Algebra Examples

$f (x) = | x |$

Step 1

The function

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

$f (x)$ .

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

Step 2

Set the argument in the absolute value equal to

$0$ to find the potential values to split the solution at.

$x = 0$

Step 3

Simplify the answer.

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Step 3.1

Create intervals around the solutions to find where

$x$ is positive and negative.

$(- \infty, 0), (0, \infty)$

Step 3.2

Substitute a value from each interval into

$x$ to figure out where the expression is positive or negative.

$\begin{array}{c} Interval & Sign on interval \\ (- \infty, 0) & - \\ (0, \infty) & + \end{array}$

Step 3.3

Integrate the argument of the absolute value.

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Step 3.3.1

Set up the integral with the argument of the absolute value.

$\int x d x$

Step 3.3.2

By the Power Rule, the integral of

$x$ with respect to

$x$ is

$\frac{1}{2} x^{2}$ .

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

Step 3.4

On the intervals where the argument is negative, multiply the solution of the integral by

$- 1$ .

${\begin{cases} - (\frac{1}{2} x^{2} + C) & x \leq 0 \\ \frac{1}{2} x^{2} + C & x > 0 \end{cases}$

Step 3.5

Combine

$\frac{1}{2}$ and

$x^{2}$ .

${\begin{cases} - (\frac{x^{2}}{2} + C) & x \leq 0 \\ \frac{1}{2} x^{2} + C & x > 0 \end{cases}$

Step 3.6

Simplify.

${\begin{cases} - \frac{x^{2}}{2} & x \leq 0 \\ \frac{x^{2}}{2} & x > 0 \end{cases} + C$

Step 3.7

Simplify.

${\begin{cases} - \frac{1}{2} x^{2} & x \leq 0 \\ \frac{1}{2} x^{2} & x > 0 \end{cases} + C$

Step 4

The function

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

$F (x) = {\begin{cases} - \frac{1}{2} x^{2} & x \leq 0 \\ \frac{1}{2} x^{2} & x > 0 \end{cases} + C$