Question

Evaluate the following integral

$\int_0^1 \: x {(1 - x)}^{n} dx$
​

Answer 1

$\large\underline{\sf{Solution-}}$

Given integral is

$\rm :\longmapsto\:\displaystyle\int_0^1\rm x {(1 - x)}^{n} dx$

Let assume that

$\rm :\longmapsto\:I \: = \: \displaystyle\int_0^1\rm x {(1 - x)}^{n} dx$

We know,

$\boxed{\tt{ \displaystyle\int_0^a\rm f(x)dx \: = \: \displaystyle\int_0^a\rm f(a - x)dx}}$

So, using this identity, we get

$\rm :\longmapsto\:I \: = \: \displaystyle\int_0^1\rm (1 - x) {([1 - (1 - x)]}^{n} dx$

$\rm :\longmapsto\:I \: = \: \displaystyle\int_0^1\rm (1 - x) {([1 - 1 + x]}^{n} dx$

$\rm :\longmapsto\:I \: = \: \displaystyle\int_0^1\rm (1 - x) {x}^{n} dx$

$\rm :\longmapsto\:I \: = \: \displaystyle\int_0^1\rm ({x}^{n} - {x}^{n + 1} )dx$

$\rm :\longmapsto\:I \: = \: \bigg[\dfrac{ {x}^{n + 1} }{n + 1} - \dfrac{ {x}^{n + 2} }{n + 2} \bigg]_0^1\rm$

$\rm :\longmapsto\:I \: = \: \dfrac{ 1}{n + 1} - \dfrac{1}{n + 2}$

$\rm :\longmapsto\:I \: = \: \dfrac{n + 2 - n - 1}{(n + 1)(n + 2)}$

$\bf :\longmapsto\:I \: = \: \dfrac{1}{(n + 1)(n + 2)}$

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MORE TO KNOW

$\boxed{\tt{ \displaystyle\int_b^a\rm f(x)dx \: = \: \displaystyle\int_b^a\rm f(y)dx}}$

$\boxed{\tt{ \displaystyle\int_b^a\rm f(x)dx \: = \: - \: \displaystyle\int_a^b\rm f(x)dx}}$

$\boxed{\tt{ \displaystyle\int_b^a\rm f(x)dx \: = \: \displaystyle\int_b^a\rm f(a + b - x)dx}}$

$\boxed{\tt{ \displaystyle\int_{ - a}^a\rm f(x)dx \: = \:2 \displaystyle\int_0^a\rm f(x)dx \: \: if \: f( - x) = f(x)}}$

$\boxed{\tt{ \displaystyle\int_{ - a}^a\rm f(x)dx \: = \: 0 \: \: if \: f( - x) = - f(x)}}$

$\boxed{\tt{ \displaystyle\int_{0}^{2a}\rm f(x)dx \: = \: 0 \: \: if \: f(2a - x) = - f(x)}}$

$\boxed{\tt{ \displaystyle\int_{0}^{2a}\rm f(x)dx \: = \:2 \displaystyle\int_0^a\rm f(x) \: \: if \: f(2a - x) = f(x)}}$