Calculus Examples

$f (x) = \frac{10}{x + 1}$ , $[0, 9]$

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 $[0, 9]$ or not, find the domain of $f (x) = \frac{10}{x + 1}$ .

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Step 1.1

Set the denominator in $\frac{10}{x + 1}$ equal to $0$ to find where the expression is undefined.

$x + 1 = 0$

Step 1.2

Subtract $1$ from 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 $[0, 9]$ .

$f (x)$ 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.

$A (x) = \frac{1}{9 - 0} (\int_{0}^{9} \frac{10}{x + 1} d x)$

Step 5

Since $10$ is constant with respect to $x$ , move $10$ out of the integral.

$A (x) = \frac{1}{9 - 0} (10 \int_{0}^{9} \frac{1}{x + 1} d x)$

Step 6

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

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Step 6.1

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

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Step 6.1.1

Differentiate $x + 1$ .

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

Step 6.1.2

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 6.1.3

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 6.1.4

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

$1 + 0$

Step 6.1.5

Add $1$ and $0$ .

$1$

Step 6.2

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

$u_{lower} = 0 + 1$

Step 6.3

Add $0$ and $1$ .

$u_{lower} = 1$

Step 6.4

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

$u_{upper} = 9 + 1$

Step 6.5

Add $9$ and $1$ .

$u_{upper} = 10$

Step 6.6

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

$u_{lower} = 1$

$u_{upper} = 10$

Step 6.7

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

$A (x) = \frac{1}{9 - 0} (10 \int_{1}^{10} \frac{1}{u} d u)$

Step 7

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

$A (x) = \frac{1}{9 - 0} (10 (\ln (| u |)]_{1}^{10}))$

Step 8

Evaluate $\ln (| u |)$ at $10$ and at $1$ .

$A (x) = \frac{1}{9 - 0} (10 (\ln (| 10 |) - \ln (| 1 |)))$

Step 9

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

$A (x) = \frac{1}{9 - 0} (10 \ln (\frac{| 10 |}{| 1 |}))$

Step 10

Simplify.

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Step 10.1

The absolute value is the distance between a number and zero. The distance between $0$ and $10$ is $10$ .

$A (x) = \frac{1}{9 - 0} (10 \ln (\frac{10}{| 1 |}))$

Step 10.2

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

$A (x) = \frac{1}{9 - 0} (10 \ln (\frac{10}{1}))$

Step 10.3

Divide $10$ by $1$ .

$A (x) = \frac{1}{9 - 0} (10 \ln (10))$

Step 11

Simplify the denominator.

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Step 11.1

Multiply $- 1$ by $0$ .

$A (x) = \frac{1}{9 + 0} \cdot (10 \ln (10))$

Step 11.2

Add $9$ and $0$ .

$A (x) = \frac{1}{9} \cdot (10 \ln (10))$

Step 12

Simplify $10 \ln (10)$ by moving $10$ inside the logarithm.

$A (x) = \frac{1}{9} \cdot \ln (10^{10})$

Step 13

Raise $10$ to the power of $10$ .

$A (x) = \frac{1}{9} \cdot \ln (10000000000)$

Step 14

Simplify $\frac{1}{9} \ln (10000000000)$ by moving $\frac{1}{9}$ inside the logarithm.

$A (x) = \ln (10000000000^{\frac{1}{9}})$

Step 15