Calculus Examples

$f (x) = \frac{1}{\sqrt{1 + x}}$ , $[0, 3]$

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 $[0, 3]$ or not, find the domain of $f (x) = \frac{1}{\sqrt{1 + x}}$ .

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Step 1.1

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

$1 + x \geq 0$

Step 1.2

Subtract $1$ from both sides of the inequality.

$x \geq - 1$

Step 1.3

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

$\sqrt{1 + x} = 0$

Step 1.4

Solve for $x$ .

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Step 1.4.1

To remove the radical on the left side of the equation, square both sides of the equation.

${\sqrt{1 + x}}^{2} = 0^{2}$

Step 1.4.2

Simplify each side of the equation.

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Step 1.4.2.1

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

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

Step 1.4.2.2

Simplify the left side.

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Step 1.4.2.2.1

Simplify ${({(1 + x)}^{\frac{1}{2}})}^{2}$ .

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Step 1.4.2.2.1.1

Multiply the exponents in ${({(1 + x)}^{\frac{1}{2}})}^{2}$ .

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Step 1.4.2.2.1.1.1

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

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

Step 1.4.2.2.1.1.2

Cancel the common factor of $2$ .

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Step 1.4.2.2.1.1.2.1

Cancel the common factor.

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

Step 1.4.2.2.1.1.2.2

Rewrite the expression.

${(1 + x)}^{1} = 0^{2}$

Step 1.4.2.2.1.2

Simplify.

$1 + x = 0^{2}$

Step 1.4.2.3

Simplify the right side.

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Step 1.4.2.3.1

Raising $0$ to any positive power yields $0$ .

$1 + x = 0$

Step 1.4.3

Subtract $1$ from both sides of the equation.

$x = - 1$

Step 1.5

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

Interval Notation:

$(- 1, \infty)$

Set-Builder Notation:

${x | x > - 1}$

Interval Notation:

$(- 1, \infty)$

Set-Builder Notation:

${x | x > - 1}$

Step 2

$f (x)$ is continuous on $[0, 3]$ .

$f (x)$ 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.

$A (x) = \frac{1}{3 - 0} (\int_{0}^{3} \frac{1}{\sqrt{1 + x}} d x)$

Step 5

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

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Step 5.1

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

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Step 5.1.1

Differentiate $1 + x$ .

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

Step 5.1.2

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

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

Step 5.1.3

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

$0 + \frac{d}{d x} [x]$

Step 5.1.4

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

$0 + 1$

Step 5.1.5

Add $0$ and $1$ .

$1$

Step 5.2

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

$u_{lower} = 1 + 0$

Step 5.3

Add $1$ and $0$ .

$u_{lower} = 1$

Step 5.4

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

$u_{upper} = 1 + 3$

Step 5.5

Add $1$ and $3$ .

$u_{upper} = 4$

Step 5.6

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

$u_{lower} = 1$

$u_{upper} = 4$

Step 5.7

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

$A (x) = \frac{1}{3 - 0} (\int_{1}^{4} \frac{1}{\sqrt{u}} d u)$

Step 6

Apply basic rules of exponents.

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Step 6.1

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

$A (x) = \frac{1}{3 - 0} (\int_{1}^{4} \frac{1}{u^{\frac{1}{2}}} d u)$

Step 6.2

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

$A (x) = \frac{1}{3 - 0} (\int_{1}^{4} {(u^{\frac{1}{2}})}^{- 1} d u)$

Step 6.3

Multiply the exponents in ${(u^{\frac{1}{2}})}^{- 1}$ .

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Step 6.3.1

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

$A (x) = \frac{1}{3 - 0} (\int_{1}^{4} u^{\frac{1}{2} \cdot - 1} d u)$

Step 6.3.2

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

$A (x) = \frac{1}{3 - 0} (\int_{1}^{4} u^{\frac{- 1}{2}} d u)$

Step 6.3.3

Move the negative in front of the fraction.

$A (x) = \frac{1}{3 - 0} (\int_{1}^{4} u^{- \frac{1}{2}} d u)$

Step 7

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

$A (x) = \frac{1}{3 - 0} (2 u^{\frac{1}{2}}]_{1}^{4})$

Step 8

Substitute and simplify.

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Step 8.1

Evaluate $2 u^{\frac{1}{2}}$ at $4$ and at $1$ .

$A (x) = \frac{1}{3 - 0} ((2 \cdot 4^{\frac{1}{2}}) - 2 \cdot 1^{\frac{1}{2}})$

Step 8.2

Simplify.

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Step 8.2.1

Rewrite $4$ as $2^{2}$ .

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

Step 8.2.2

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

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

Step 8.2.3

Cancel the common factor of $2$ .

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Step 8.2.3.1

Cancel the common factor.

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

Step 8.2.3.2

Rewrite the expression.

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

Step 8.2.4

Evaluate the exponent.

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

Step 8.2.5

Multiply $2$ by $2$ .

$A (x) = \frac{1}{3 - 0} (4 - 2 \cdot 1^{\frac{1}{2}})$

Step 8.2.6

One to any power is one.

$A (x) = \frac{1}{3 - 0} (4 - 2 \cdot 1)$

Step 8.2.7

Multiply $- 2$ by $1$ .

$A (x) = \frac{1}{3 - 0} (4 - 2)$

Step 8.2.8

Subtract $2$ from $4$ .

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

Step 9

Simplify the denominator.

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Step 9.1

Multiply $- 1$ by $0$ .

$A (x) = \frac{1}{3 + 0} \cdot 2$

Step 9.2

Add $3$ and $0$ .

$A (x) = \frac{1}{3} \cdot 2$

Step 10

Combine $\frac{1}{3}$ and $2$ .

$A (x) = \frac{2}{3}$

Step 11