Find the Function f(x)=|x-1|

$f (x) = | x - 1 |$

The function $F (x)$ can be found by evaluating the indefinite integral of the derivative $f (x)$ .

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

Set the argument in the absolute value equal to $0$ to find the potential values to split the solution at.

$x - 1 = 0$

Simplify the answer.

Solve the equation for $x$ .

$x = 1$

Create intervals around the solutions to find where $x - 1$ is positive and negative.

$(- \infty, 1), (1, \infty)$

Substitute a value from each interval into $x - 1$ to figure out where the expression is positive or negative.

$\begin{array}{c} Interval & Sign on interval \\ (- \infty, 1) & - \\ (1, \infty) & + \end{array}$

Integrate the argument of the absolute value.

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

$\int x - 1 d x$

Split the single integral into multiple integrals.

$\int x d x + \int - 1 d x$

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

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

Apply the constant rule.

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

Simplify.

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

On the intervals where the argument is negative, multiply the solution of the integral by $- 1$ .

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

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

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

Simplify.

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

Simplify.

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

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

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