Calculus Examples

$f (t) = t e^{- t^{2}}$ , $[0, 7]$

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:

${t | t \in ℝ}$

Step 2

$f (t)$ is continuous on $[0, 7]$ .

$f (t)$ is continuous

Step 3

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

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

Step 4

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

$A (t) = \frac{1}{7 - 0} (\int_{0}^{7} t e^{- t^{2}} d t)$

Step 5

Let $u = - t^{2}$ . Then $d u = - 2 t d t$ , so $- \frac{1}{2} d u = t d t$ . Rewrite using $u$ and $d$ $u$ .

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Step 5.1

Let $u = - t^{2}$ . Find $\frac{d u}{d t}$ .

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Step 5.1.1

Differentiate $- t^{2}$ .

$\frac{d}{d t} [- t^{2}]$

Step 5.1.2

Since $- 1$ is constant with respect to $t$ , the derivative of $- t^{2}$ with respect to $t$ is $- \frac{d}{d t} [t^{2}]$ .

$- \frac{d}{d t} [t^{2}]$

Step 5.1.3

Differentiate using the Power Rule which states that $\frac{d}{d t} [t^{n}]$ is $n t^{n - 1}$ where $n = 2$ .

$- (2 t)$

Step 5.1.4

Multiply $2$ by $- 1$ .

$- 2 t$

Step 5.2

Substitute the lower limit in for $t$ in $u = - t^{2}$ .

$u_{lower} = - 0^{2}$

Step 5.3

Simplify.

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Step 5.3.1

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

$u_{lower} = - 0$

Step 5.3.2

Multiply $- 1$ by $0$ .

$u_{lower} = 0$

Step 5.4

Substitute the upper limit in for $t$ in $u = - t^{2}$ .

$u_{upper} = - 7^{2}$

Step 5.5

Simplify.

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Step 5.5.1

Raise $7$ to the power of $2$ .

$u_{upper} = - 1 \cdot 49$

Step 5.5.2

Multiply $- 1$ by $49$ .

$u_{upper} = - 49$

Step 5.6

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

$u_{lower} = 0$

$u_{upper} = - 49$

Step 5.7

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

$A (t) = \frac{1}{7 - 0} (\int_{0}^{- 49} e^{u} (\frac{1}{- 2}) d u)$

Step 6

Simplify.

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Step 6.1

Move the negative in front of the fraction.

$A (t) = \frac{1}{7 - 0} (\int_{0}^{- 49} e^{u} (- \frac{1}{2}) d u)$

Step 6.2

Combine $e^{u}$ and $\frac{1}{2}$ .

$A (t) = \frac{1}{7 - 0} (\int_{0}^{- 49} - \frac{e^{u}}{2} d u)$

Step 7

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

$A (t) = \frac{1}{7 - 0} (- \int_{0}^{- 49} \frac{e^{u}}{2} d u)$

Step 8

Since $\frac{1}{2}$ is constant with respect to $u$ , move $\frac{1}{2}$ out of the integral.

$A (t) = \frac{1}{7 - 0} (- (\frac{1}{2} \cdot \int_{0}^{- 49} e^{u} d u))$

Step 9

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

$A (t) = \frac{1}{7 - 0} (- \frac{1}{2} \cdot e^{u}]_{0}^{- 49})$

Step 10

Substitute and simplify.

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Step 10.1

Evaluate $e^{u}$ at $- 49$ and at $0$ .

$A (t) = \frac{1}{7 - 0} (- \frac{1}{2} \cdot ((e^{- 49}) - e^{0}))$

Step 10.2

Simplify.

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Step 10.2.1

Anything raised to $0$ is $1$ .

$A (t) = \frac{1}{7 - 0} (- \frac{1}{2} \cdot (e^{- 49} - 1 \cdot 1))$

Step 10.2.2

Multiply $- 1$ by $1$ .

$A (t) = \frac{1}{7 - 0} (- \frac{1}{2} \cdot (e^{- 49} - 1))$

Step 11

Simplify.

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Step 11.1

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

$A (t) = \frac{1}{7 - 0} (- \frac{1}{2} \cdot (\frac{1}{e^{49}} - 1))$

Step 11.2

Apply the distributive property.

$A (t) = \frac{1}{7 - 0} (- \frac{1}{2} \cdot \frac{1}{e^{49}} - \frac{1}{2} \cdot - 1)$

Step 11.3

Multiply $\frac{1}{e^{49}}$ by $\frac{1}{2}$ .

$A (t) = \frac{1}{7 - 0} (- \frac{1}{e^{49} \cdot 2} - \frac{1}{2} \cdot - 1)$

Step 11.4

Multiply $- \frac{1}{2} \cdot - 1$ .

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Step 11.4.1

Multiply $- 1$ by $- 1$ .

$A (t) = \frac{1}{7 - 0} (- \frac{1}{e^{49} \cdot 2} + 1 (\frac{1}{2}))$

Step 11.4.2

Multiply $\frac{1}{2}$ by $1$ .

$A (t) = \frac{1}{7 - 0} (- \frac{1}{e^{49} \cdot 2} + \frac{1}{2})$

Step 11.5

Move $2$ to the left of $e^{49}$ .

$A (t) = \frac{1}{7 - 0} (- \frac{1}{2 e^{49}} + \frac{1}{2})$

Step 12

Simplify the denominator.

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Step 12.1

Multiply $- 1$ by $0$ .

$A (t) = \frac{1}{7 + 0} \cdot (- \frac{1}{2 e^{49}} + \frac{1}{2})$

Step 12.2

Add $7$ and $0$ .

$A (t) = \frac{1}{7} \cdot (- \frac{1}{2 e^{49}} + \frac{1}{2})$

Step 13

Apply the distributive property.

$A (t) = \frac{1}{7} \cdot (- \frac{1}{2 e^{49}}) + \frac{1}{7} \cdot \frac{1}{2}$

Step 14

Multiply $\frac{1}{7} (- \frac{1}{2 e^{49}})$ .

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Step 14.1

Multiply $\frac{1}{7}$ by $\frac{1}{2 e^{49}}$ .

$A (t) = - \frac{1}{7 (2 e^{49})} + \frac{1}{7} \cdot \frac{1}{2}$

Step 14.2

Multiply $2$ by $7$ .

$A (t) = - \frac{1}{14 e^{49}} + \frac{1}{7} \cdot \frac{1}{2}$

Step 15

Multiply $\frac{1}{7} \cdot \frac{1}{2}$ .

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Step 15.1

Multiply $\frac{1}{7}$ by $\frac{1}{2}$ .

$A (t) = - \frac{1}{14 e^{49}} + \frac{1}{7 \cdot 2}$

Step 15.2

Multiply $7$ by $2$ .

$A (t) = - \frac{1}{14 e^{49}} + \frac{1}{14}$

Step 16