Question

If $m,n\in N$ then the value of ${\int }_a^b(x-a{)}^m(b-x{)}^{n}dx$ is equal to

• A. $\frac{(b-a{)}^{m+n}.m!n!}{(m+n)!}$
• B. $\frac{(b-a{)}^{m+n+1}.m!n!}{(m+n+1)!}$
• C. $\frac{(b-a{)}^m.m!}{m!}$
• D. $Noneofthese$

Answer 1

Answer: (A)

$\frac{(b-a{)}^{m+n}.m!n!}{(m+n)!}$

Put $x-a=t⟹dx=dt$

Limits as $x$ from $a\to b$

$t$ from $0\to b-a$

$⟹{\int }_0^{b-a}{t}^m(b-a-t{)}^{n}dt$

$⟹(b-a{)}^n{\int }_0^{b-a}{t}^m{(1-\frac{t}{b-a})}^{n}dt$

Put $p=\frac{t}{b-a}⟹dt=(b-a)dp$

Limits as $t$ from $0\to b-a$

$p$ from $0\to 1$

$⟹(b-a{)}^{m+n}{\int }_0^1{p}^m{(1-p)}^{n}dp$

$⟹(b-a{)}^{m+n}\frac{m!.n!}{(m+n)!}$