Calculus Examples

$f (x) = 3 x - 6$ , $(0, 4)$

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.

$(- \infty, \infty)$

${x | x \in ℝ}$

$f (x)$ is continuous on $[0, 4]$ .

$f (x)$ is continuous

The average value of function $f$ over the interval $[a, b]$ is defined as $A (x) = \frac{1}{b - a} \int_{a}^{b} f (x) d x$ .

$A (x) = \frac{1}{b - a} \int_{a}^{b} f (x) d x$

Substitute the actual values into the formula for the average value of a function.

$A (x) = \frac{1}{4 - 0} (\int_{0}^{4} 3 x - 6 d x)$

Since integration is linear, the integral of $3 x - 6$ with respect to $x$ is $\int_{0}^{4} 3 x d x + \int_{0}^{4} - 6 d x$ .

$A (x) = \frac{1}{4 - 0} (\int_{0}^{4} 3 x d x + \int_{0}^{4} - 6 d x)$

Since $3$ is constant with respect to $x$ , the integral of $3 x$ with respect to $x$ is $3 \int_{0}^{4} x d x$ .

$A (x) = \frac{1}{4 - 0} (3 \int_{0}^{4} x d x + \int_{0}^{4} - 6 d x)$

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

$A (x) = \frac{1}{4 - 0} (3 (\frac{1}{2} x^{2}]_{0}^{4}) + \int_{0}^{4} - 6 d x)$

Combine fractions.

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Write $x^{2}$ as a fraction with denominator $1$ .

$A (x) = \frac{1}{4 - 0} (3 (\frac{1}{2} \cdot \frac{x^{2}}{1}]_{0}^{4}) + \int_{0}^{4} - 6 d x)$

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

$A (x) = \frac{1}{4 - 0} (3 (\frac{x^{2}}{2}]_{0}^{4}) + \int_{0}^{4} - 6 d x)$

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

$A (x) = \frac{1}{4 - 0} (3 (\frac{x^{2}}{2}]_{0}^{4}) + - 6 x]_{0}^{4})$

Simplify the answer.

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

$A (x) = \frac{1}{4 - 0} (3 ((\frac{4^{2}}{2}) - \frac{0^{2}}{2}) + - 6 x]_{0}^{4})$

Evaluate $- 6 x$ at $4$ and at $0$ .

$A (x) = \frac{1}{4 - 0} (3 (\frac{4^{2}}{2} - \frac{0^{2}}{2}) + (- 6 \cdot 4 + 6 \cdot 0))$

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

$A (x) = \frac{1}{4 - 0} (3 (\frac{16}{2} - \frac{0^{2}}{2}) + (- 6 \cdot 4 + 6 \cdot 0))$

Reduce the expression by cancelling the common factors.

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Factor $2$ out of $2$ .

$A (x) = \frac{1}{4 - 0} (3 (\frac{2 \cdot 8}{2 (1)} - \frac{0^{2}}{2}) + (- 6 \cdot 4 + 6 \cdot 0))$

Cancel the common factor.

$A (x) = \frac{1}{4 - 0} (3 (\frac{2 \cdot 8}{2 \cdot 1} - \frac{0^{2}}{2}) + (- 6 \cdot 4 + 6 \cdot 0))$

Rewrite the expression.

$A (x) = \frac{1}{4 - 0} (3 (\frac{8}{1} - \frac{0^{2}}{2}) + (- 6 \cdot 4 + 6 \cdot 0))$

Divide $8$ by $1$ .

$A (x) = \frac{1}{4 - 0} (3 (8 - \frac{0^{2}}{2}) + (- 6 \cdot 4 + 6 \cdot 0))$

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

$A (x) = \frac{1}{4 - 0} (3 (8 - \frac{0}{2}) + (- 6 \cdot 4 + 6 \cdot 0))$

Reduce the expression by cancelling the common factors.

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Factor $2$ out of $2$ .

$A (x) = \frac{1}{4 - 0} (3 (8 - \frac{2 \cdot 0}{2 \cdot 1}) + (- 6 \cdot 4 + 6 \cdot 0))$

Cancel the common factor.

$A (x) = \frac{1}{4 - 0} (3 (8 - \frac{2 \cdot 0}{2 \cdot 1}) + (- 6 \cdot 4 + 6 \cdot 0))$

Rewrite the expression.

$A (x) = \frac{1}{4 - 0} (3 (8 - \frac{0}{1}) + (- 6 \cdot 4 + 6 \cdot 0))$

Divide $0$ by $1$ .

$A (x) = \frac{1}{4 - 0} (3 (8 - 0) + (- 6 \cdot 4 + 6 \cdot 0))$

Multiply $- 1$ by $0$ .

$A (x) = \frac{1}{4 - 0} (3 (8 + 0) + (- 6 \cdot 4 + 6 \cdot 0))$

Add $8$ and $0$ .

$A (x) = \frac{1}{4 - 0} (3 \cdot 8 + (- 6 \cdot 4 + 6 \cdot 0))$

Multiply $3$ by $8$ .

$A (x) = \frac{1}{4 - 0} (24 + (- 6 \cdot 4 + 6 \cdot 0))$

Multiply $- 6$ by $4$ .

$A (x) = \frac{1}{4 - 0} (24 + (- 24 + 6 \cdot 0))$

Multiply $6$ by $0$ .

$A (x) = \frac{1}{4 - 0} (24 + (- 24 + 0))$

Add $- 24$ and $0$ .

$A (x) = \frac{1}{4 - 0} (24 - 24)$

Subtract $24$ from $24$ .

$A (x) = \frac{1}{4 - 0} (0)$

Simplify the denominator.

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Multiply $- 1$ by $0$ .

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

Add $4$ and $0$ .

$A (x) = \frac{1}{4} \cdot 0$

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

$A (x) = 0$

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