Calculus Examples

$g (x) = {(x - 6)}^{2}$ , $[5, 8]$

Step 1

The domain of the expression is all real numbers except where the expression is undefined. In this case, there is no real number that makes the expression undefined.

Interval Notation:

$(- \infty, \infty)$

Set-Builder Notation:

${x | x \in ℝ}$

Step 2

$g (x)$ is continuous on $[5, 8]$ .

$g (x)$ is continuous

Step 3

The average value of function $g$ 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.

$A (x) = \frac{1}{8 - 5} (\int_{5}^{8} {(x - 6)}^{2} d x)$

Step 5

Let $u = x - 6$ . Then $d u = d x$ . Rewrite using $u$ and $d$ $u$ .

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Step 5.1

Let $u = x - 6$ . Find $\frac{d u}{d x}$ .

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Step 5.1.1

Differentiate $x - 6$ .

$\frac{d}{d x} [x - 6]$

Step 5.1.2

By the Sum Rule, the derivative of $x - 6$ with respect to $x$ is $\frac{d}{d x} [x] + \frac{d}{d x} [- 6]$ .

$\frac{d}{d x} [x] + \frac{d}{d x} [- 6]$

Step 5.1.3

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 1$ .

$1 + \frac{d}{d x} [- 6]$

Step 5.1.4

Since $- 6$ is constant with respect to $x$ , the derivative of $- 6$ with respect to $x$ is $0$ .

$1 + 0$

Step 5.1.5

Add $1$ and $0$ .

$1$

Step 5.2

Substitute the lower limit in for $x$ in $u = x - 6$ .

$u_{lower} = 5 - 6$

Step 5.3

Subtract $6$ from $5$ .

$u_{lower} = - 1$

Step 5.4

Substitute the upper limit in for $x$ in $u = x - 6$ .

$u_{upper} = 8 - 6$

Step 5.5

Subtract $6$ from $8$ .

$u_{upper} = 2$

Step 5.6

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

$u_{lower} = - 1$

$u_{upper} = 2$

Step 5.7

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

$A (x) = \frac{1}{8 - 5} (\int_{- 1}^{2} u^{2} d u)$

Step 6

By the Power Rule, the integral of $u^{2}$ with respect to $u$ is $\frac{1}{3} u^{3}$ .

$A (x) = \frac{1}{8 - 5} (\frac{1}{3} u^{3}]_{- 1}^{2})$

Step 7

Substitute and simplify.

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Step 7.1

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

$A (x) = \frac{1}{8 - 5} ((\frac{1}{3} \cdot 2^{3}) - \frac{1}{3} \cdot {(- 1)}^{3})$

Step 7.2

Simplify.

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Step 7.2.1

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

$A (x) = \frac{1}{8 - 5} (\frac{1}{3} \cdot 8 - \frac{1}{3} \cdot {(- 1)}^{3})$

Step 7.2.2

Combine $\frac{1}{3}$ and $8$ .

$A (x) = \frac{1}{8 - 5} (\frac{8}{3} - \frac{1}{3} \cdot {(- 1)}^{3})$

Step 7.2.3

Raise $- 1$ to the power of $3$ .

$A (x) = \frac{1}{8 - 5} (\frac{8}{3} - \frac{1}{3} \cdot - 1)$

Step 7.2.4

Multiply $- 1$ by $- 1$ .

$A (x) = \frac{1}{8 - 5} (\frac{8}{3} + 1 (\frac{1}{3}))$

Step 7.2.5

Multiply $\frac{1}{3}$ by $1$ .

$A (x) = \frac{1}{8 - 5} (\frac{8}{3} + \frac{1}{3})$

Step 7.2.6

Combine the numerators over the common denominator.

$A (x) = \frac{1}{8 - 5} (\frac{8 + 1}{3})$

Step 7.2.7

Add $8$ and $1$ .

$A (x) = \frac{1}{8 - 5} (\frac{9}{3})$

Step 7.2.8

Cancel the common factor of $9$ and $3$ .

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Step 7.2.8.1

Factor $3$ out of $9$ .

$A (x) = \frac{1}{8 - 5} (\frac{3 \cdot 3}{3})$

Step 7.2.8.2

Cancel the common factors.

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Step 7.2.8.2.1

Factor $3$ out of $3$ .

$A (x) = \frac{1}{8 - 5} (\frac{3 \cdot 3}{3 (1)})$

Step 7.2.8.2.2

Cancel the common factor.

$A (x) = \frac{1}{8 - 5} (\frac{3 \cdot 3}{3 \cdot 1})$

Step 7.2.8.2.3

Rewrite the expression.

$A (x) = \frac{1}{8 - 5} (\frac{3}{1})$

Step 7.2.8.2.4

Divide $3$ by $1$ .

$A (x) = \frac{1}{8 - 5} (3)$

Step 8

Subtract $5$ from $8$ .

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

Step 9

Cancel the common factor of $3$ .

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Step 9.1

Cancel the common factor.

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

Step 9.2

Rewrite the expression.

$A (x) = 1$

Step 10