Algebra Examples

f (x) = {(x - 1)}^{3} + 2

$f (x) = {(x - 1)}^{3} + 2$

Step 1

The function

F (x)

$F (x)$ can be found by evaluating 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

Split the single integral into multiple integrals.

\int {(x - 1)}^{3} d x + \int 2 d x

$\int {(x - 1)}^{3} d x + \int 2 d x$

Step 3

Let

u = x - 1

$u = x - 1$ . Then

d u = d x

$d u = d x$ . Rewrite using

u

$u$ and

d

$d$

u

$u$ .

Tap for more steps...

Step 3.1

Let

u = x - 1

$u = x - 1$ . Find

\frac{d u}{d x}

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

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Step 3.1.1

Differentiate

x - 1

$x - 1$ .

\frac{d}{d x} [x - 1]

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

Step 3.1.2

By the Sum Rule, the derivative of

x - 1

$x - 1$ with respect to

x

$x$ is

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

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

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

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

Step 3.1.3

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}

$n x^{n - 1}$ where

n = 1

$n = 1$ .

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

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

Step 3.1.4

Since

- 1

$- 1$ is constant with respect to

x

$x$ , the derivative of

- 1

$- 1$ with respect to

x

$x$ is

0

$0$ .

1 + 0

$1 + 0$

Step 3.1.5

Add

1

$1$ and

0

$0$ .

1

$1$

1

$1$

Step 3.2

Rewrite the problem using

u

$u$ and

d u

$d u$ .

\int u^{3} d u + \int 2 d x

$\int u^{3} d u + \int 2 d x$

\int u^{3} d u + \int 2 d x

$\int u^{3} d u + \int 2 d x$

Step 4

By the Power Rule, the integral of

u^{3}

$u^{3}$ with respect to

u

$u$ is

\frac{1}{4} u^{4}

$\frac{1}{4} u^{4}$ .

\frac{1}{4} u^{4} + C + \int 2 d x

$\frac{1}{4} u^{4} + C + \int 2 d x$

Step 5

Apply the constant rule.

\frac{1}{4} u^{4} + C + 2 x + C

$\frac{1}{4} u^{4} + C + 2 x + C$

Step 6

Simplify.

\frac{1}{4} u^{4} + 2 x + C

$\frac{1}{4} u^{4} + 2 x + C$

Step 7

Replace all occurrences of

u

$u$ with

x - 1

$x - 1$ .

\frac{1}{4} {(x - 1)}^{4} + 2 x + C

$\frac{1}{4} {(x - 1)}^{4} + 2 x + C$

Step 8

The function

F

$F$ if derived from the integral of the derivative of the function. This is valid by the fundamental theorem of calculus.

F (x) = \frac{1}{4} \cdot {(x - 1)}^{4} + 2 x + C

$F (x) = \frac{1}{4} \cdot {(x - 1)}^{4} + 2 x + C$