Calculus Examples

$f (x) = - \frac{4}{x - 2}$ , $[0, 1]$

Step 1

To find the average value of a function, the function should be continuous on the closed interval $[a, b]$ . To find whether $f (x)$ is continuous on $[0, 1]$ or not, find the domain of $f (x) = - \frac{4}{x - 2}$ .

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Step 1.1

Set the denominator in $\frac{4}{x - 2}$ equal to $0$ to find where the expression is undefined.

$x - 2 = 0$

Step 1.2

Add $2$ to both sides of the equation.

$x = 2$

Step 1.3

The domain is all values of $x$ that make the expression defined.

Interval Notation:

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

Set-Builder Notation:

${x | x \neq 2}$

Interval Notation:

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

Set-Builder Notation:

${x | x \neq 2}$

Step 2

$f (x)$ is continuous on $[0, 1]$ .

$f (x)$ is continuous

Step 3

The average value of function $f$ over the interval $[a, b]$ is defined as $A (x) = \frac{1}{b - a} \int_{a}^{b} f (x) d x$ .

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

Step 4

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

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

Step 5

Since $- 1$ is constant with respect to $x$ , move $- 1$ out of the integral.

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

Step 6

Since $4$ is constant with respect to $x$ , move $4$ out of the integral.

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

Step 7

Multiply $4$ by $- 1$ .

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

Step 8

Let $u = x - 2$ . Then $d u = d x$ . Rewrite using $u$ and $d$ $u$ .

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Step 8.1

Let $u = x - 2$ . Find $\frac{d u}{d x}$ .

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Step 8.1.1

Differentiate $x - 2$ .

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

Step 8.1.2

By the Sum Rule, the derivative of $x - 2$ with respect to $x$ is $\frac{d}{d x} [x] + \frac{d}{d x} [- 2]$ .

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

Step 8.1.3

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 1$ .

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

Step 8.1.4

Since $- 2$ is constant with respect to $x$ , the derivative of $- 2$ with respect to $x$ is $0$ .

$1 + 0$

Step 8.1.5

Add $1$ and $0$ .

$1$

Step 8.2

Substitute the lower limit in for $x$ in $u = x - 2$ .

$u_{lower} = 0 - 2$

Step 8.3

Subtract $2$ from $0$ .

$u_{lower} = - 2$

Step 8.4

Substitute the upper limit in for $x$ in $u = x - 2$ .

$u_{upper} = 1 - 2$

Step 8.5

Subtract $2$ from $1$ .

$u_{upper} = - 1$

Step 8.6

The values found for $u_{lower}$ and $u_{upper}$ will be used to evaluate the definite integral.

$u_{lower} = - 2$

$u_{upper} = - 1$

Step 8.7

Rewrite the problem using $u$ , $d u$ , and the new limits of integration.

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

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