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