Calculus Examples

$\int | 1 - x | d x$

Step 1

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

$1 - x = 0$

Step 2

Simplify the answer.

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Step 2.1

Solve the equation for $x$ .

$x = 1$

Step 2.2

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

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

Step 2.3

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

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

Step 2.4

Integrate the argument of the absolute value.

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Step 2.4.1

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

$\int 1 - x d x$

Step 2.4.2

Split the single integral into multiple integrals.

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

Step 2.4.3

Apply the constant rule.

$x + C + \int - x d x$

Step 2.4.4

Since $- 1$ is constant with respect to $x$ , move $- 1$ out of the integral.

$x + C - \int x d x$

Step 2.4.5

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

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

Step 2.4.6

Simplify.

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

Step 2.5

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

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

Step 2.6

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

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

Step 2.7

Simplify.

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

Step 2.8

Simplify.

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