Algebra Examples

$f (x) = | - 2 x - 3 |$

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.

$- 2 x - 3 = 0$

Step 3

Solve the equation for $x$ .

$x = - \frac{3}{2}$

Step 4

Create intervals around the solutions to find where $- 2 x - 3$ is positive and negative.

$(- \infty, - \frac{3}{2}), (- \frac{3}{2}, \infty)$

Step 5

Substitute a value from each interval into $- 2 x - 3$ to figure out where the expression is positive or negative.

$\begin{array}{c} Interval & Sign on interval \\ (- \infty, - \frac{3}{2}) & + \\ (- \frac{3}{2}, \infty) & - \end{array}$

Step 6

Integrate the argument of the absolute value.

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Step 6.1

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

$\int - 2 x - 3 d x$

Step 6.2

Split the single integral into multiple integrals.

$\int - 2 x d x + \int - 3 d x$

Step 6.3

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

$- 2 \int x d x + \int - 3 d x$

Step 6.4

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

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

Step 6.5

Apply the constant rule.

$- 2 (\frac{1}{2} x^{2} + C) - 3 x + C$

Step 6.6

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

$- 2 (\frac{x^{2}}{2} + C) - 3 x + C$

Step 6.7

Simplify.

$- x^{2} - 3 x + C$

Step 7

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

${\begin{cases} - x^{2} - 3 x + C & x \leq - \frac{3}{2} \\ - (- x^{2} - 3 x + C) & x > - \frac{3}{2} \end{cases}$

Step 8

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

$F (x) = {\begin{cases} - x^{2} - 3 x + C & x \leq - \frac{3}{2} \\ - (- x^{2} - 3 x + C) & x > - \frac{3}{2} \end{cases}$