Calculus Examples

$f (x) = \frac{1}{x^{2}}$ , $(1, 2)$

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 $[1, 2]$ or not, find the domain of $f (x) = \frac{1}{x^{2}}$ .

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Step 1.1

Set the denominator in $\frac{1}{x^{2}}$ equal to $0$ to find where the expression is undefined.

$x^{2} = 0$

Step 1.2

Solve for $x$ .

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Step 1.2.1

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

$x = \pm \sqrt{0}$

Step 1.2.2

Simplify $\pm \sqrt{0}$ .

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Step 1.2.2.1

Rewrite $0$ as $0^{2}$ .

$x = \pm \sqrt{0^{2}}$

Step 1.2.2.2

Pull terms out from under the radical, assuming positive real numbers.

$x = \pm 0$

Step 1.2.2.3

Plus or minus $0$ is $0$ .

$x = 0$

Step 1.3

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

Interval Notation:

$(- \infty, 0) \cup (0, \infty)$

Set-Builder Notation:

${x | x \neq 0}$

Interval Notation:

$(- \infty, 0) \cup (0, \infty)$

Set-Builder Notation:

${x | x \neq 0}$

Step 2

$f (x)$ is continuous on $[1, 2]$ .

$f (x)$ 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.

$A (x) = \frac{1}{2 - 1} (\int_{1}^{2} \frac{1}{x^{2}} d x)$

Step 5

Apply basic rules of exponents.

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Step 5.1

Move $x^{2}$ out of the denominator by raising it to the $- 1$ power.

$A (x) = \frac{1}{2 - 1} (\int_{1}^{2} {(x^{2})}^{- 1} d x)$

Step 5.2

Multiply the exponents in ${(x^{2})}^{- 1}$ .

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Step 5.2.1

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$A (x) = \frac{1}{2 - 1} (\int_{1}^{2} x^{2 \cdot - 1} d x)$

Step 5.2.2

Multiply $2$ by $- 1$ .

$A (x) = \frac{1}{2 - 1} (\int_{1}^{2} x^{- 2} d x)$

Step 6

By the Power Rule, the integral of $x^{- 2}$ with respect to $x$ is $- x^{- 1}$ .

$A (x) = \frac{1}{2 - 1} (- x^{- 1}]_{1}^{2})$

Step 7

Substitute and simplify.

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Step 7.1

Evaluate $- x^{- 1}$ at $2$ and at $1$ .

$A (x) = \frac{1}{2 - 1} ((- 2^{- 1}) + 1^{- 1})$

Step 7.2

Simplify.

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Step 7.2.1

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

$A (x) = \frac{1}{2 - 1} (- \frac{1}{2} + 1^{- 1})$

Step 7.2.2

One to any power is one.

$A (x) = \frac{1}{2 - 1} (- \frac{1}{2} + 1)$

Step 7.2.3

Write $1$ as a fraction with a common denominator.

$A (x) = \frac{1}{2 - 1} (- \frac{1}{2} + \frac{2}{2})$

Step 7.2.4

Combine the numerators over the common denominator.

$A (x) = \frac{1}{2 - 1} (\frac{- 1 + 2}{2})$

Step 7.2.5

Add $- 1$ and $2$ .

$A (x) = \frac{1}{2 - 1} (\frac{1}{2})$

Step 8

Subtract $1$ from $2$ .

$A (x) = \frac{1}{1} \cdot \frac{1}{2}$

Step 9

Cancel the common factor of $1$ .

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Step 9.1

Cancel the common factor.

$A (x) = \frac{1}{1} \cdot \frac{1}{2}$

Step 9.2

Rewrite the expression.

$A (x) = 1 \cdot \frac{1}{2}$

Step 10

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

$A (x) = \frac{1}{2}$

Step 11