Calculus Examples

$f (x) = \sqrt{x + 1}$ , $[0, 24]$

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 $[0, 24]$ or not, find the domain of $f (x) = \sqrt{x + 1}$ .

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Step 1.1

Set the radicand in $\sqrt{x + 1}$ greater than or equal to $0$ to find where the expression is defined.

$x + 1 \geq 0$

Step 1.2

Subtract $1$ from both sides of the inequality.

$x \geq - 1$

Step 1.3

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

Interval Notation:

$[- 1, \infty)$

Set-Builder Notation:

${x | x \geq - 1}$

Interval Notation:

$[- 1, \infty)$

Set-Builder Notation:

${x | x \geq - 1}$

Step 2

$f (x)$ is continuous on $[0, 24]$ .

$f (x)$ 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.

$A (x) = \frac{1}{24 - 0} (\int_{0}^{24} \sqrt{x + 1} d x)$

Step 5

Let $u = x + 1$ . Then $d u = d x$ . Rewrite using $u$ and $d$ $u$ .

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Step 5.1

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

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Step 5.1.1

Differentiate $x + 1$ .

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

Step 5.1.2

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

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

Step 5.1.3

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

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

Step 5.1.4

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

$1 + 0$

Step 5.1.5

Add $1$ and $0$ .

$1$

Step 5.2

Substitute the lower limit in for $x$ in $u = x + 1$ .

$u_{lower} = 0 + 1$

Step 5.3

Add $0$ and $1$ .

$u_{lower} = 1$

Step 5.4

Substitute the upper limit in for $x$ in $u = x + 1$ .

$u_{upper} = 24 + 1$

Step 5.5

Add $24$ and $1$ .

$u_{upper} = 25$

Step 5.6

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

$u_{lower} = 1$

$u_{upper} = 25$

Step 5.7

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

$A (x) = \frac{1}{24 - 0} (\int_{1}^{25} \sqrt{u} d u)$

Step 6

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{u}$ as $u^{\frac{1}{2}}$ .

$A (x) = \frac{1}{24 - 0} (\int_{1}^{25} u^{\frac{1}{2}} d u)$

Step 7

By the Power Rule, the integral of $u^{\frac{1}{2}}$ with respect to $u$ is $\frac{2}{3} u^{\frac{3}{2}}$ .

$A (x) = \frac{1}{24 - 0} (\frac{2}{3} u^{\frac{3}{2}}]_{1}^{25})$

Step 8

Substitute and simplify.

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Step 8.1

Evaluate $\frac{2}{3} u^{\frac{3}{2}}$ at $25$ and at $1$ .

$A (x) = \frac{1}{24 - 0} ((\frac{2}{3} \cdot 25^{\frac{3}{2}}) - \frac{2}{3} \cdot 1^{\frac{3}{2}})$

Step 8.2

Simplify.

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Step 8.2.1

Rewrite $25$ as $5^{2}$ .

$A (x) = \frac{1}{24 - 0} (\frac{2}{3} \cdot {(5^{2})}^{\frac{3}{2}} - \frac{2}{3} \cdot 1^{\frac{3}{2}})$

Step 8.2.2

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

$A (x) = \frac{1}{24 - 0} (\frac{2}{3} \cdot 5^{2 (\frac{3}{2})} - \frac{2}{3} \cdot 1^{\frac{3}{2}})$

Step 8.2.3

Cancel the common factor of $2$ .

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Step 8.2.3.1

Cancel the common factor.

$A (x) = \frac{1}{24 - 0} (\frac{2}{3} \cdot 5^{2 (\frac{3}{2})} - \frac{2}{3} \cdot 1^{\frac{3}{2}})$

Step 8.2.3.2

Rewrite the expression.

$A (x) = \frac{1}{24 - 0} (\frac{2}{3} \cdot 5^{3} - \frac{2}{3} \cdot 1^{\frac{3}{2}})$

Step 8.2.4

Raise $5$ to the power of $3$ .

$A (x) = \frac{1}{24 - 0} (\frac{2}{3} \cdot 125 - \frac{2}{3} \cdot 1^{\frac{3}{2}})$

Step 8.2.5

Combine $\frac{2}{3}$ and $125$ .

$A (x) = \frac{1}{24 - 0} (\frac{2 \cdot 125}{3} - \frac{2}{3} \cdot 1^{\frac{3}{2}})$

Step 8.2.6

Multiply $2$ by $125$ .

$A (x) = \frac{1}{24 - 0} (\frac{250}{3} - \frac{2}{3} \cdot 1^{\frac{3}{2}})$

Step 8.2.7

One to any power is one.

$A (x) = \frac{1}{24 - 0} (\frac{250}{3} - \frac{2}{3} \cdot 1)$

Step 8.2.8

Multiply $- 1$ by $1$ .

$A (x) = \frac{1}{24 - 0} (\frac{250}{3} - \frac{2}{3})$

Step 8.2.9

Combine the numerators over the common denominator.

$A (x) = \frac{1}{24 - 0} (\frac{250 - 2}{3})$

Step 8.2.10

Subtract $2$ from $250$ .

$A (x) = \frac{1}{24 - 0} (\frac{248}{3})$

Step 9

Simplify the denominator.

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Step 9.1

Multiply $- 1$ by $0$ .

$A (x) = \frac{1}{24 + 0} \cdot \frac{248}{3}$

Step 9.2

Add $24$ and $0$ .

$A (x) = \frac{1}{24} \cdot \frac{248}{3}$

Step 10

Simplify terms.

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Step 10.1

Cancel the common factor of $8$ .

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Step 10.1.1

Factor $8$ out of $24$ .

$A (x) = \frac{1}{8 (3)} \cdot \frac{248}{3}$

Step 10.1.2

Factor $8$ out of $248$ .

$A (x) = \frac{1}{8 \cdot 3} \cdot \frac{8 \cdot 31}{3}$

Step 10.1.3

Cancel the common factor.

$A (x) = \frac{1}{8 \cdot 3} \cdot \frac{8 \cdot 31}{3}$

Step 10.1.4

Rewrite the expression.

$A (x) = \frac{1}{3} \cdot \frac{31}{3}$

Step 10.2

Multiply $\frac{1}{3}$ by $\frac{31}{3}$ .

$A (x) = \frac{31}{3 \cdot 3}$

Step 10.3

Multiply $3$ by $3$ .

$A (x) = \frac{31}{9}$

Step 11