Calculus Examples

$\int_{0}^{\sqrt{15}} \frac{8 x}{\sqrt{x^{2} + 1}} d x$

Step 1

Since $8$ is constant with respect to $x$ , move $8$ out of the integral.

$8 \int_{0}^{\sqrt{15}} \frac{x}{\sqrt{x^{2} + 1}} d x$

Step 2

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 2.1

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

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Step 2.1.1

Differentiate $x^{2} + 1$ .

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

Step 2.1.2

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 2.1.3

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 2.1.4

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

$2 x + 0$

Step 2.1.5

Add $2 x$ and $0$ .

$2 x$

Step 2.2

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

$u_{lower} = 0^{2} + 1$

Step 2.3

Simplify.

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Step 2.3.1

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

$u_{lower} = 0 + 1$

Step 2.3.2

Add $0$ and $1$ .

$u_{lower} = 1$

Step 2.4

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

$u_{upper} = {\sqrt{15}}^{2} + 1$

Step 2.5

Simplify.

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Step 2.5.1

Rewrite ${\sqrt{15}}^{2}$ as $15$ .

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Step 2.5.1.1

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{15}$ as $15^{\frac{1}{2}}$ .

$u_{upper} = {(15^{\frac{1}{2}})}^{2} + 1$

Step 2.5.1.2

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

$u_{upper} = 15^{\frac{1}{2} \cdot 2} + 1$

Step 2.5.1.3

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

$u_{upper} = 15^{\frac{2}{2}} + 1$

Step 2.5.1.4

Cancel the common factor of $2$ .

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Step 2.5.1.4.1

Cancel the common factor.

$u_{upper} = 15^{\frac{2}{2}} + 1$

Step 2.5.1.4.2

Rewrite the expression.

$u_{upper} = 15^{1} + 1$

Step 2.5.1.5

Evaluate the exponent.

$u_{upper} = 15 + 1$

Step 2.5.2

Add $15$ and $1$ .

$u_{upper} = 16$

Step 2.6

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

$u_{lower} = 1$

$u_{upper} = 16$

Step 2.7

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

$8 \int_{1}^{16} \frac{1}{\sqrt{u}} \cdot \frac{1}{2} d u$

Step 3

Simplify.

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Step 3.1

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

$8 \int_{1}^{16} \frac{1}{\sqrt{u} \cdot 2} d u$

Step 3.2

Move $2$ to the left of $\sqrt{u}$ .

$8 \int_{1}^{16} \frac{1}{2 \sqrt{u}} d u$

Step 4

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

$8 (\frac{1}{2} \int_{1}^{16} \frac{1}{\sqrt{u}} d u)$

Step 5

Simplify the expression.

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Step 5.1

Simplify.

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Step 5.1.1

Combine $\frac{1}{2}$ and $8$ .

$\frac{8}{2} \int_{1}^{16} \frac{1}{\sqrt{u}} d u$

Step 5.1.2

Cancel the common factor of $8$ and $2$ .

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Step 5.1.2.1

Factor $2$ out of $8$ .

$\frac{2 \cdot 4}{2} \int_{1}^{16} \frac{1}{\sqrt{u}} d u$

Step 5.1.2.2

Cancel the common factors.

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Step 5.1.2.2.1

Factor $2$ out of $2$ .

$\frac{2 \cdot 4}{2 (1)} \int_{1}^{16} \frac{1}{\sqrt{u}} d u$

Step 5.1.2.2.2

Cancel the common factor.

$\frac{2 \cdot 4}{2 \cdot 1} \int_{1}^{16} \frac{1}{\sqrt{u}} d u$

Step 5.1.2.2.3

Rewrite the expression.

$\frac{4}{1} \int_{1}^{16} \frac{1}{\sqrt{u}} d u$

Step 5.1.2.2.4

Divide $4$ by $1$ .

$4 \int_{1}^{16} \frac{1}{\sqrt{u}} d u$

Step 5.2

Apply basic rules of exponents.

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Step 5.2.1

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

$4 \int_{1}^{16} \frac{1}{u^{\frac{1}{2}}} d u$

Step 5.2.2

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

$4 \int_{1}^{16} {(u^{\frac{1}{2}})}^{- 1} d u$

Step 5.2.3

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

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Step 5.2.3.1

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

$4 \int_{1}^{16} u^{\frac{1}{2} \cdot - 1} d u$

Step 5.2.3.2

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

$4 \int_{1}^{16} u^{\frac{- 1}{2}} d u$

Step 5.2.3.3

Move the negative in front of the fraction.

$4 \int_{1}^{16} u^{- \frac{1}{2}} d u$

Step 6

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

$4 (2 u^{\frac{1}{2}}]_{1}^{16})$

Step 7

Substitute and simplify.

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Step 7.1

Evaluate $2 u^{\frac{1}{2}}$ at $16$ and at $1$ .

$4 ((2 \cdot 16^{\frac{1}{2}}) - 2 \cdot 1^{\frac{1}{2}})$

Step 7.2

Simplify.

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Step 7.2.1

Rewrite $16$ as $4^{2}$ .

$4 (2 \cdot {(4^{2})}^{\frac{1}{2}} - 2 \cdot 1^{\frac{1}{2}})$

Step 7.2.2

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

$4 (2 \cdot 4^{2 (\frac{1}{2})} - 2 \cdot 1^{\frac{1}{2}})$

Step 7.2.3

Cancel the common factor of $2$ .

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Step 7.2.3.1

Cancel the common factor.

$4 (2 \cdot 4^{2 (\frac{1}{2})} - 2 \cdot 1^{\frac{1}{2}})$

Step 7.2.3.2

Rewrite the expression.

$4 (2 \cdot 4^{1} - 2 \cdot 1^{\frac{1}{2}})$

Step 7.2.4

Evaluate the exponent.

$4 (2 \cdot 4 - 2 \cdot 1^{\frac{1}{2}})$

Step 7.2.5

Multiply $2$ by $4$ .

$4 (8 - 2 \cdot 1^{\frac{1}{2}})$

Step 7.2.6

One to any power is one.

$4 (8 - 2 \cdot 1)$

Step 7.2.7

Multiply $- 2$ by $1$ .

$4 (8 - 2)$

Step 7.2.8

Subtract $2$ from $8$ .

$4 \cdot 6$

Step 7.2.9

Multiply $4$ by $6$ .

$24$

Step 8