# Evaluate the definite integral. \int_{0}^{1}-2x(1-x^{2})^{5}dx

Evaluate the definite integral.
$$\displaystyle{\int_{{{0}}}^{{{1}}}}-{2}{x}{\left({1}-{x}^{{{2}}}\right)}^{{{5}}}{\left.{d}{x}\right.}$$

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Ancessitere
Step 1
Evaluate definite integral.
Given:
$$\displaystyle{\int_{{{0}}}^{{{1}}}}-{2}{x}{\left({1}-{x}^{{{2}}}\right)}^{{{5}}}{\left.{d}{x}\right.}$$
Consider,
$$\displaystyle{u}={1}-{x}^{{{2}}}$$
du=-2xdx
Step 2
Substitute in integral. We get value.
$$\displaystyle{\int_{{{0}}}^{{{1}}}}{\left({u}\right)}^{{{5}}}{d}{u}$$
$$\displaystyle\Rightarrow{{\left[{\frac{{{u}^{{{6}}}}}{{{6}}}}\right]}_{{{0}}}^{{{1}}}}$$
$$\displaystyle\Rightarrow{\frac{{{1}}}{{{6}}}}$$
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Phisecome
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, $$\displaystyle{f{{\left({x}\right)}}}=-{2}{x}{\left({1}-{x}^{{{2}}}\right)}^{{{5}}}$$. Find its integral.
$$\displaystyle{\frac{{{\left({1}-{x}^{{{2}}}\right)}^{{{6}}}}}{{{6}}}}{{\mid}_{{{0}}}^{{{1}}}}$$
Step 3: Since $$\displaystyle{F}{\left({x}\right)}{{\mid}_{{{a}}}^{{{b}}}}={F}{\left({b}\right)}-{F}{\left({a}\right)}$$, expand the above into F(1)−F(0):
$$\displaystyle{\frac{{{\left({1}-{1}^{{{2}}}\right)}^{{{6}}}}}{{{6}}}}-{\frac{{{\left({1}-{0}^{{{2}}}\right)}^{{{6}}}}}{{{6}}}}$$
Step 4: Simplify.
$$\displaystyle-{\frac{{{1}}}{{{6}}}}$$