Calculus Examples

$f (x) = 4 x - 2$ , $(1, 3)$

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:

${x | x \in ℝ}$

$f (x)$ is continuous on $[1, 3]$ .

$f (x)$ 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.

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

Split the single integral into multiple integrals.

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

Since $4$ is constant with respect to $x$ , move $4$ out of the integral.

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

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

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

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

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

Since $- 2$ is constant with respect to $x$ , move $- 2$ out of the integral.

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

Substitute and simplify.

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Evaluate $\frac{x^{2}}{2}$ at $3$ and at $1$ .

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

Evaluate $- 2 x$ at $3$ and at $1$ .

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

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

$A (x) = \frac{1}{3 - 1} (4 (\frac{9}{2} - \frac{1^{2}}{2}) + (- 2 \cdot 3 + 2 \cdot 1))$

One to any power is one.

$A (x) = \frac{1}{3 - 1} (4 (\frac{9}{2} - \frac{1}{2}) + (- 2 \cdot 3 + 2 \cdot 1))$

Combine the numerators over the common denominator.

$A (x) = \frac{1}{3 - 1} (4 (\frac{9 - 1}{2}) + (- 2 \cdot 3 + 2 \cdot 1))$

Subtract $1$ from $9$ .

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

Reduce the expression by cancelling the common factors.

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Factor $2$ out of $8$ .

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

Cancel the common factors.

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Factor $2$ out of $2$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Divide $4$ by $1$ .

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

Multiply $4$ by $4$ .

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

Multiply $- 2$ by $3$ .

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

Multiply $2$ by $1$ .

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

Add $- 6$ and $2$ .

$A (x) = \frac{1}{3 - 1} (16 - 4)$

Subtract $4$ from $16$ .

$A (x) = \frac{1}{3 - 1} (12)$

Subtract $1$ from $3$ .

$A (x) = \frac{1}{2} \cdot 12$

Cancel the common factor of $2$ .

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Write $12$ as a fraction with denominator $1$ .

$A (x) = \frac{1}{2} \cdot \frac{12}{1}$

Factor out the greatest common factor $2$ .

$A (x) = \frac{1}{2 \cdot 1} \cdot \frac{2 \cdot 6}{1}$

Cancel the common factor.

$A (x) = \frac{1}{2 \cdot 1} \cdot \frac{2 \cdot 6}{1}$

Rewrite the expression.

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

Simplify.

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Multiply $\frac{1}{1}$ and $\frac{6}{1}$ .

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

Divide $6$ by $1$ .

$A (x) = 6$

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