Calculus Examples

$f (x) = \sqrt{x}$ , $(1, 1)$

Step 1

To find the average value of a function, the function should be continuous on the closed interval $[a, b]$ . To find whether $f (x)$ is continuous on $[1, 1]$ or not, find the domain of $f (x) = \sqrt{x}$ .

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Set the radicand in $\sqrt{x}$ greater than or equal to $0$ to find where the expression is defined.

$x \geq 0$

The domain is all values of $x$ that make the expression defined.

Interval Notation:

$[0, \infty)$

Set-Builder Notation:

${x | x \geq 0}$

Interval Notation:

$[0, \infty)$

Set-Builder Notation:

${x | x \geq 0}$

Step 2

$f (x)$ is continuous on $[1, 1]$ .

$f (x)$ 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.

$A (x) = \frac{1}{1 - 1} (\int_{1}^{1} \sqrt{x} d x)$

Step 5

Since the bounds are the same, the value of the definite integral is $0$ .

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

Step 6

Subtract $1$ from $1$ .

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

Step 7

The expression contains a division by $0$ . The expression is undefined.

Undefined