Calculus Examples

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

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 $[2, 4]$ or not, find the domain of $f (x) = \frac{1}{x - 1}$ .

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Step 1.1

Set the denominator in $\frac{1}{x - 1}$ equal to $0$ to find where the expression is undefined.

$x - 1 = 0$

Step 1.2

Add $1$ to both sides of the equation.

$x = 1$

Step 1.3

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

Interval Notation:

$(- \infty, 1) \cup (1, \infty)$

Set-Builder Notation:

${x | x \neq 1}$

Interval Notation:

$(- \infty, 1) \cup (1, \infty)$

Set-Builder Notation:

${x | x \neq 1}$

Step 2

$f (x)$ is continuous on $[2, 4]$ .

$f (x)$ 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.

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

Step 5

Let $u = x - 1$ . Then $d u = d x$ . Rewrite using $u$ and $d$ $u$ .

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Step 5.1

Let $u = x - 1$ . Find $\frac{d u}{d x}$ .

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Step 5.1.1

Differentiate $x - 1$ .

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

Step 5.1.2

By the Sum Rule, the derivative of $x - 1$ with respect to $x$ is $\frac{d}{d x} [x] + \frac{d}{d x} [- 1]$ .

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

Step 5.1.3

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

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

Step 5.1.4

Since $- 1$ is constant with respect to $x$ , the derivative of $- 1$ with respect to $x$ is $0$ .

$1 + 0$

Step 5.1.5

Add $1$ and $0$ .

$1$

Step 5.2

Substitute the lower limit in for $x$ in $u = x - 1$ .

$u_{lower} = 2 - 1$

Step 5.3

Subtract $1$ from $2$ .

$u_{lower} = 1$

Step 5.4

Substitute the upper limit in for $x$ in $u = x - 1$ .

$u_{upper} = 4 - 1$

Step 5.5

Subtract $1$ from $4$ .

$u_{upper} = 3$

Step 5.6

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

$u_{lower} = 1$

$u_{upper} = 3$

Step 5.7

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

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

Step 6

The integral of $\frac{1}{u}$ with respect to $u$ is $\ln (| u |)$ .

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

Step 7

Evaluate $\ln (| u |)$ at $3$ and at $1$ .

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

Step 8

Use the quotient property of logarithms, $\log_{b} (x) - \log_{b} (y) = \log_{b} (\frac{x}{y})$ .

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

Step 9

Simplify.

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Step 9.1

The absolute value is the distance between a number and zero. The distance between $0$ and $3$ is $3$ .

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

Step 9.2

The absolute value is the distance between a number and zero. The distance between $0$ and $1$ is $1$ .

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

Step 9.3

Divide $3$ by $1$ .

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

Step 10

Subtract $2$ from $4$ .

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

Step 11

Simplify $\frac{1}{2} \ln (3)$ by moving $\frac{1}{2}$ inside the logarithm.

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

Step 12