Calculus Examples

$y = e^{- x}$ ; $[0, 5]$

Step 1

Write $y = e^{- x}$ as a function.

$f (x) = e^{- x}$

Step 2

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 3

$f (x)$ is continuous on $[0, 5]$ .

$f (x)$ is continuous

Step 4

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 5

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 6

Let $u = - x$ . Then $d u = - d x$ , so $- d u = d x$ . Rewrite using $u$ and $d$ $u$ .

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Step 6.1

Let $u = - x$ . Find $\frac{d u}{d x}$ .

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Step 6.1.1

Differentiate $- x$ .

$\frac{d}{d x} [- x]$

Step 6.1.2

Since $- 1$ is constant with respect to $x$ , the derivative of $- x$ with respect to $x$ is $- \frac{d}{d x} [x]$ .

$- \frac{d}{d x} [x]$

Step 6.1.3

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 1$ .

$- 1 \cdot 1$

Step 6.1.4

Multiply $- 1$ by $1$ .

$- 1$

Step 6.2

Substitute the lower limit in for $x$ in $u = - x$ .

$u_{lower} = - 0$

Step 6.3

Multiply $- 1$ by $0$ .

$u_{lower} = 0$

Step 6.4

Substitute the upper limit in for $x$ in $u = - x$ .

$u_{upper} = - 1 \cdot 5$

Step 6.5

Multiply $- 1$ by $5$ .

$u_{upper} = - 5$

Step 6.6

The values found for $u_{lower}$ and $u_{upper}$ will be used to evaluate the definite integral.

$u_{lower} = 0$

$u_{upper} = - 5$

Step 6.7

Rewrite the problem using $u$ , $d u$ , and the new limits of integration.

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

Step 7

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

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

Step 8

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

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

Step 9

Substitute and simplify.

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Step 9.1

Evaluate $e^{u}$ at $- 5$ and at $0$ .

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

Step 9.2

Simplify.

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Step 9.2.1

Anything raised to $0$ is $1$ .

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

Step 9.2.2

Multiply $- 1$ by $1$ .

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

Step 10

Simplify.

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Step 10.1

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

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

Step 10.2

Apply the distributive property.

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

Step 10.3

Multiply $- 1$ by $- 1$ .

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

Step 11

Simplify the denominator.

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Step 11.1

Multiply $- 1$ by $0$ .

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

Step 11.2

Add $5$ and $0$ .

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

Step 12

Simplify terms.

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Step 12.1

Apply the distributive property.

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

Step 12.2

Multiply $\frac{1}{5}$ by $\frac{1}{e^{5}}$ .

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

Step 12.3

Multiply $\frac{1}{5}$ by $1$ .

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

Step 13