Calculus Examples

$h (x) = 6 \cos^{4} (x) \sin (x)$ , $[0, π]$

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

$h (x)$ is continuous on $[0, π]$ .

$h (x)$ is continuous

Step 3

The average value of function $h$ 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.

$A (x) = \frac{1}{π - 0} (\int_{0}^{π} 6 \cos^{4} (x) \sin (x) d x)$

Step 5

Since $6$ is constant with respect to $x$ , move $6$ out of the integral.

$A (x) = \frac{1}{π - 0} (6 \int_{0}^{π} \cos^{4} (x) \sin (x) d x)$

Step 6

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

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Step 6.1

Let $u = \cos (x)$ . Find $\frac{d u}{d x}$ .

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Step 6.1.1

Differentiate $\cos (x)$ .

$\frac{d}{d x} [\cos (x)]$

Step 6.1.2

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

$- \sin (x)$

Step 6.2

Substitute the lower limit in for $x$ in $u = \cos (x)$ .

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

Step 6.3

The exact value of $\cos (0)$ is $1$ .

$u_{lower} = 1$

Step 6.4

Substitute the upper limit in for $x$ in $u = \cos (x)$ .

$u_{upper} = \cos (π)$

Step 6.5

Simplify.

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Step 6.5.1

Apply the reference angle by finding the angle with equivalent trig values in the first quadrant. Make the expression negative because cosine is negative in the second quadrant.

$u_{upper} = - \cos (0)$

Step 6.5.2

The exact value of $\cos (0)$ is $1$ .

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

Step 6.5.3

Multiply $- 1$ by $1$ .

$u_{upper} = - 1$

Step 6.6

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

$u_{lower} = 1$

$u_{upper} = - 1$

Step 6.7

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

$A (x) = \frac{1}{π - 0} (6 \int_{1}^{- 1} - u^{4} d u)$

Step 7

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

$A (x) = \frac{1}{π - 0} (6 (- \int_{1}^{- 1} u^{4} d u))$

Step 8

Multiply $- 1$ by $6$ .

$A (x) = \frac{1}{π - 0} (- 6 \int_{1}^{- 1} u^{4} d u)$

Step 9

By the Power Rule, the integral of $u^{4}$ with respect to $u$ is $\frac{1}{5} u^{5}$ .

$A (x) = \frac{1}{π - 0} (- 6 (\frac{1}{5} u^{5}]_{1}^{- 1}))$

Step 10

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

$A (x) = \frac{1}{π - 0} (- 6 (\frac{u^{5}}{5}]_{1}^{- 1}))$

Step 11

Substitute and simplify.

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Step 11.1

Evaluate $\frac{u^{5}}{5}$ at $- 1$ and at $1$ .

$A (x) = \frac{1}{π - 0} (- 6 ((\frac{{(- 1)}^{5}}{5}) - \frac{1^{5}}{5}))$

Step 11.2

Simplify.

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Step 11.2.1

Raise $- 1$ to the power of $5$ .

$A (x) = \frac{1}{π - 0} (- 6 (\frac{- 1}{5} - \frac{1^{5}}{5}))$

Step 11.2.2

Move the negative in front of the fraction.

$A (x) = \frac{1}{π - 0} (- 6 (- \frac{1}{5} - \frac{1^{5}}{5}))$

Step 11.2.3

One to any power is one.

$A (x) = \frac{1}{π - 0} (- 6 (- \frac{1}{5} - \frac{1}{5}))$

Step 11.2.4

Subtract $\frac{1}{5}$ from $- \frac{1}{5}$ .

$A (x) = \frac{1}{π - 0} (- 6 (- 2 (\frac{1}{5})))$

Step 11.2.5

Combine $- 2$ and $\frac{1}{5}$ .

$A (x) = \frac{1}{π - 0} (- 6 (\frac{- 2}{5}))$

Step 11.2.6

Move the negative in front of the fraction.

$A (x) = \frac{1}{π - 0} (- 6 (- \frac{2}{5}))$

Step 11.2.7

Multiply $- 1$ by $- 6$ .

$A (x) = \frac{1}{π - 0} (6 (\frac{2}{5}))$

Step 11.2.8

Combine $6$ and $\frac{2}{5}$ .

$A (x) = \frac{1}{π - 0} (\frac{6 \cdot 2}{5})$

Step 11.2.9

Multiply $6$ by $2$ .

$A (x) = \frac{1}{π - 0} (\frac{12}{5})$

Step 12

Simplify the denominator.

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Step 12.1

Multiply $- 1$ by $0$ .

$A (x) = \frac{1}{π + 0} \cdot \frac{12}{5}$

Step 12.2

Add $π$ and $0$ .

$A (x) = \frac{1}{π} \cdot \frac{12}{5}$

Step 13

Combine fractions.

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Step 13.1

Multiply $\frac{1}{π}$ by $\frac{12}{5}$ .

$A (x) = \frac{12}{π \cdot 5}$

Step 13.2

Move $5$ to the left of $π$ .

$A (x) = \frac{12}{5 π}$

Step 14