# Determine the following definite integrals: \int_{-1}^{3}|2x-x^{2}|dx

Determine the following definite integrals:
${\int }_{-1}^{3}|2x-{x}^{2}|dx$
You can still ask an expert for help

• Questions are typically answered in as fast as 30 minutes

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

avortarF
Step 1
Evaluate the integrals:
$f\left(x\right)={\int }_{-1}^{3}|2x-{x}^{2}|dx$
Eliminate the absolutes,

$f\left(x\right)={\int }_{-1}^{0}\left(-2x+{x}^{2}\right)dx+{\int }_{0}^{2}\left(2x-{x}^{2}\right)dx+{\int }_{2}^{3}\left(-2x+{x}^{2}\right)dx$
$={\left[-2\left(\frac{{x}^{2}}{2}\right)+\frac{{x}^{3}}{3}\right]}_{-1}^{0}+{\left[2\left(\frac{{x}^{2}}{2}\right)-\frac{{x}^{3}}{3}\right]}_{0}^{2}+{\left[-2\left(\frac{{x}^{2}}{2}\right)+\frac{{x}^{3}}{3}\right]}_{2}^{3}$
$={\left[-{x}^{2}+\frac{{x}^{3}}{3}\right]}_{-1}^{0}+{\left[{x}^{2}-\frac{{x}^{3}}{3}\right]}_{0}^{2}+{\left[-{x}^{2}+\frac{{x}^{3}}{3}\right]}_{2}^{3}$
$=\left[0-\left(-{\left(-1\right)}^{2}+\frac{{\left(-1\right)}^{3}}{3}\right)\right]+\left[{2}^{2}-\frac{{2}^{3}}{3}-0\right]+\left[\left(-{3}^{2}+\frac{{3}^{3}}{3}\right)-\left(-{2}^{2}+\frac{{2}^{3}}{3}\right)\right]$
Jeffrey Jordon