JEE MAIN - Mathematics (2003 - No. 21)
The value of the integral $$I = \int\limits_0^1 {x{{\left( {1 - x} \right)}^n}dx} $$ is
$${1 \over {n + 1}} + {1 \over {n + 2}}$$
$${1 \over {n + 1}}$$
$${1 \over {n + 2}}$$
$${1 \over {n + 1}} - {1 \over {n + 2}}$$
Explanation
$$I = \int\limits_0^1 {x{{\left( {1 - x} \right)}^n}} dx$$
$$ = \int\limits_0^1 {\left( {1 - x} \right){{\left( {1 - 1 + x} \right)}^n}dx} $$
$$ = \int\limits_0^1 {\left( {1 - x} \right)} {x^n}dx$$
$$ = \left[ {{{{x^{n + 1}}} \over {n + 1}} - {{{x^{n + 2}}} \over {n + 2}}} \right]_0^1$$
$$ = {1 \over {n + 1}} - {1 \over {n + 2}}$$
$$ = \int\limits_0^1 {\left( {1 - x} \right){{\left( {1 - 1 + x} \right)}^n}dx} $$
$$ = \int\limits_0^1 {\left( {1 - x} \right)} {x^n}dx$$
$$ = \left[ {{{{x^{n + 1}}} \over {n + 1}} - {{{x^{n + 2}}} \over {n + 2}}} \right]_0^1$$
$$ = {1 \over {n + 1}} - {1 \over {n + 2}}$$
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