Calculus Examples

$f (x) = \frac{5}{x^{2} + 1}$ ; $[- 1, 1]$

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 $f (x) = \frac{5}{x^{2} + 1}$ .

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Step 1.1

Set the denominator in $\frac{5}{x^{2} + 1}$ equal to $0$ to find where the expression is undefined.

$x^{2} + 1 = 0$

Step 1.2

Solve for $x$ .

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Step 1.2.1

Subtract $1$ from both sides of the equation.

$x^{2} = - 1$

Step 1.2.2

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

$x = \pm \sqrt{- 1}$

Step 1.2.3

Rewrite $\sqrt{- 1}$ as $i$ .

$x = \pm i$

Step 1.2.4

The complete solution is the result of both the positive and negative portions of the solution.

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Step 1.2.4.1

First, use the positive value of the $\pm$ to find the first solution.

$x = i$

Step 1.2.4.2

Next, use the negative value of the $\pm$ to find the second solution.

$x = - i$

Step 1.2.4.3

The complete solution is the result of both the positive and negative portions of the solution.

$x = i, - i$

Step 1.3

The domain is all real numbers.

Interval Notation:

$(- \infty, \infty)$

Set-Builder Notation:

${x | x \in ℝ}$

Interval Notation:

$(- \infty, \infty)$

Set-Builder Notation:

${x | x \in ℝ}$

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.

$A (x) = \frac{1}{1 + 1} (\int_{- 1}^{1} \frac{5}{x^{2} + 1} d x)$

Step 5

Since $5$ is constant with respect to $x$ , move $5$ out of the integral.

$A (x) = \frac{1}{1 + 1} (5 \int_{- 1}^{1} \frac{1}{x^{2} + 1} d x)$

Step 6

Simplify the expression.

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Step 6.1

Reorder $x^{2}$ and $1$ .

$A (x) = \frac{1}{1 + 1} (5 \int_{- 1}^{1} \frac{1}{1 + x^{2}} d x)$

Step 6.2

Rewrite $1$ as $1^{2}$ .

$A (x) = \frac{1}{1 + 1} (5 \int_{- 1}^{1} \frac{1}{1^{2} + x^{2}} d x)$

Step 7

The integral of $\frac{1}{1^{2} + x^{2}}$ with respect to $x$ is $\arctan (x)]_{- 1}^{1}$ .

$A (x) = \frac{1}{1 + 1} (5 (\arctan (x)]_{- 1}^{1}))$

Step 8

Evaluate $\arctan (x)$ at $1$ and at $- 1$ .

$A (x) = \frac{1}{1 + 1} (5 (\arctan (1) - \arctan (- 1)))$

Step 9

Simplify.

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Step 9.1

Simplify each term.

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Step 9.1.1

The exact value of $\arctan (1)$ is $\frac{π}{4}$ .

$A (x) = \frac{1}{1 + 1} (5 (\frac{π}{4} - \arctan (- 1)))$

Step 9.1.2

The exact value of $\arctan (- 1)$ is $- \frac{π}{4}$ .

$A (x) = \frac{1}{1 + 1} (5 (\frac{π}{4} + \frac{π}{4}))$

Step 9.1.3

Multiply $- - \frac{π}{4}$ .

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Step 9.1.3.1

Multiply $- 1$ by $- 1$ .

$A (x) = \frac{1}{1 + 1} (5 (\frac{π}{4} + 1 (\frac{π}{4})))$

Step 9.1.3.2

Multiply $\frac{π}{4}$ by $1$ .

$A (x) = \frac{1}{1 + 1} (5 (\frac{π}{4} + \frac{π}{4}))$

Step 9.2

Combine the numerators over the common denominator.

$A (x) = \frac{1}{1 + 1} (5 (\frac{π + π}{4}))$

Step 9.3

Add $π$ and $π$ .

$A (x) = \frac{1}{1 + 1} (5 (\frac{2 π}{4}))$

Step 9.4

Cancel the common factor of $2$ and $4$ .

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Step 9.4.1

Factor $2$ out of $2 π$ .

$A (x) = \frac{1}{1 + 1} (5 (\frac{2 (π)}{4}))$

Step 9.4.2

Cancel the common factors.

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Step 9.4.2.1

Factor $2$ out of $4$ .

$A (x) = \frac{1}{1 + 1} (5 (\frac{2 π}{2 \cdot 2}))$

Step 9.4.2.2

Cancel the common factor.

$A (x) = \frac{1}{1 + 1} (5 (\frac{2 π}{2 \cdot 2}))$

Step 9.4.2.3

Rewrite the expression.

$A (x) = \frac{1}{1 + 1} (5 (\frac{π}{2}))$

Step 9.5

Combine $5$ and $\frac{π}{2}$ .

$A (x) = \frac{1}{1 + 1} (\frac{5 π}{2})$

Step 10

Add $1$ and $1$ .

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

Step 11

Multiply $\frac{1}{2} \cdot \frac{5 π}{2}$ .

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Step 11.1

Multiply $\frac{1}{2}$ by $\frac{5 π}{2}$ .

$A (x) = \frac{5 π}{2 \cdot 2}$

Step 11.2

Multiply $2$ by $2$ .

$A (x) = \frac{5 π}{4}$

Step 12