Calculus Examples

$\int | 2 x - 1 | d x$

Step 1

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

$2 x - 1 = 0$

Step 2

Solve the equation for $x$ .

$x = \frac{1}{2}$

Step 3

Create intervals around the solutions to find where $2 x - 1$ is positive and negative.

$(- \infty, \frac{1}{2}), (\frac{1}{2}, \infty)$

Step 4

Substitute a value from each interval into $2 x - 1$ to figure out where the expression is positive or negative.

$\begin{array}{c} Interval & Sign on interval \\ (- \infty, \frac{1}{2}) & - \\ (\frac{1}{2}, \infty) & + \end{array}$

Step 5

Integrate the argument of the absolute value.

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Step 5.1

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

$\int 2 x - 1 d x$

Step 5.2

Split the single integral into multiple integrals.

$\int 2 x d x + \int - 1 d x$

Step 5.3

Since $2$ is constant with respect to $x$ , move $2$ out of the integral.

$2 \int x d x + \int - 1 d x$

Step 5.4

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

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

Step 5.5

Apply the constant rule.

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

Step 5.6

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

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

Step 5.7

Simplify.

$x^{2} - x + C$

Step 6

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

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