Calculus Examples

$f (x) = 4 - x^{2}$ , $[- 2, 2]$

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 $[- 2, 2]$ .

$f (x)$ 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.

$A (x) = \frac{1}{2 + 2} (\int_{- 2}^{2} 4 - x^{2} d x)$

Step 5

Split the single integral into multiple integrals.

$A (x) = \frac{1}{2 + 2} (\int_{- 2}^{2} 4 d x + \int_{- 2}^{2} - x^{2} d x)$

Step 6

Apply the constant rule.

$A (x) = \frac{1}{2 + 2} (4 x]_{- 2}^{2} + \int_{- 2}^{2} - x^{2} d x)$

Step 7

Since $- 1$ is constant with respect to $x$ , move $- 1$ out of the integral.

$A (x) = \frac{1}{2 + 2} (4 x]_{- 2}^{2} - \int_{- 2}^{2} x^{2} d x)$

Step 8

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

$A (x) = \frac{1}{2 + 2} (4 x]_{- 2}^{2} - (\frac{1}{3} x^{3}]_{- 2}^{2}))$

Step 9

Simplify the answer.

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Step 9.1

Combine $\frac{1}{3}$ and $x^{3}$ .

$A (x) = \frac{1}{2 + 2} (4 x]_{- 2}^{2} - (\frac{x^{3}}{3}]_{- 2}^{2}))$

Step 9.2

Substitute and simplify.

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Step 9.2.1

Evaluate $4 x$ at $2$ and at $- 2$ .

$A (x) = \frac{1}{2 + 2} ((4 \cdot 2) - 4 \cdot - 2 - (\frac{x^{3}}{3}]_{- 2}^{2}))$

Step 9.2.2

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

$A (x) = \frac{1}{2 + 2} (4 \cdot 2 - 4 \cdot - 2 - (\frac{2^{3}}{3} - \frac{{(- 2)}^{3}}{3}))$

Step 9.2.3

Simplify.

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Step 9.2.3.1

Multiply $4$ by $2$ .

$A (x) = \frac{1}{2 + 2} (8 - 4 \cdot - 2 - (\frac{2^{3}}{3} - \frac{{(- 2)}^{3}}{3}))$

Step 9.2.3.2

Multiply $- 4$ by $- 2$ .

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

Step 9.2.3.3

Add $8$ and $8$ .

$A (x) = \frac{1}{2 + 2} (16 - (\frac{2^{3}}{3} - \frac{{(- 2)}^{3}}{3}))$

Step 9.2.3.4

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

$A (x) = \frac{1}{2 + 2} (16 - (\frac{8}{3} - \frac{{(- 2)}^{3}}{3}))$

Step 9.2.3.5

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

$A (x) = \frac{1}{2 + 2} (16 - (\frac{8}{3} - \frac{- 8}{3}))$

Step 9.2.3.6

Move the negative in front of the fraction.

$A (x) = \frac{1}{2 + 2} (16 - (\frac{8}{3} + \frac{8}{3}))$

Step 9.2.3.7

Multiply $- 1$ by $- 1$ .

$A (x) = \frac{1}{2 + 2} (16 - (\frac{8}{3} + 1 (\frac{8}{3})))$

Step 9.2.3.8

Multiply $\frac{8}{3}$ by $1$ .

$A (x) = \frac{1}{2 + 2} (16 - (\frac{8}{3} + \frac{8}{3}))$

Step 9.2.3.9

Combine the numerators over the common denominator.

$A (x) = \frac{1}{2 + 2} (16 - \frac{8 + 8}{3})$

Step 9.2.3.10

Add $8$ and $8$ .

$A (x) = \frac{1}{2 + 2} (16 - \frac{16}{3})$

Step 9.2.3.11

To write $16$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$A (x) = \frac{1}{2 + 2} (16 \cdot \frac{3}{3} - \frac{16}{3})$

Step 9.2.3.12

Combine $16$ and $\frac{3}{3}$ .

$A (x) = \frac{1}{2 + 2} (\frac{16 \cdot 3}{3} - \frac{16}{3})$

Step 9.2.3.13

Combine the numerators over the common denominator.

$A (x) = \frac{1}{2 + 2} (\frac{16 \cdot 3 - 16}{3})$

Step 9.2.3.14

Simplify the numerator.

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Step 9.2.3.14.1

Multiply $16$ by $3$ .

$A (x) = \frac{1}{2 + 2} (\frac{48 - 16}{3})$

Step 9.2.3.14.2

Subtract $16$ from $48$ .

$A (x) = \frac{1}{2 + 2} (\frac{32}{3})$

Step 10

Add $2$ and $2$ .

$A (x) = \frac{1}{4} \cdot \frac{32}{3}$

Step 11

Cancel the common factor of $4$ .

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Step 11.1

Factor $4$ out of $32$ .

$A (x) = \frac{1}{4} \cdot \frac{4 (8)}{3}$

Step 11.2

Cancel the common factor.

$A (x) = \frac{1}{4} \cdot \frac{4 \cdot 8}{3}$

Step 11.3

Rewrite the expression.

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

Step 12