Calculus Examples

$f (x) = e^{x}$ , $[0, 5]$

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 $[0, 5]$ .

$f (x)$ 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.

$A (x) = \frac{1}{5 - 0} (\int_{0}^{5} e^{x} d x)$

Step 5

The integral of $e^{x}$ with respect to $x$ is $e^{x}$ .

$A (x) = \frac{1}{5 - 0} (e^{x}]_{0}^{5})$

Step 6

Substitute and simplify.

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Step 6.1

Evaluate $e^{x}$ at $5$ and at $0$ .

$A (x) = \frac{1}{5 - 0} ((e^{5}) - e^{0})$

Step 6.2

Simplify.

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Step 6.2.1

Anything raised to $0$ is $1$ .

$A (x) = \frac{1}{5 - 0} (e^{5} - 1 \cdot 1)$

Step 6.2.2

Multiply $- 1$ by $1$ .

$A (x) = \frac{1}{5 - 0} (e^{5} - 1)$

Step 7

Simplify the denominator.

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Step 7.1

Multiply $- 1$ by $0$ .

$A (x) = \frac{1}{5 + 0} \cdot (e^{5} - 1)$

Step 7.2

Add $5$ and $0$ .

$A (x) = \frac{1}{5} \cdot (e^{5} - 1)$

Step 8

Simplify terms.

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Step 8.1

Apply the distributive property.

$A (x) = \frac{1}{5} \cdot e^{5} + \frac{1}{5} \cdot - 1$

Step 8.2

Combine $\frac{1}{5}$ and $e^{5}$ .

$A (x) = \frac{e^{5}}{5} + \frac{1}{5} \cdot - 1$

Step 8.3

Combine $\frac{1}{5}$ and $- 1$ .

$A (x) = \frac{e^{5}}{5} + \frac{- 1}{5}$

Step 8.4

Move the negative in front of the fraction.

$A (x) = \frac{e^{5}}{5} - \frac{1}{5}$

Step 9