Calculus Examples

$f (x) = \sin (x)$ , $[0, π]$

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

$f (x)$ is continuous on $[0, π]$ .

$f (x)$ 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.

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

Step 5

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

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

Step 6

Simplify the answer.

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Step 6.1

Evaluate $- \cos (x)$ at $π$ and at $0$ .

$A (x) = \frac{1}{π - 0} (- \cos (π) + \cos (0))$

Step 6.2

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

$A (x) = \frac{1}{π - 0} (- \cos (π) + 1)$

Step 6.3

Simplify.

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Step 6.3.1

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.

$A (x) = \frac{1}{π - 0} (\cos (0) + 1)$

Step 6.3.2

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

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

Step 6.3.3

Multiply $- 1$ by $1$ .

$A (x) = \frac{1}{π - 0} (1 + 1)$

Step 6.3.4

Multiply $- 1$ by $- 1$ .

$A (x) = \frac{1}{π - 0} (1 + 1)$

Step 6.3.5

Add $1$ and $1$ .

$A (x) = \frac{1}{π - 0} (2)$

Step 7

Simplify the denominator.

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Step 7.1

Multiply $- 1$ by $0$ .

$A (x) = \frac{1}{π + 0} \cdot 2$

Step 7.2

Add $π$ and $0$ .

$A (x) = \frac{1}{π} \cdot 2$

Step 8

Combine $\frac{1}{π}$ and $2$ .

$A (x) = \frac{2}{π}$

Step 9