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

Determine the following definite integrals:
$$\displaystyle{\int_{{-{1}}}^{{{3}}}}{\left|{2}{x}-{x}^{{{2}}}\right|}{\left.{d}{x}\right.}$$

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Step 1
Evaluate the integrals:
$$\displaystyle{f{{\left({x}\right)}}}={\int_{{-{1}}}^{{{3}}}}{\left|{2}{x}-{x}^{{{2}}}\right|}{\left.{d}{x}\right.}$$
Eliminate the absolutes,
$$\displaystyle-{1}\le{x}\le{0}\ {f{{\left({x}\right)}}}={\left(-{2}{x}+{x}^{{{2}}}\right)}$$
$$\displaystyle{0}\le{x}\le{2}\ {f{{\left({x}\right)}}}={\left({2}{x}-{x}^{{{2}}}\right)}$$
$$\displaystyle{2}\le{x}\le{3}\ {f{{\left({x}\right)}}}={\left(-{2}{x}+{x}^{{{2}}}\right)}$$
$$\displaystyle{f{{\left({x}\right)}}}={\int_{{-{1}}}^{{{0}}}}{\left(-{2}{x}+{x}^{{{2}}}\right)}{\left.{d}{x}\right.}+{\int_{{{0}}}^{{{2}}}}{\left({2}{x}-{x}^{{{2}}}\right)}{\left.{d}{x}\right.}+{\int_{{{2}}}^{{{3}}}}{\left(-{2}{x}+{x}^{{{2}}}\right)}{\left.{d}{x}\right.}$$
$$\displaystyle={{\left[-{2}{\left({\frac{{{x}^{{{2}}}}}{{{2}}}}\right)}+{\frac{{{x}^{{{3}}}}}{{{3}}}}\right]}_{{-{1}}}^{{{0}}}}+{{\left[{2}{\left({\frac{{{x}^{{{2}}}}}{{{2}}}}\right)}-{\frac{{{x}^{{{3}}}}}{{{3}}}}\right]}_{{{0}}}^{{{2}}}}+{{\left[-{2}{\left({\frac{{{x}^{{{2}}}}}{{{2}}}}\right)}+{\frac{{{x}^{{{3}}}}}{{{3}}}}\right]}_{{{2}}}^{{{3}}}}$$
$$\displaystyle={{\left[-{x}^{{{2}}}+{\frac{{{x}^{{{3}}}}}{{{3}}}}\right]}_{{-{1}}}^{{{0}}}}+{{\left[{x}^{{{2}}}-{\frac{{{x}^{{{3}}}}}{{{3}}}}\right]}_{{{0}}}^{{{2}}}}+{{\left[-{x}^{{{2}}}+{\frac{{{x}^{{{3}}}}}{{{3}}}}\right]}_{{{2}}}^{{{3}}}}$$
$$\displaystyle={\left[{0}-{\left(-{\left(-{1}\right)}^{{{2}}}+{\frac{{{\left(-{1}\right)}^{{{3}}}}}{{{3}}}}\right)}\right]}+{\left[{2}^{{{2}}}-{\frac{{{2}^{{{3}}}}}{{{3}}}}-{0}\right]}+{\left[{\left(-{3}^{{{2}}}+{\frac{{{3}^{{{3}}}}}{{{3}}}}\right)}-{\left(-{2}^{{{2}}}+{\frac{{{2}^{{{3}}}}}{{{3}}}}\right)}\right]}$$
$$\displaystyle={\left[-{\left(-{1}+{\frac{{{\left(-{1}\right)}}}{{{3}}}}\right)}\right]}+{\left[{4}-{\frac{{{8}}}{{{3}}}}\right]}+{\left[{\left(-{9}+{\frac{{{27}}}{{{3}}}}\right)}-{\left(-{4}+{\frac{{{8}}}{{{3}}}}\right)}\right]}$$
$$\displaystyle={\frac{{{4}}}{{{3}}}}+{\frac{{{4}}}{{{3}}}}+{0}-{\left(-{\frac{{{4}}}{{{3}}}}\right)}$$
$$\displaystyle={\frac{{{4}}}{{{3}}}}+{\frac{{{4}}}{{{3}}}}+{\frac{{{4}}}{{{3}}}}$$
$$\displaystyle={\frac{{{12}}}{{{3}}}}$$
=4
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