Q. If $I(m,n) = \int_{0}^{1}{t^{m}(1 + t)^{n}}dt$, then the expression for $I(m,n)$ in terms of $I(m + 1,\ n + 1)$ is
A) $\frac{2^{n}}{m + 1} - \frac{n}{m + 1}(m + 1,\ n - 1)$
B) $\frac{n}{m + 1}I(m + 1,\ n - 1)$
C) $\frac{2^{n}}{m + 1} + \frac{n}{m + 1}I(m + 1,n - 1)$
D) $\frac{m}{m + 1}I(m + 1,n - 1)$
Explanation
(a)
Here, $I(m,n) = \int_{0}^{1}{t^{m}(1 + t)^{n}dt}$
$\Rightarrow I(m,n) = \left\{ (1 + t)^{n}.\frac{t^{m + 1}}{m + 1} \right\}_{0}^{1} - \int_{0}^{1}{n(1 + t)^{n - 1}.\frac{t^{m + 1}}{m + 1}dt}$
$= \frac{2^{n}}{m + 1} - \frac{n}{m + 1}\int_{0}^{1}{(1 + t)^{n - 1}.t^{m + 1}dt}$
$\therefore I(m,n) = \frac{2^{n}}{m + 1} - \frac{n}{m + 1}.I(m + 1,n - 1)$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.