Calculus Examples

$f (x) = 5 x - 3$ , $[0, 6]$

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

$f (x)$ is continuous on $[0, 6]$ .

$f (x)$ 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.

$A (x) = \frac{1}{6 - 0} (\int_{0}^{6} 5 x - 3 d x)$

Step 5

Split the single integral into multiple integrals.

$A (x) = \frac{1}{6 - 0} (\int_{0}^{6} 5 x d x + \int_{0}^{6} - 3 d x)$

Step 6

Since $5$ is constant with respect to $x$ , move $5$ out of the integral.

$A (x) = \frac{1}{6 - 0} (5 \int_{0}^{6} x d x + \int_{0}^{6} - 3 d x)$

Step 7

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

$A (x) = \frac{1}{6 - 0} (5 (\frac{1}{2} x^{2}]_{0}^{6}) + \int_{0}^{6} - 3 d x)$

Step 8

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

$A (x) = \frac{1}{6 - 0} (5 (\frac{x^{2}}{2}]_{0}^{6}) + \int_{0}^{6} - 3 d x)$

Step 9

Apply the constant rule.

$A (x) = \frac{1}{6 - 0} (5 (\frac{x^{2}}{2}]_{0}^{6}) + - 3 x]_{0}^{6})$

Step 10

Substitute and simplify.

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Evaluate $\frac{x^{2}}{2}$ at $6$ and at $0$ .

$A (x) = \frac{1}{6 - 0} (5 ((\frac{6^{2}}{2}) - \frac{0^{2}}{2}) + - 3 x]_{0}^{6})$

Evaluate $- 3 x$ at $6$ and at $0$ .

$A (x) = \frac{1}{6 - 0} (5 (\frac{6^{2}}{2} - \frac{0^{2}}{2}) - 3 \cdot 6 + 3 \cdot 0)$

Simplify.

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Raise $6$ to the power of $2$ .

$A (x) = \frac{1}{6 - 0} (5 (\frac{36}{2} - \frac{0^{2}}{2}) - 3 \cdot 6 + 3 \cdot 0)$

Cancel the common factor of $36$ and $2$ .

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Factor $2$ out of $36$ .

$A (x) = \frac{1}{6 - 0} (5 (\frac{2 \cdot 18}{2} - \frac{0^{2}}{2}) - 3 \cdot 6 + 3 \cdot 0)$

Cancel the common factors.

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Factor $2$ out of $2$ .

$A (x) = \frac{1}{6 - 0} (5 (\frac{2 \cdot 18}{2 (1)} - \frac{0^{2}}{2}) - 3 \cdot 6 + 3 \cdot 0)$

Cancel the common factor.

$A (x) = \frac{1}{6 - 0} (5 (\frac{2 \cdot 18}{2 \cdot 1} - \frac{0^{2}}{2}) - 3 \cdot 6 + 3 \cdot 0)$

Rewrite the expression.

$A (x) = \frac{1}{6 - 0} (5 (\frac{18}{1} - \frac{0^{2}}{2}) - 3 \cdot 6 + 3 \cdot 0)$

Divide $18$ by $1$ .

$A (x) = \frac{1}{6 - 0} (5 (18 - \frac{0^{2}}{2}) - 3 \cdot 6 + 3 \cdot 0)$

Raising $0$ to any positive power yields $0$ .

$A (x) = \frac{1}{6 - 0} (5 (18 - \frac{0}{2}) - 3 \cdot 6 + 3 \cdot 0)$

Cancel the common factor of $0$ and $2$ .

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Factor $2$ out of $0$ .

$A (x) = \frac{1}{6 - 0} (5 (18 - \frac{2 (0)}{2}) - 3 \cdot 6 + 3 \cdot 0)$

Cancel the common factors.

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Factor $2$ out of $2$ .

$A (x) = \frac{1}{6 - 0} (5 (18 - \frac{2 \cdot 0}{2 \cdot 1}) - 3 \cdot 6 + 3 \cdot 0)$

Cancel the common factor.

$A (x) = \frac{1}{6 - 0} (5 (18 - \frac{2 \cdot 0}{2 \cdot 1}) - 3 \cdot 6 + 3 \cdot 0)$

Rewrite the expression.

$A (x) = \frac{1}{6 - 0} (5 (18 - \frac{0}{1}) - 3 \cdot 6 + 3 \cdot 0)$

Divide $0$ by $1$ .

$A (x) = \frac{1}{6 - 0} (5 (18 - 0) - 3 \cdot 6 + 3 \cdot 0)$

Multiply $- 1$ by $0$ .

$A (x) = \frac{1}{6 - 0} (5 (18 + 0) - 3 \cdot 6 + 3 \cdot 0)$

Add $18$ and $0$ .

$A (x) = \frac{1}{6 - 0} (5 \cdot 18 - 3 \cdot 6 + 3 \cdot 0)$

Multiply $5$ by $18$ .

$A (x) = \frac{1}{6 - 0} (90 - 3 \cdot 6 + 3 \cdot 0)$

Multiply $- 3$ by $6$ .

$A (x) = \frac{1}{6 - 0} (90 - 18 + 3 \cdot 0)$

Multiply $3$ by $0$ .

$A (x) = \frac{1}{6 - 0} (90 - 18 + 0)$

Add $- 18$ and $0$ .

$A (x) = \frac{1}{6 - 0} (90 - 18)$

Subtract $18$ from $90$ .

$A (x) = \frac{1}{6 - 0} (72)$

Step 11

Simplify the denominator.

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Multiply $- 1$ by $0$ .

$A (x) = \frac{1}{6 + 0} \cdot 72$

Add $6$ and $0$ .

$A (x) = \frac{1}{6} \cdot 72$

Step 12

Cancel the common factor of $6$ .

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Factor $6$ out of $72$ .

$A (x) = \frac{1}{6} \cdot (6 (12))$

Cancel the common factor.

$A (x) = \frac{1}{6} \cdot (6 \cdot 12)$

Rewrite the expression.

$A (x) = 12$

Step 13

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