Calculus Examples

$f (X) = x^{2} + x - 9$ ; $[0, 18]$

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 $[0, 18]$ .

$f (X)$ 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.

$A (X) = \frac{1}{18 - 0} (\int_{0}^{18} x^{2} + x - 9 d X)$

Step 5

Apply the constant rule.

$A (X) = \frac{1}{18 - 0} ((x^{2} + x - 9) X]_{0}^{18})$

Step 6

Substitute and simplify.

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Step 6.1

Evaluate $(x^{2} + x - 9) X$ at $18$ and at $0$ .

$A (X) = \frac{1}{18 - 0} (((x^{2} + x - 9) \cdot 18) - (x^{2} + x - 9) \cdot 0)$

Step 6.2

Simplify.

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Step 6.2.1

Move $18$ to the left of $x^{2} + x - 9$ .

$A (X) = \frac{1}{18 - 0} (18 \cdot (x^{2} + x - 9) - (x^{2} + x - 9) \cdot 0)$

Step 6.2.2

Multiply $0$ by $- 1$ .

$A (X) = \frac{1}{18 - 0} (18 (x^{2} + x - 9) + 0 (x^{2} + x - 9))$

Step 6.2.3

Multiply $0$ by $x^{2} + x - 9$ .

$A (X) = \frac{1}{18 - 0} (18 (x^{2} + x - 9) + 0)$

Step 6.2.4

Add $18 (x^{2} + x - 9)$ and $0$ .

$A (X) = \frac{1}{18 - 0} (18 (x^{2} + x - 9))$

Step 7

Simplify the denominator.

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Step 7.1

Multiply $- 1$ by $0$ .

$A (X) = \frac{1}{18 + 0} \cdot (18 (x^{2} + x - 9))$

Step 7.2

Add $18$ and $0$ .

$A (X) = \frac{1}{18} \cdot (18 (x^{2} + x - 9))$

Step 8

Cancel the common factor of $18$ .

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Step 8.1

Cancel the common factor.

$A (X) = \frac{1}{18} \cdot (18 (x^{2} + x - 9))$

Step 8.2

Rewrite the expression.

$A (X) = x^{2} + x - 9$

Step 9