Calculus Examples

$f (t) = e^{\sin (t)} \cos (t)$ , $[0, \frac{π}{2}]$

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 $[0, \frac{π}{2}]$ .

$f (t)$ 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.

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

Step 5

Let $u = \sin (t)$ . Then $d u = \cos (t) d t$ , so $\frac{1}{\cos (t)} d u = d t$ . Rewrite using $u$ and $d$ $u$ .

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Step 5.1

Let $u = \sin (t)$ . Find $\frac{d u}{d t}$ .

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Step 5.1.1

Differentiate $\sin (t)$ .

$\frac{d}{d t} [\sin (t)]$

Step 5.1.2

The derivative of $\sin (t)$ with respect to $t$ is $\cos (t)$ .

$\cos (t)$

Step 5.2

Substitute the lower limit in for $t$ in $u = \sin (t)$ .

$u_{lower} = \sin (0)$

Step 5.3

The exact value of $\sin (0)$ is $0$ .

$u_{lower} = 0$

Step 5.4

Substitute the upper limit in for $t$ in $u = \sin (t)$ .

$u_{upper} = \sin (\frac{π}{2})$

Step 5.5

The exact value of $\sin (\frac{π}{2})$ is $1$ .

$u_{upper} = 1$

Step 5.6

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

$u_{lower} = 0$

$u_{upper} = 1$

Step 5.7

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

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

Step 6

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

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

Step 7

Substitute and simplify.

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Step 7.1

Evaluate $e^{u}$ at $1$ and at $0$ .

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

Step 7.2

Simplify.

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Step 7.2.1

Simplify.

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

Step 7.2.2

Anything raised to $0$ is $1$ .

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

Step 7.2.3

Multiply $- 1$ by $1$ .

$A (t) = \frac{1}{\frac{π}{2} - 0} (e - 1)$

Step 8

Simplify the denominator.

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Step 8.1

Multiply $- 1$ by $0$ .

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

Step 8.2

Add $\frac{π}{2}$ and $0$ .

$A (t) = \frac{1}{\frac{π}{2}} \cdot (e - 1)$

Step 9

Multiply the numerator by the reciprocal of the denominator.

$A (t) = 1 (\frac{2}{π}) \cdot (e - 1)$

Step 10

Multiply $\frac{2}{π}$ by $1$ .

$A (t) = \frac{2}{π} \cdot (e - 1)$

Step 11

Apply the distributive property.

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

Step 12

Combine $\frac{2}{π}$ and $e$ .

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

Step 13

Multiply $\frac{2}{π} \cdot - 1$ .

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Step 13.1

Combine $\frac{2}{π}$ and $- 1$ .

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

Step 13.2

Multiply $2$ by $- 1$ .

$A (t) = \frac{2 e}{π} + \frac{- 2}{π}$

Step 14

Move the negative in front of the fraction.

$A (t) = \frac{2 e}{π} - \frac{2}{π}$

Step 15