Question

If m, n are positive integers then ${\int }_u^b{(x-a)}^m{(b-x)}^{n}dx$ equals

• A. $\frac{m!n!}{(m+n)!}$
• B. $\frac{m!n!{(b-a)}^{m+n}}{(m+n+1)!}$
• C. $\frac{m!n!{(b-a)}^{m+n+1}}{(m+n+1)!}$
• D. None of these

Answer 1

The correct option is C $\frac{m!n!{(b-a)}^{m+n+1}}{(m+n+1)!}$

$I_{m,n}={\int }_a^b(x-a{)}^m(b-x{)}^{n}dx=(b-a{)}^{m+n+1}{\int }_0^1{t}^m(1-t{)}^{n}dt$

Putting $t={\sin }^2\theta \therefore dt=2\sin \theta \cos \theta andift=0,\theta =0$also for $t=I,\theta =\frac{\pi }{2}$

$I=(b-a{)}^{n+n+1}\times 2{\int }_0^{\frac{\pi }{2}}{\sin }^{2m+1}\theta {\cos }^{2n+1}\theta d\theta$

$=(b-a{)}^{m+n+1}\frac{\left[\right(2m\left)\right(2m-2).....2]\left[\right(2n\left)\right(2n-2)......2]}{(2m+2n+2)(2m+2n)......6\cdot 4\cdot 2}$

$=(b-a{)}^{m+n+1}\times 2$

$\frac{2^m\left[m\right(m-1).....3\cdot 1]2n\left[n\right(n-1\left)\right(n-2).....3\cdot 1]}{2^{m+n+1}(m+n+1)(m+n)(m+n-1)....3\cdot 2\cdot 1}$

$=\frac{(b-a{)}^{m+n+1}m!n!}{(m+n+1)!}$