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Question: The value of the integral \(I = \int _ { 0 } ^ { 1 } x ( 1 - x ) ^ { n } d x\) is...

The value of the integral I=∫01x(1−x)ndxI = \int _ { 0 } ^ { 1 } x ( 1 - x ) ^ { n } d x is

A

1n+1\frac { 1 } { n + 1 }

B

1n+2\frac { 1 } { n + 2 }

C

1n+1−1n+2\frac { 1 } { n + 1 } - \frac { 1 } { n + 2 }

D

1n+1+1n+2\frac { 1 } { n + 1 } + \frac { 1 } { n + 2 }

Answer

1n+1−1n+2\frac { 1 } { n + 1 } - \frac { 1 } { n + 2 }

Explanation

Solution

I=∫01x(1−x)ndxI = \int _ { 0 } ^ { 1 } x ( 1 - x ) ^ { n } d x

−I=∫01−x(1−x)ndx=∫01(1−x−1)(1−x)ndx- I = \int _ { 0 } ^ { 1 } - x ( 1 - x ) ^ { n } d x = \int _ { 0 } ^ { 1 } ( 1 - x - 1 ) ( 1 - x ) ^ { n } d x

=∫01(1−x)n+1dx−∫01(1−x)ndx= \int _ { 0 } ^ { 1 } ( 1 - x ) ^ { n + 1 } d x - \int _ { 0 } ^ { 1 } ( 1 - x ) ^ { n } d x

=[(1−x)n+2−(n+2)]01−[(1−x)n+1−(n+1)]01=1n+2−1n+1= \left[ \frac { ( 1 - x ) ^ { n + 2 } } { - ( n + 2 ) } \right] _ { 0 } ^ { 1 } - \left[ \frac { ( 1 - x ) ^ { n + 1 } } { - ( n + 1 ) } \right] _ { 0 } ^ { 1 } = \frac { 1 } { n + 2 } - \frac { 1 } { n + 1 }

⇒I=1n+1−1n+2\Rightarrow I = \frac { 1 } { n + 1 } - \frac { 1 } { n + 2 }