Question

Find the value of 4 0 ò (1 - x) .dx

Answer 1

Given that,

Suppose the function is

$I=\int_{0}^{4}{x(1-x)^n}dx$

We need to calculate the value of I

Using the given function

$I=\int_{0}^{4}{x(1-x)^n}dx$....(i)

We know that,

$\int_{0}^{a}{f(x)}dx=\int_{0}^{a}f(a-x)dx$

From equation (i)

$I=\int_{0}^{4}{x(1-x)^n}dx$

$I=\int_{0}^{4}{(4-x)(4-(1-x))^n}dx$

$I=\int_{0}^{4}{(4-x)x^n}dx$

$I=\int_{0}^{4}{4x^n-x^{n+1}}dx$

On integrating

$I=(4(\dfrac{x^{n+1}}{n+1})-\dfrac{x^{n+2}}{n+2})_{0}^{4}$

$I= 4(\dfrac{4^{n+1}}{n+1}-\dfrac{4^{n+2}}{n+2})-0$

$I=16(\dfrac{1^{n+1}}{n+1}-\dfrac{1^{n+2}}{n+2})$

$I=16(\dfrac{1}{n+1}-\dfrac{1}{n+2})$

$I=16(\dfrac{n+2-n-1}{(n+1)(n+2)})$

$I=\dfrac{16}{(n+1)(n+2)}$

Hence, The value of $\int_{0}^{4}{x(1-x)^n dx}$ is $\dfrac{16}{(n+1)(n+2)}$