Calculus Examples

$f (x) = x + 12 x^{- 1}$ , $[- 4, - 3]$

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 $[- 4, - 3]$ or not, find the domain of $f (x) = x + 12 x^{- 1}$ .

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Step 1.1

Set the base in $x^{- 1}$ equal to $0$ to find where the expression is undefined.

$x = 0$

Step 1.2

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

Interval Notation:

$(- \infty, 0) \cup (0, \infty)$

Set-Builder Notation:

${x | x \neq 0}$

Interval Notation:

$(- \infty, 0) \cup (0, \infty)$

Set-Builder Notation:

${x | x \neq 0}$

Step 2

$f (x)$ is continuous on $[- 4, - 3]$ .

$f (x)$ 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.

$A (x) = \frac{1}{- 3 + 4} (\int_{- 4}^{- 3} x + 12 x^{- 1} d x)$

Step 5

Split the single integral into multiple integrals.

$A (x) = \frac{1}{- 3 + 4} (\int_{- 4}^{- 3} x d x + \int_{- 4}^{- 3} 12 x^{- 1} d x)$

Step 6

By the Power Rule, the integral of $x$ with respect to $x$ is $\frac{1}{2} x^{2}$ .

$A (x) = \frac{1}{- 3 + 4} (\frac{1}{2} x^{2}]_{- 4}^{- 3} + \int_{- 4}^{- 3} 12 x^{- 1} d x)$

Step 7

Since $12$ is constant with respect to $x$ , move $12$ out of the integral.

$A (x) = \frac{1}{- 3 + 4} (\frac{1}{2} x^{2}]_{- 4}^{- 3} + 12 \int_{- 4}^{- 3} x^{- 1} d x)$

Step 8

The integral of $x^{- 1}$ with respect to $x$ is $\ln (| x |)$ .

$A (x) = \frac{1}{- 3 + 4} (\frac{1}{2} x^{2}]_{- 4}^{- 3} + 12 (\ln (| x |)]_{- 4}^{- 3}))$

Step 9

Simplify the answer.

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Step 9.1

Substitute and simplify.

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Step 9.1.1

Evaluate $\frac{1}{2} x^{2}$ at $- 3$ and at $- 4$ .

$A (x) = \frac{1}{- 3 + 4} ((\frac{1}{2} \cdot {(- 3)}^{2}) - \frac{1}{2} \cdot {(- 4)}^{2} + 12 (\ln (| x |)]_{- 4}^{- 3}))$

Step 9.1.2

Evaluate $\ln (| x |)$ at $- 3$ and at $- 4$ .

$A (x) = \frac{1}{- 3 + 4} (\frac{1}{2} \cdot {(- 3)}^{2} - \frac{1}{2} \cdot {(- 4)}^{2} + 12 (\ln (| - 3 |) - \ln (| - 4 |)))$

Step 9.1.3

Simplify.

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Step 9.1.3.1

Raise $- 3$ to the power of $2$ .

$A (x) = \frac{1}{- 3 + 4} (\frac{1}{2} \cdot 9 - \frac{1}{2} \cdot {(- 4)}^{2} + 12 (\ln (| - 3 |) - \ln (| - 4 |)))$

Step 9.1.3.2

Combine $\frac{1}{2}$ and $9$ .

$A (x) = \frac{1}{- 3 + 4} (\frac{9}{2} - \frac{1}{2} \cdot {(- 4)}^{2} + 12 (\ln (| - 3 |) - \ln (| - 4 |)))$

Step 9.1.3.3

Raise $- 4$ to the power of $2$ .

$A (x) = \frac{1}{- 3 + 4} (\frac{9}{2} - \frac{1}{2} \cdot 16 + 12 (\ln (| - 3 |) - \ln (| - 4 |)))$

Step 9.1.3.4

Multiply $16$ by $- 1$ .

$A (x) = \frac{1}{- 3 + 4} (\frac{9}{2} - 16 (\frac{1}{2}) + 12 (\ln (| - 3 |) - \ln (| - 4 |)))$

Step 9.1.3.5

Combine $- 16$ and $\frac{1}{2}$ .

$A (x) = \frac{1}{- 3 + 4} (\frac{9}{2} + \frac{- 16}{2} + 12 (\ln (| - 3 |) - \ln (| - 4 |)))$

Step 9.1.3.6

Cancel the common factor of $- 16$ and $2$ .

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Step 9.1.3.6.1

Factor $2$ out of $- 16$ .

$A (x) = \frac{1}{- 3 + 4} (\frac{9}{2} + \frac{2 \cdot - 8}{2} + 12 (\ln (| - 3 |) - \ln (| - 4 |)))$

Step 9.1.3.6.2

Cancel the common factors.

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Step 9.1.3.6.2.1

Factor $2$ out of $2$ .

$A (x) = \frac{1}{- 3 + 4} (\frac{9}{2} + \frac{2 \cdot - 8}{2 (1)} + 12 (\ln (| - 3 |) - \ln (| - 4 |)))$

Step 9.1.3.6.2.2

Cancel the common factor.

$A (x) = \frac{1}{- 3 + 4} (\frac{9}{2} + \frac{2 \cdot - 8}{2 \cdot 1} + 12 (\ln (| - 3 |) - \ln (| - 4 |)))$

Step 9.1.3.6.2.3

Rewrite the expression.

$A (x) = \frac{1}{- 3 + 4} (\frac{9}{2} + \frac{- 8}{1} + 12 (\ln (| - 3 |) - \ln (| - 4 |)))$

Step 9.1.3.6.2.4

Divide $- 8$ by $1$ .

$A (x) = \frac{1}{- 3 + 4} (\frac{9}{2} - 8 + 12 (\ln (| - 3 |) - \ln (| - 4 |)))$

Step 9.1.3.7

To write $- 8$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$A (x) = \frac{1}{- 3 + 4} (\frac{9}{2} - 8 \cdot \frac{2}{2} + 12 (\ln (| - 3 |) - \ln (| - 4 |)))$

Step 9.1.3.8

Combine $- 8$ and $\frac{2}{2}$ .

$A (x) = \frac{1}{- 3 + 4} (\frac{9}{2} + \frac{- 8 \cdot 2}{2} + 12 (\ln (| - 3 |) - \ln (| - 4 |)))$

Step 9.1.3.9

Combine the numerators over the common denominator.

$A (x) = \frac{1}{- 3 + 4} (\frac{9 - 8 \cdot 2}{2} + 12 (\ln (| - 3 |) - \ln (| - 4 |)))$

Step 9.1.3.10

Simplify the numerator.

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Step 9.1.3.10.1

Multiply $- 8$ by $2$ .

$A (x) = \frac{1}{- 3 + 4} (\frac{9 - 16}{2} + 12 (\ln (| - 3 |) - \ln (| - 4 |)))$

Step 9.1.3.10.2

Subtract $16$ from $9$ .

$A (x) = \frac{1}{- 3 + 4} (\frac{- 7}{2} + 12 (\ln (| - 3 |) - \ln (| - 4 |)))$

Step 9.1.3.11

Move the negative in front of the fraction.

$A (x) = \frac{1}{- 3 + 4} (- \frac{7}{2} + 12 (\ln (| - 3 |) - \ln (| - 4 |)))$

Step 9.1.3.12

To write $12 (\ln (| - 3 |) - \ln (| - 4 |))$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$A (x) = \frac{1}{- 3 + 4} (- \frac{7}{2} + 12 (\ln (| - 3 |) - \ln (| - 4 |)) \cdot \frac{2}{2})$

Step 9.1.3.13

Combine $12 (\ln (| - 3 |) - \ln (| - 4 |))$ and $\frac{2}{2}$ .

$A (x) = \frac{1}{- 3 + 4} (- \frac{7}{2} + \frac{12 (\ln (| - 3 |) - \ln (| - 4 |)) \cdot 2}{2})$

Step 9.1.3.14

Combine the numerators over the common denominator.

$A (x) = \frac{1}{- 3 + 4} (\frac{- 7 + 12 (\ln (| - 3 |) - \ln (| - 4 |)) \cdot 2}{2})$

Step 9.1.3.15

Multiply $2$ by $12$ .

$A (x) = \frac{1}{- 3 + 4} (\frac{- 7 + 24 (\ln (| - 3 |) - \ln (| - 4 |))}{2})$

Step 9.2

Simplify.

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Step 9.2.1

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

$A (x) = \frac{1}{- 3 + 4} (\frac{- 7 + 24 \ln (\frac{| - 3 |}{| - 4 |})}{2})$

Step 9.2.2

Rewrite $- 7$ as $- 1 (7)$ .

$A (x) = \frac{1}{- 3 + 4} (\frac{- 1 \cdot 7 + 24 \ln (\frac{| - 3 |}{| - 4 |})}{2})$

Step 9.2.3

Factor $- 1$ out of $24 \ln (\frac{| - 3 |}{| - 4 |})$ .

$A (x) = \frac{1}{- 3 + 4} (\frac{- 1 \cdot 7 - (- 24 \ln (\frac{| - 3 |}{| - 4 |}))}{2})$

Step 9.2.4

Factor $- 1$ out of $- 1 (7) - (- 24 \ln (\frac{| - 3 |}{| - 4 |}))$ .

$A (x) = \frac{1}{- 3 + 4} (\frac{- 1 (7 - 24 \ln (\frac{| - 3 |}{| - 4 |}))}{2})$

Step 9.2.5

Move the negative in front of the fraction.

$A (x) = \frac{1}{- 3 + 4} (- \frac{7 - 24 \ln (\frac{| - 3 |}{| - 4 |})}{2})$

Step 9.3

Simplify.

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Step 9.3.1

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

$A (x) = \frac{1}{- 3 + 4} (- \frac{7 - 24 \ln (\frac{3}{| - 4 |})}{2})$

Step 9.3.2

The absolute value is the distance between a number and zero. The distance between $- 4$ and $0$ is $4$ .

$A (x) = \frac{1}{- 3 + 4} (- \frac{7 - 24 \ln (\frac{3}{4})}{2})$

Step 10

Reduce the expression by cancelling the common factors.

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Step 10.1

Add $- 3$ and $4$ .

$A (x) = \frac{1}{1} \cdot (- \frac{7 - 24 \ln (\frac{3}{4})}{2})$

Step 10.2

Cancel the common factor of $1$ .

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Step 10.2.1

Cancel the common factor.

$A (x) = \frac{1}{1} \cdot (- \frac{7 - 24 \ln (\frac{3}{4})}{2})$

Step 10.2.2

Rewrite the expression.

$A (x) = 1 \cdot (- \frac{7 - 24 \ln (\frac{3}{4})}{2})$

Step 10.3

Multiply $- \frac{7 - 24 \ln (\frac{3}{4})}{2}$ by $1$ .

$A (x) = - \frac{7 - 24 \ln (\frac{3}{4})}{2}$

Step 11

Simplify the numerator.

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Step 11.1

Simplify $- 24 \ln (\frac{3}{4})$ by moving $24$ inside the logarithm.

$A (x) = - \frac{7 - \ln ({(\frac{3}{4})}^{24})}{2}$

Step 11.2

Apply the product rule to $\frac{3}{4}$ .

$A (x) = - \frac{7 - \ln (\frac{3^{24}}{4^{24}})}{2}$

Step 11.3

Raise $3$ to the power of $24$ .

$A (x) = - \frac{7 - \ln (\frac{282429536481}{4^{24}})}{2}$

Step 11.4

Raise $4$ to the power of $24$ .

$A (x) = - \frac{7 - \ln (\frac{282429536481}{281474976710656})}{2}$

Step 12