Determine the following definite integrals:

${\int}_{-1}^{3}|2x-{x}^{2}|dx$

aflacatn
2021-09-16
Answered

Determine the following definite integrals:

${\int}_{-1}^{3}|2x-{x}^{2}|dx$

You can still ask an expert for help

avortarF

Answered 2021-09-17
Author has **113** answers

Step 1

Evaluate the integrals:

$f\left(x\right)={\int}_{-1}^{3}|2x-{x}^{2}|dx$

Eliminate the absolutes,

$-1\le x\le 0\text{}f\left(x\right)=(-2x+{x}^{2})$

$0\le x\le 2\text{}f\left(x\right)=(2x-{x}^{2})$

$2\le x\le 3\text{}f\left(x\right)=(-2x+{x}^{2})$

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

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

$={[-{x}^{2}+\frac{{x}^{3}}{3}]}_{-1}^{0}+{[{x}^{2}-\frac{{x}^{3}}{3}]}_{0}^{2}+{[-{x}^{2}+\frac{{x}^{3}}{3}]}_{2}^{3}$

$=[0-(-{(-1)}^{2}+\frac{{(-1)}^{3}}{3})]+[{2}^{2}-\frac{{2}^{3}}{3}-0]+[(-{3}^{2}+\frac{{3}^{3}}{3})-(-{2}^{2}+\frac{{2}^{3}}{3})]$

Evaluate the integrals:

Eliminate the absolutes,

Jeffrey Jordon

Answered 2021-11-08
Author has **2027** answers

Answer is given below (on video)

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To do this, I let

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