0-n-x-2-n-x-p-dx-for-p-gt-0-

Question Number 17604 by tawa tawa last updated on 08/Jul/17

\int_{0}^{n} x^{2} {(n - x)}^{p} dx for p > 0

Answered by sma3l2996 last updated on 08/Jul/17

t = n - x \Rightarrow d t = - d x

I = \int_{0}^{n} x^{2} {(n - x)}^{p} d x = \int_{0}^{n} {(n - t)}^{2} t^{p} d t = \int_{0}^{n} (n^{2} t^{p} - 2 {n t}^{p + 1} + t^{p + 2}) d t

= {[\frac{n^{2}}{p + 1} t^{p + 1} - \frac{2 n}{p + 2} t^{p + 2} + \frac{t^{p + 3}}{p + 3}]}_{0}^{n}

= \frac{n^{2}}{p + 1} n^{p + 1} - \frac{2 n}{p + 2} n^{p + 2} + \frac{n^{p + 3}}{p + 3}

= n^{p + 3} (\frac{1}{p + 1} - \frac{2}{p + 2} + \frac{1}{p + 3}) = n^{p + 3} (\frac{(p + 3) (p + 2) - 2 (p + 1) (p + 3) + (p + 1) (p + 2)}{(p + 1) (p + 2) (p + 3)})

= n^{p + 3} (\frac{p^{2} + 5 p + 6 - 2 p^{2} - 8 p - 6 + p^{2} + 3 p + 2}{(p + 1) (p + 2) (p + 3)})

I = \frac{2 n^{p + 3}}{(p + 1) (p + 2) (p + 3)}

Commented by tawa tawa last updated on 08/Jul/17

God bless you sir .

Answered by alex041103 last updated on 08/Jul/17

We use the substitution

u = n - x \Rightarrow - d u = d x and x^{2} = {(n - u)}^{2}

And we change the limits of integration

x = 0 x = n

u = n u = 0

\Rightarrow \underset{0}{\int^{n}} x^{2} {(n - x)}^{p} d x = - \underset{n}{\int^{0}} {(n - u)}^{2} u^{p} d u

= \underset{0}{\int^{n}} (n^{2} - 2 n u + u^{2}) u^{p} d u

= n^{2} \underset{0}{\int^{n}} u^{p} d u - 2 n \underset{0}{\int^{n}} u^{p + 1} d u + \underset{0}{\int^{n}} u^{p + 2} d u

= n^{2} \frac{n^{p + 1} - 0}{p + 1} - 2 n \frac{n^{p + 2} - 0}{p + 2} + \frac{n^{p + 3} - 0}{p + 3}

= \frac{n^{p + 3}}{p + 1} - 2 \frac{n^{p + 3}}{p + 2} + \frac{n^{p + 3}}{p + 3}

= n^{p + 3} (\frac{1}{p + 1} - \frac{2}{p + 2} + \frac{1}{p + 3})

This can be expressed as

\underset{0}{\int^{n}} x^{2} {(n - x)}^{p} d x = \frac{2 n^{p + 3}}{(p + 1) (p + 2) (p + 3)}

Commented by tawa tawa last updated on 08/Jul/17

God bless you sir .