Calculus Examples

$f (x) = x + \frac{96}{x}$ ,

$[6, 16]$

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

$[6, 16]$ or not, find the domain of

$f (x) = x + \frac{96}{x}$ .

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Step 1.1

Set the denominator in

$\frac{96}{x}$ 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

$[6, 16]$ .

$f (x)$ 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.

$A (x) = \frac{1}{16 - 6} (\int_{6}^{16} x + \frac{96}{x} d x)$

Step 5

Split the single integral into multiple integrals.

$A (x) = \frac{1}{16 - 6} (\int_{6}^{16} x d x + \int_{6}^{16} \frac{96}{x} d x)$

Step 6

By the Power Rule, the integral of

$x$ with respect to

$x$ is

$\frac{1}{2} x^{2}$ .

$A (x) = \frac{1}{16 - 6} (\frac{1}{2} x^{2}]_{6}^{16} + \int_{6}^{16} \frac{96}{x} d x)$

Step 7

Since

$96$ is constant with respect to

$x$ , move

$96$ out of the integral.

$A (x) = \frac{1}{16 - 6} (\frac{1}{2} x^{2}]_{6}^{16} + 96 \int_{6}^{16} \frac{1}{x} d x)$

Step 8

The integral of

$\frac{1}{x}$ with respect to

$x$ is

$\ln (| x |)$ .

$A (x) = \frac{1}{16 - 6} (\frac{1}{2} x^{2}]_{6}^{16} + 96 (\ln (| x |)]_{6}^{16}))$

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

$16$ and at

$6$ .

$A (x) = \frac{1}{16 - 6} ((\frac{1}{2} \cdot 16^{2}) - \frac{1}{2} \cdot 6^{2} + 96 (\ln (| x |)]_{6}^{16}))$

Step 9.1.2

Evaluate

$\ln (| x |)$ at

$16$ and at

$6$ .

$A (x) = \frac{1}{16 - 6} (\frac{1}{2} \cdot 16^{2} - \frac{1}{2} \cdot 6^{2} + 96 (\ln (| 16 |) - \ln (| 6 |)))$

Step 9.1.3

Simplify.

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Step 9.1.3.1

Raise

$16$ to the power of

$2$ .

$A (x) = \frac{1}{16 - 6} (\frac{1}{2} \cdot 256 - \frac{1}{2} \cdot 6^{2} + 96 (\ln (| 16 |) - \ln (| 6 |)))$

Step 9.1.3.2

Combine

$\frac{1}{2}$ and

$256$ .

$A (x) = \frac{1}{16 - 6} (\frac{256}{2} - \frac{1}{2} \cdot 6^{2} + 96 (\ln (| 16 |) - \ln (| 6 |)))$

Step 9.1.3.3

Cancel the common factor of

$256$ and

$2$ .

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Step 9.1.3.3.1

Factor

$2$ out of

$256$ .

$A (x) = \frac{1}{16 - 6} (\frac{2 \cdot 128}{2} - \frac{1}{2} \cdot 6^{2} + 96 (\ln (| 16 |) - \ln (| 6 |)))$

Step 9.1.3.3.2

Cancel the common factors.

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Step 9.1.3.3.2.1

Factor

$2$ out of

$2$ .

$A (x) = \frac{1}{16 - 6} (\frac{2 \cdot 128}{2 (1)} - \frac{1}{2} \cdot 6^{2} + 96 (\ln (| 16 |) - \ln (| 6 |)))$

Step 9.1.3.3.2.2

Cancel the common factor.

$A (x) = \frac{1}{16 - 6} (\frac{2 \cdot 128}{2 \cdot 1} - \frac{1}{2} \cdot 6^{2} + 96 (\ln (| 16 |) - \ln (| 6 |)))$

Step 9.1.3.3.2.3

Rewrite the expression.

$A (x) = \frac{1}{16 - 6} (\frac{128}{1} - \frac{1}{2} \cdot 6^{2} + 96 (\ln (| 16 |) - \ln (| 6 |)))$

Step 9.1.3.3.2.4

Divide

$128$ by

$1$ .

$A (x) = \frac{1}{16 - 6} (128 - \frac{1}{2} \cdot 6^{2} + 96 (\ln (| 16 |) - \ln (| 6 |)))$

Step 9.1.3.4

Raise

$6$ to the power of

$2$ .

$A (x) = \frac{1}{16 - 6} (128 - \frac{1}{2} \cdot 36 + 96 (\ln (| 16 |) - \ln (| 6 |)))$

Step 9.1.3.5

Multiply

$36$ by

$- 1$ .

$A (x) = \frac{1}{16 - 6} (128 - 36 (\frac{1}{2}) + 96 (\ln (| 16 |) - \ln (| 6 |)))$

Step 9.1.3.6

Combine

$- 36$ and

$\frac{1}{2}$ .

$A (x) = \frac{1}{16 - 6} (128 + \frac{- 36}{2} + 96 (\ln (| 16 |) - \ln (| 6 |)))$

Step 9.1.3.7

Cancel the common factor of

$- 36$ and

$2$ .

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Step 9.1.3.7.1

Factor

$2$ out of

$- 36$ .

$A (x) = \frac{1}{16 - 6} (128 + \frac{2 \cdot - 18}{2} + 96 (\ln (| 16 |) - \ln (| 6 |)))$

Step 9.1.3.7.2

Cancel the common factors.

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Step 9.1.3.7.2.1

Factor

$2$ out of

$2$ .

$A (x) = \frac{1}{16 - 6} (128 + \frac{2 \cdot - 18}{2 (1)} + 96 (\ln (| 16 |) - \ln (| 6 |)))$

Step 9.1.3.7.2.2

Cancel the common factor.

$A (x) = \frac{1}{16 - 6} (128 + \frac{2 \cdot - 18}{2 \cdot 1} + 96 (\ln (| 16 |) - \ln (| 6 |)))$

Step 9.1.3.7.2.3

Rewrite the expression.

$A (x) = \frac{1}{16 - 6} (128 + \frac{- 18}{1} + 96 (\ln (| 16 |) - \ln (| 6 |)))$

Step 9.1.3.7.2.4

Divide

$- 18$ by

$1$ .

$A (x) = \frac{1}{16 - 6} (128 - 18 + 96 (\ln (| 16 |) - \ln (| 6 |)))$

Step 9.1.3.8

Subtract

$18$ from

$128$ .

$A (x) = \frac{1}{16 - 6} (110 + 96 (\ln (| 16 |) - \ln (| 6 |)))$

Step 9.2

Use the quotient property of logarithms,

$\log_{b} (x) - \log_{b} (y) = \log_{b} (\frac{x}{y})$ .

$A (x) = \frac{1}{16 - 6} (110 + 96 \ln (\frac{| 16 |}{| 6 |}))$

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

$0$ and

$16$ is

$16$ .

$A (x) = \frac{1}{16 - 6} (110 + 96 \ln (\frac{16}{| 6 |}))$

Step 9.3.2

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

$0$ and

$6$ is

$6$ .

$A (x) = \frac{1}{16 - 6} (110 + 96 \ln (\frac{16}{6}))$

Step 9.3.3

Cancel the common factor of

$16$ and

$6$ .

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Step 9.3.3.1

Factor

$2$ out of

$16$ .

$A (x) = \frac{1}{16 - 6} (110 + 96 \ln (\frac{2 (8)}{6}))$

Step 9.3.3.2

Cancel the common factors.

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Step 9.3.3.2.1

Factor

$2$ out of

$6$ .

$A (x) = \frac{1}{16 - 6} (110 + 96 \ln (\frac{2 \cdot 8}{2 \cdot 3}))$

Step 9.3.3.2.2

Cancel the common factor.

$A (x) = \frac{1}{16 - 6} (110 + 96 \ln (\frac{2 \cdot 8}{2 \cdot 3}))$

Step 9.3.3.2.3

Rewrite the expression.

$A (x) = \frac{1}{16 - 6} (110 + 96 \ln (\frac{8}{3}))$

Step 10

Subtract

$6$ from

$16$ .

$A (x) = \frac{1}{10} \cdot (110 + 96 \ln (\frac{8}{3}))$

Step 11

Simplify each term.

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Step 11.1

Simplify

$96 \ln (\frac{8}{3})$ by moving

$96$ inside the logarithm.

$A (x) = \frac{1}{10} \cdot (110 + \ln ({(\frac{8}{3})}^{96}))$

Step 11.2

Apply the product rule to

$\frac{8}{3}$ .

$A (x) = \frac{1}{10} \cdot (110 + \ln (\frac{8^{96}}{3^{96}}))$

Step 12

Simplify terms.

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Step 12.1

Apply the distributive property.

$A (x) = \frac{1}{10} \cdot 110 + \frac{1}{10} \cdot \ln (\frac{8^{96}}{3^{96}})$

Step 12.2

Cancel the common factor of

$10$ .

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Step 12.2.1

Factor

$10$ out of

$110$ .

$A (x) = \frac{1}{10} \cdot (10 (11)) + \frac{1}{10} \cdot \ln (\frac{8^{96}}{3^{96}})$

Step 12.2.2

Cancel the common factor.

$A (x) = \frac{1}{10} \cdot (10 \cdot 11) + \frac{1}{10} \cdot \ln (\frac{8^{96}}{3^{96}})$

Step 12.2.3

Rewrite the expression.

$A (x) = 11 + \frac{1}{10} \cdot \ln (\frac{8^{96}}{3^{96}})$

Step 13

Simplify

$\frac{1}{10} \ln (\frac{8^{96}}{3^{96}})$ by moving

$\frac{1}{10}$ inside the logarithm.

$A (x) = 11 + \ln ({(\frac{8^{96}}{3^{96}})}^{\frac{1}{10}})$

Step 14

Simplify each term.

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Step 14.1

Apply the product rule to

$\frac{8^{96}}{3^{96}}$ .

$A (x) = 11 + \ln (\frac{{(8^{96})}^{\frac{1}{10}}}{{(3^{96})}^{\frac{1}{10}}})$

Step 14.2

Multiply the exponents in

${(8^{96})}^{\frac{1}{10}}$ .

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Step 14.2.1

Apply the power rule and multiply exponents,

${(a^{m})}^{n} = a^{m n}$ .

$A (x) = 11 + \ln (\frac{8^{96 (\frac{1}{10})}}{{(3^{96})}^{\frac{1}{10}}})$

Step 14.2.2

Cancel the common factor of

$2$ .

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Step 14.2.2.1

Factor

$2$ out of

$96$ .

$A (x) = 11 + \ln (\frac{8^{2 (48) (\frac{1}{10})}}{{(3^{96})}^{\frac{1}{10}}})$

Step 14.2.2.2

Factor

$2$ out of

$10$ .

$A (x) = 11 + \ln (\frac{8^{2 \cdot (48 (\frac{1}{2 \cdot 5}))}}{{(3^{96})}^{\frac{1}{10}}})$

Step 14.2.2.3

Cancel the common factor.

$A (x) = 11 + \ln (\frac{8^{2 \cdot (48 (\frac{1}{2 \cdot 5}))}}{{(3^{96})}^{\frac{1}{10}}})$

Step 14.2.2.4

Rewrite the expression.

$A (x) = 11 + \ln (\frac{8^{48 (\frac{1}{5})}}{{(3^{96})}^{\frac{1}{10}}})$

Step 14.2.3

Combine

$48$ and

$\frac{1}{5}$ .

$A (x) = 11 + \ln (\frac{8^{\frac{48}{5}}}{{(3^{96})}^{\frac{1}{10}}})$

Step 14.3

Multiply the exponents in

${(3^{96})}^{\frac{1}{10}}$ .

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Step 14.3.1

Apply the power rule and multiply exponents,

${(a^{m})}^{n} = a^{m n}$ .

$A (x) = 11 + \ln (\frac{8^{\frac{48}{5}}}{3^{96 (\frac{1}{10})}})$

Step 14.3.2

Cancel the common factor of

$2$ .

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Step 14.3.2.1

Factor

$2$ out of

$96$ .

$A (x) = 11 + \ln (\frac{8^{\frac{48}{5}}}{3^{2 (48) (\frac{1}{10})}})$

Step 14.3.2.2

Factor

$2$ out of

$10$ .

$A (x) = 11 + \ln (\frac{8^{\frac{48}{5}}}{3^{2 \cdot (48 (\frac{1}{2 \cdot 5}))}})$

Step 14.3.2.3

Cancel the common factor.

$A (x) = 11 + \ln (\frac{8^{\frac{48}{5}}}{3^{2 \cdot (48 (\frac{1}{2 \cdot 5}))}})$

Step 14.3.2.4

Rewrite the expression.

$A (x) = 11 + \ln (\frac{8^{\frac{48}{5}}}{3^{48 (\frac{1}{5})}})$

Step 14.3.3

Combine

$48$ and

$\frac{1}{5}$ .

$A (x) = 11 + \ln (\frac{8^{\frac{48}{5}}}{3^{\frac{48}{5}}})$

Step 15