Calculus Examples

$f (x) = \frac{15 x}{x^{2} + 1}$ , $[- 2, 2]$

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 $[- 2, 2]$ or not, find the domain of $f (x) = \frac{15 x}{x^{2} + 1}$ .

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Step 1.1

Set the denominator in $\frac{15 x}{x^{2} + 1}$ 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 $[- 2, 2]$ .

$f (x)$ 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.

$A (x) = \frac{1}{2 + 2} (\int_{- 2}^{2} \frac{15 x}{x^{2} + 1} d x)$

Step 5

Since $15$ is constant with respect to $x$ , move $15$ out of the integral.

$A (x) = \frac{1}{2 + 2} (15 \int_{- 2}^{2} \frac{x}{x^{2} + 1} d x)$

Step 6

Let $u = x^{2} + 1$ . Then $d u = 2 x d x$ , so $\frac{1}{2} d u = x d x$ . Rewrite using $u$ and $d$ $u$ .

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Step 6.1

Let $u = x^{2} + 1$ . Find $\frac{d u}{d x}$ .

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Step 6.1.1

Differentiate $x^{2} + 1$ .

$\frac{d}{d x} [x^{2} + 1]$

Step 6.1.2

By the Sum Rule, the derivative of $x^{2} + 1$ with respect to $x$ is $\frac{d}{d x} [x^{2}] + \frac{d}{d x} [1]$ .

$\frac{d}{d x} [x^{2}] + \frac{d}{d x} [1]$

Step 6.1.3

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 2$ .

$2 x + \frac{d}{d x} [1]$

Step 6.1.4

Since $1$ is constant with respect to $x$ , the derivative of $1$ with respect to $x$ is $0$ .

$2 x + 0$

Step 6.1.5

Add $2 x$ and $0$ .

$2 x$

Step 6.2

Substitute the lower limit in for $x$ in $u = x^{2} + 1$ .

$u_{lower} = {(- 2)}^{2} + 1$

Step 6.3

Simplify.

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Step 6.3.1

Raise $- 2$ to the power of $2$ .

$u_{lower} = 4 + 1$

Step 6.3.2

Add $4$ and $1$ .

$u_{lower} = 5$

Step 6.4

Substitute the upper limit in for $x$ in $u = x^{2} + 1$ .

$u_{upper} = 2^{2} + 1$

Step 6.5

Simplify.

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Step 6.5.1

Raise $2$ to the power of $2$ .

$u_{upper} = 4 + 1$

Step 6.5.2

Add $4$ and $1$ .

$u_{upper} = 5$

Step 6.6

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

$u_{lower} = 5$

$u_{upper} = 5$

Step 6.7

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

$A (x) = \frac{1}{2 + 2} (15 \int_{5}^{5} \frac{1}{u} \cdot \frac{1}{2} d u)$

Step 7

Simplify.

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Step 7.1

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

$A (x) = \frac{1}{2 + 2} (15 \int_{5}^{5} \frac{1}{u \cdot 2} d u)$

Step 7.2

Move $2$ to the left of $u$ .

$A (x) = \frac{1}{2 + 2} (15 \int_{5}^{5} \frac{1}{2 u} d u)$

Step 8

Since $\frac{1}{2}$ is constant with respect to $u$ , move $\frac{1}{2}$ out of the integral.

$A (x) = \frac{1}{2 + 2} (15 (\frac{1}{2} \cdot \int_{5}^{5} \frac{1}{u} d u))$

Step 9

Combine $\frac{1}{2}$ and $15$ .

$A (x) = \frac{1}{2 + 2} (\frac{15}{2} \cdot \int_{5}^{5} \frac{1}{u} d u)$

Step 10

The integral of $\frac{1}{u}$ with respect to $u$ is $\ln (| u |)$ .

$A (x) = \frac{1}{2 + 2} (\frac{15}{2} \cdot \ln (| u |)]_{5}^{5})$

Step 11

Substitute and simplify.

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Step 11.1

Evaluate $\ln (| u |)$ at $5$ and at $5$ .

$A (x) = \frac{1}{2 + 2} (\frac{15}{2} \cdot ((\ln (| 5 |)) - \ln (| 5 |)))$

Step 11.2

Simplify.

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Step 11.2.1

Subtract $\ln (| 5 |)$ from $\ln (| 5 |)$ .

$A (x) = \frac{1}{2 + 2} (\frac{15}{2} \cdot 0)$

Step 11.2.2

Multiply $\frac{15}{2}$ by $0$ .

$A (x) = \frac{1}{2 + 2} (0)$

Step 12

Add $2$ and $2$ .

$A (x) = \frac{1}{4} \cdot 0$

Step 13

Multiply $\frac{1}{4}$ by $0$ .

$A (x) = 0$

Step 14