Calculus Examples

$f (x) = 1 x^{3} - 1 x^{2} + 4$ , $[- 2, 0]$

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 $[- 2, 0]$ .

$f (x)$ 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.

$A (x) = \frac{1}{0 + 2} (\int_{- 2}^{0} 1 x^{3} - 1 x^{2} + 4 d x)$

Step 5

Simplify.

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Step 5.1

Multiply $x^{3}$ by $1$ .

$A (x) = \frac{1}{0 + 2} (\int_{- 2}^{0} x^{3} - 1 x^{2} + 4 d x)$

Step 5.2

Rewrite $- 1 x^{2}$ as $- x^{2}$ .

$A (x) = \frac{1}{0 + 2} (\int_{- 2}^{0} x^{3} - x^{2} + 4 d x)$

Step 6

Split the single integral into multiple integrals.

$A (x) = \frac{1}{0 + 2} (\int_{- 2}^{0} x^{3} d x + \int_{- 2}^{0} - x^{2} d x + \int_{- 2}^{0} 4 d x)$

Step 7

By the Power Rule, the integral of $x^{3}$ with respect to $x$ is $\frac{1}{4} x^{4}$ .

$A (x) = \frac{1}{0 + 2} (\frac{1}{4} x^{4}]_{- 2}^{0} + \int_{- 2}^{0} - x^{2} d x + \int_{- 2}^{0} 4 d x)$

Step 8

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

$A (x) = \frac{1}{0 + 2} (\frac{1}{4} x^{4}]_{- 2}^{0} - \int_{- 2}^{0} x^{2} d x + \int_{- 2}^{0} 4 d x)$

Step 9

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

$A (x) = \frac{1}{0 + 2} (\frac{1}{4} x^{4}]_{- 2}^{0} - (\frac{1}{3} x^{3}]_{- 2}^{0}) + \int_{- 2}^{0} 4 d x)$

Step 10

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

$A (x) = \frac{1}{0 + 2} (\frac{1}{4} x^{4}]_{- 2}^{0} - (\frac{x^{3}}{3}]_{- 2}^{0}) + \int_{- 2}^{0} 4 d x)$

Step 11

Apply the constant rule.

$A (x) = \frac{1}{0 + 2} (\frac{1}{4} x^{4}]_{- 2}^{0} - (\frac{x^{3}}{3}]_{- 2}^{0}) + 4 x]_{- 2}^{0})$

Step 12

Simplify the answer.

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Step 12.1

Combine $\frac{1}{4} x^{4}]_{- 2}^{0}$ and $4 x]_{- 2}^{0}$ .

$A (x) = \frac{1}{0 + 2} (\frac{1}{4} x^{4} + 4 x]_{- 2}^{0} - (\frac{x^{3}}{3}]_{- 2}^{0}))$

Step 12.2

Substitute and simplify.

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Step 12.2.1

Evaluate $\frac{1}{4} x^{4} + 4 x$ at $0$ and at $- 2$ .

$A (x) = \frac{1}{0 + 2} ((\frac{1}{4} \cdot 0^{4} + 4 \cdot 0) - (\frac{1}{4} \cdot {(- 2)}^{4} + 4 \cdot - 2) - (\frac{x^{3}}{3}]_{- 2}^{0}))$

Step 12.2.2

Evaluate $\frac{x^{3}}{3}$ at $0$ and at $- 2$ .

$A (x) = \frac{1}{0 + 2} ((\frac{1}{4} \cdot 0^{4} + 4 \cdot 0) - (\frac{1}{4} \cdot {(- 2)}^{4} + 4 \cdot - 2) - ((\frac{0^{3}}{3}) - \frac{{(- 2)}^{3}}{3}))$

Step 12.2.3

Simplify.

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Step 12.2.3.1

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

$A (x) = \frac{1}{0 + 2} (\frac{1}{4} \cdot 0 + 4 \cdot 0 - (\frac{1}{4} \cdot {(- 2)}^{4} + 4 \cdot - 2) - ((\frac{0^{3}}{3}) - \frac{{(- 2)}^{3}}{3}))$

Step 12.2.3.2

Multiply $\frac{1}{4}$ by $0$ .

$A (x) = \frac{1}{0 + 2} (0 + 4 \cdot 0 - (\frac{1}{4} \cdot {(- 2)}^{4} + 4 \cdot - 2) - ((\frac{0^{3}}{3}) - \frac{{(- 2)}^{3}}{3}))$

Step 12.2.3.3

Multiply $4$ by $0$ .

$A (x) = \frac{1}{0 + 2} (0 + 0 - (\frac{1}{4} \cdot {(- 2)}^{4} + 4 \cdot - 2) - ((\frac{0^{3}}{3}) - \frac{{(- 2)}^{3}}{3}))$

Step 12.2.3.4

Add $0$ and $0$ .

$A (x) = \frac{1}{0 + 2} (0 - (\frac{1}{4} \cdot {(- 2)}^{4} + 4 \cdot - 2) - ((\frac{0^{3}}{3}) - \frac{{(- 2)}^{3}}{3}))$

Step 12.2.3.5

Raise $- 2$ to the power of $4$ .

$A (x) = \frac{1}{0 + 2} (0 - (\frac{1}{4} \cdot 16 + 4 \cdot - 2) - ((\frac{0^{3}}{3}) - \frac{{(- 2)}^{3}}{3}))$

Step 12.2.3.6

Combine $\frac{1}{4}$ and $16$ .

$A (x) = \frac{1}{0 + 2} (0 - (\frac{16}{4} + 4 \cdot - 2) - ((\frac{0^{3}}{3}) - \frac{{(- 2)}^{3}}{3}))$

Step 12.2.3.7

Cancel the common factor of $16$ and $4$ .

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Step 12.2.3.7.1

Factor $4$ out of $16$ .

$A (x) = \frac{1}{0 + 2} (0 - (\frac{4 \cdot 4}{4} + 4 \cdot - 2) - ((\frac{0^{3}}{3}) - \frac{{(- 2)}^{3}}{3}))$

Step 12.2.3.7.2

Cancel the common factors.

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Step 12.2.3.7.2.1

Factor $4$ out of $4$ .

$A (x) = \frac{1}{0 + 2} (0 - (\frac{4 \cdot 4}{4 (1)} + 4 \cdot - 2) - ((\frac{0^{3}}{3}) - \frac{{(- 2)}^{3}}{3}))$

Step 12.2.3.7.2.2

Cancel the common factor.

$A (x) = \frac{1}{0 + 2} (0 - (\frac{4 \cdot 4}{4 \cdot 1} + 4 \cdot - 2) - ((\frac{0^{3}}{3}) - \frac{{(- 2)}^{3}}{3}))$

Step 12.2.3.7.2.3

Rewrite the expression.

$A (x) = \frac{1}{0 + 2} (0 - (\frac{4}{1} + 4 \cdot - 2) - ((\frac{0^{3}}{3}) - \frac{{(- 2)}^{3}}{3}))$

Step 12.2.3.7.2.4

Divide $4$ by $1$ .

$A (x) = \frac{1}{0 + 2} (0 - (4 + 4 \cdot - 2) - ((\frac{0^{3}}{3}) - \frac{{(- 2)}^{3}}{3}))$

Step 12.2.3.8

Multiply $4$ by $- 2$ .

$A (x) = \frac{1}{0 + 2} (0 - (4 - 8) - ((\frac{0^{3}}{3}) - \frac{{(- 2)}^{3}}{3}))$

Step 12.2.3.9

Subtract $8$ from $4$ .

$A (x) = \frac{1}{0 + 2} (0 + 4 - ((\frac{0^{3}}{3}) - \frac{{(- 2)}^{3}}{3}))$

Step 12.2.3.10

Multiply $- 1$ by $- 4$ .

$A (x) = \frac{1}{0 + 2} (0 + 4 - ((\frac{0^{3}}{3}) - \frac{{(- 2)}^{3}}{3}))$

Step 12.2.3.11

Add $0$ and $4$ .

$A (x) = \frac{1}{0 + 2} (4 - ((\frac{0^{3}}{3}) - \frac{{(- 2)}^{3}}{3}))$

Step 12.2.3.12

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

$A (x) = \frac{1}{0 + 2} (4 - (\frac{0}{3} - \frac{{(- 2)}^{3}}{3}))$

Step 12.2.3.13

Cancel the common factor of $0$ and $3$ .

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Step 12.2.3.13.1

Factor $3$ out of $0$ .

$A (x) = \frac{1}{0 + 2} (4 - (\frac{3 (0)}{3} - \frac{{(- 2)}^{3}}{3}))$

Step 12.2.3.13.2

Cancel the common factors.

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Step 12.2.3.13.2.1

Factor $3$ out of $3$ .

$A (x) = \frac{1}{0 + 2} (4 - (\frac{3 \cdot 0}{3 \cdot 1} - \frac{{(- 2)}^{3}}{3}))$

Step 12.2.3.13.2.2

Cancel the common factor.

$A (x) = \frac{1}{0 + 2} (4 - (\frac{3 \cdot 0}{3 \cdot 1} - \frac{{(- 2)}^{3}}{3}))$

Step 12.2.3.13.2.3

Rewrite the expression.

$A (x) = \frac{1}{0 + 2} (4 - (\frac{0}{1} - \frac{{(- 2)}^{3}}{3}))$

Step 12.2.3.13.2.4

Divide $0$ by $1$ .

$A (x) = \frac{1}{0 + 2} (4 - (0 - \frac{{(- 2)}^{3}}{3}))$

Step 12.2.3.14

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

$A (x) = \frac{1}{0 + 2} (4 - (0 - \frac{- 8}{3}))$

Step 12.2.3.15

Move the negative in front of the fraction.

$A (x) = \frac{1}{0 + 2} (4 - (0 + \frac{8}{3}))$

Step 12.2.3.16

Multiply $- 1$ by $- 1$ .

$A (x) = \frac{1}{0 + 2} (4 - (0 + 1 (\frac{8}{3})))$

Step 12.2.3.17

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

$A (x) = \frac{1}{0 + 2} (4 - (0 + \frac{8}{3}))$

Step 12.2.3.18

Add $0$ and $\frac{8}{3}$ .

$A (x) = \frac{1}{0 + 2} (4 - \frac{8}{3})$

Step 12.2.3.19

To write $4$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$A (x) = \frac{1}{0 + 2} (4 \cdot \frac{3}{3} - \frac{8}{3})$

Step 12.2.3.20

Combine $4$ and $\frac{3}{3}$ .

$A (x) = \frac{1}{0 + 2} (\frac{4 \cdot 3}{3} - \frac{8}{3})$

Step 12.2.3.21

Combine the numerators over the common denominator.

$A (x) = \frac{1}{0 + 2} (\frac{4 \cdot 3 - 8}{3})$

Step 12.2.3.22

Simplify the numerator.

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Step 12.2.3.22.1

Multiply $4$ by $3$ .

$A (x) = \frac{1}{0 + 2} (\frac{12 - 8}{3})$

Step 12.2.3.22.2

Subtract $8$ from $12$ .

$A (x) = \frac{1}{0 + 2} (\frac{4}{3})$

Step 13

Add $0$ and $2$ .

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

Step 14

Cancel the common factor of $2$ .

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Step 14.1

Factor $2$ out of $4$ .

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

Step 14.2

Cancel the common factor.

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

Step 14.3

Rewrite the expression.

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

Step 15