Calculus Examples

$y = \frac{3}{x - 2}$ , $[4, 7]$

Step 1

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

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

Step 2

To find the average value of a function, the function should be continuous on the closed interval $[a, b]$ . To find whether $f (x)$ is continuous on $[4, 7]$ or not, find the domain of $f (x) = \frac{3}{x - 2}$ .

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Step 2.1

Set the denominator in $\frac{3}{x - 2}$ equal to $0$ to find where the expression is undefined.

$x - 2 = 0$

Step 2.2

Add $2$ to both sides of the equation.

$x = 2$

Step 2.3

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 3

$f (x)$ is continuous on $[4, 7]$ .

$f (x)$ is continuous

Step 4

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 5

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

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

Step 6

Since $3$ is constant with respect to $x$ , move $3$ out of the integral.

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

Step 7

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

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Step 7.1

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

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Step 7.1.1

Differentiate $x - 2$ .

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

Step 7.1.2

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 7.1.3

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 7.1.4

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

$1 + 0$

Step 7.1.5

Add $1$ and $0$ .

$1$

Step 7.2

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

$u_{lower} = 4 - 2$

Step 7.3

Subtract $2$ from $4$ .

$u_{lower} = 2$

Step 7.4

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

$u_{upper} = 7 - 2$

Step 7.5

Subtract $2$ from $7$ .

$u_{upper} = 5$

Step 7.6

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

$u_{lower} = 2$

$u_{upper} = 5$

Step 7.7

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

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

Step 8

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

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

Step 9

Evaluate $\ln (| u |)$ at $5$ and at $2$ .

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

Step 10

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

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

Step 11

Simplify.

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Step 11.1

The absolute value is the distance between a number and zero. The distance between $0$ and $5$ is $5$ .

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

Step 11.2

The absolute value is the distance between a number and zero. The distance between $0$ and $2$ is $2$ .

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

Step 12

Subtract $4$ from $7$ .

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

Step 13

Cancel the common factor of $3$ .

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Step 13.1

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

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

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