Calculus Examples

f (x) = {(\sqrt{2} - 6 x)}^{5}

$f (x) = {(\sqrt{2} - 6 x)}^{5}$

Step 1

The function

F (x)

$F (x)$ can be found by finding the indefinite integral of the derivative

f (x)

$f (x)$ .

F (x) = \int f (x) d x

$F (x) = \int f (x) d x$

Step 2

Set up the integral to solve.

F (x) = \int {(\sqrt{2} - 6 x)}^{5} d x

$F (x) = \int {(\sqrt{2} - 6 x)}^{5} d x$

Step 3

Let

u = \sqrt{2} - 6 x

$u = \sqrt{2} - 6 x$ . Then

d u = - 6 d x

$d u = - 6 d x$ , so

- \frac{1}{6} d u = d x

$- \frac{1}{6} d u = d x$ . Rewrite using

u

$u$ and

d

$d$

u

$u$ .

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Step 3.1

Let

u = \sqrt{2} - 6 x

$u = \sqrt{2} - 6 x$ . Find

\frac{d u}{d x}

$\frac{d u}{d x}$ .

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Step 3.1.1

Differentiate

\sqrt{2} - 6 x

$\sqrt{2} - 6 x$ .

\frac{d}{d x} [\sqrt{2} - 6 x]

$\frac{d}{d x} [\sqrt{2} - 6 x]$

Step 3.1.2

Differentiate.

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Step 3.1.2.1

By the Sum Rule, the derivative of

\sqrt{2} - 6 x

$\sqrt{2} - 6 x$ with respect to

x

$x$ is

\frac{d}{d x} [\sqrt{2}] + \frac{d}{d x} [- 6 x]

$\frac{d}{d x} [\sqrt{2}] + \frac{d}{d x} [- 6 x]$ .

\frac{d}{d x} [\sqrt{2}] + \frac{d}{d x} [- 6 x]

$\frac{d}{d x} [\sqrt{2}] + \frac{d}{d x} [- 6 x]$

Step 3.1.2.2

Since

\sqrt{2}

$\sqrt{2}$ is constant with respect to

x

$x$ , the derivative of

\sqrt{2}

$\sqrt{2}$ with respect to

x

$x$ is

0

$0$ .

0 + \frac{d}{d x} [- 6 x]

$0 + \frac{d}{d x} [- 6 x]$

0 + \frac{d}{d x} [- 6 x]

$0 + \frac{d}{d x} [- 6 x]$

Step 3.1.3

Evaluate

\frac{d}{d x} [- 6 x]

$\frac{d}{d x} [- 6 x]$ .

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Step 3.1.3.1

Since

- 6

$- 6$ is constant with respect to

x

$x$ , the derivative of

- 6 x

$- 6 x$ with respect to

x

$x$ is

- 6 \frac{d}{d x} [x]

$- 6 \frac{d}{d x} [x]$ .

0 - 6 \frac{d}{d x} [x]

$0 - 6 \frac{d}{d x} [x]$

Step 3.1.3.2

Differentiate using the Power Rule which states that

\frac{d}{d x} [x^{n}]

$\frac{d}{d x} [x^{n}]$ is

$n x^{n - 1}$ where

$n = 1$ .

$0 - 6 \cdot 1$

Step 3.1.3.3

Multiply

$- 6$ by

$1$ .

$0 - 6$

$0 - 6$

Step 3.1.4

Subtract

$6$ from

$0$ .

$- 6$

$- 6$

Step 3.2

Rewrite the problem using

$u$ and

$d u$ .

$\int u^{5} \frac{1}{- 6} d u$

$\int u^{5} \frac{1}{- 6} d u$

Step 4

Simplify.

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Step 4.1

Move the negative in front of the fraction.

$\int u^{5} (- \frac{1}{6}) d u$

Step 4.2

Combine

$u^{5}$ and

$\frac{1}{6}$ .

$\int - \frac{u^{5}}{6} d u$

$\int - \frac{u^{5}}{6} d u$

Step 5

Since

$- 1$ is constant with respect to

$u$ , move

$- 1$ out of the integral.

$- \int \frac{u^{5}}{6} d u$

Step 6

Since

$\frac{1}{6}$ is constant with respect to

$u$ , move

$\frac{1}{6}$ out of the integral.

$- (\frac{1}{6} \int u^{5} d u)$

Step 7

By the Power Rule, the integral of

$u^{5}$ with respect to

$u$ is

$\frac{1}{6} u^{6}$ .

$- \frac{1}{6} (\frac{1}{6} u^{6} + C)$

Step 8

Simplify.

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Step 8.1

Rewrite

$- \frac{1}{6} (\frac{1}{6} u^{6} + C)$ as

$- \frac{1}{6} \cdot \frac{1}{6} u^{6} + C$ .

$- \frac{1}{6} \cdot \frac{1}{6} u^{6} + C$

Step 8.2

Simplify.

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Step 8.2.1

Multiply

$\frac{1}{6}$ by

$\frac{1}{6}$ .

$- \frac{1}{6 \cdot 6} u^{6} + C$

Step 8.2.2

Multiply

$6$ by

$6$ .

$- \frac{1}{36} u^{6} + C$

$- \frac{1}{36} u^{6} + C$

$- \frac{1}{36} u^{6} + C$

Step 9

Replace all occurrences of

$u$ with

$\sqrt{2} - 6 x$ .

$- \frac{1}{36} {(\sqrt{2} - 6 x)}^{6} + C$

Step 10

The answer is the antiderivative of the function

$f (x) = {(\sqrt{2} - 6 x)}^{5}$ .

$F (x) =$

$- \frac{1}{36} {(\sqrt{2} - 6 x)}^{6} + C$

$[\begin{array}{c} x^{2} & \frac{1}{2} \\ \sqrt{π} & \int x d x \end{array}]$