# Evaluate the definite integral. \int_{0}^{-2}(5x^{2}-4x+2)dx

Evaluate the definite integral.
$$\displaystyle{\int_{{{0}}}^{{-{2}}}}{\left({5}{x}^{{{2}}}-{4}{x}+{2}\right)}{\left.{d}{x}\right.}$$

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Anot1954
Step 1
Here, we use the result of integration that the integral of $$\displaystyle{x}^{{{n}}}\ {i}{s}\ {x}^{{\frac{{{n}+{1}}}{{{n}+{1}}}}}$$. and the solve the definite integral as shown on board.
$$\displaystyle{\int_{{{0}}}^{{-{2}}}}{\left({5}{x}^{{{2}}}-{4}{x}+{2}\right)}{\left.{d}{x}\right.}={{\left|{5}{\frac{{{x}^{{{3}}}}}{{{3}}}}-{2}{x}^{{{2}}}+{2}{x}\right|}_{{{0}}}^{{-{2}}}}$$
$$\displaystyle={\left({5}{\frac{{{\left(-{2}\right)}^{{{3}}}}}{{{3}}}}-{2}{\left(-{2}\right)}^{{{2}}}+{2}{\left(-{2}\right)}\right)}-{\left({5}{\frac{{{0}^{{{3}}}}}{{{3}}}}-{2}{\left({0}\right)}^{{{2}}}+{2}{\left({0}\right)}\right)}$$
$$\displaystyle=-{\frac{{{40}}}{{{3}}}}-{8}-{4}-{0}=-{\frac{{{76}}}{{{3}}}}$$
Step 2
Ans: -76/3
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Nancy Johnson
Step 1: If f(x) is a continuous function from a to b, and if F(x) is its integral, then:
$$\displaystyle{\int_{{{a}}}^{{{b}}}}{f{{\left({x}\right)}}}{\left.{d}{x}\right.}={F}{\left({x}\right)}{{\mid}_{{{a}}}^{{{b}}}}={F}{\left({b}\right)}-{F}{\left({a}\right)}$$
Step 2: In this case, $$\displaystyle{f{{\left({x}\right)}}}={\left({5}{x}^{{{2}}}-{4}{x}+{2}\right)}$$. Find its integral.
$$\displaystyle{\frac{{{5}{x}^{{{3}}}}}{{{3}}}}-{2}{x}^{{{2}}}+{2}{x}{{\mid}_{{{0}}}^{{-{2}}}}$$
Step 3: Since $$\displaystyle{F}{\left({x}\right)}{{\mid}_{{{a}}}^{{{b}}}}={F}{\left({b}\right)}-{F}{\left({a}\right)}$$, expand the above into F(−2)−F(0):
$$\displaystyle{\left({\frac{{{5}{\left(-{2}\right)}^{{{3}}}}}{{{3}}}}-{2}{\left(-{2}\right)}^{{{2}}}+{2}\times-{2}\right)}-{\left({\frac{{{5}\times{0}^{{{3}}}}}{{{3}}}}-{2}\times{0}^{{{2}}}+{2}\times{0}\right)}$$
Step 4: Simplify.
$$\displaystyle=-{\frac{{{76}}}{{{3}}}}$$