = \frac{1}{7 - 4} (3 ln (\frac{5}{2}))

$A (x) = \frac{1}{7 - 4} (3 \ln (\frac{5}{2}))$

A (x) = \frac{1}{7 - 4} (3 ln (\frac{5}{2}))

$A (x) = \frac{1}{7 - 4} (3 \ln (\frac{5}{2}))$

Step 12

Subtract

4

$4$ from

7

$7$ .

A (x) = \frac{1}{3} \cdot (3 ln (\frac{5}{2}))

$A (x) = \frac{1}{3} \cdot (3 \ln (\frac{5}{2}))$

Step 13

Cancel the common factor of

3

$3$ .

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Step 13.1

Factor

3

$3$ out of

3 ln (\frac{5}{2})

$3 \ln (\frac{5}{2})$ .

A (x) = \frac{1}{3} \cdot (3 (ln (\frac{5}{2})))

$A (x) = \frac{1}{3} \cdot (3 (\ln (\frac{5}{2})))$

Step 13.2

Cancel the common factor.

$A (x) = \frac{1}{3} \cdot (3 \ln (\frac{5}{2}))$

Step 13.3

Rewrite the expression.

$A (x) = \ln (\frac{5}{2})$

Step 14