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) $\frac{2^{n}}{m + 1} - \frac{n}{m + 1}\ I(m + 1,\ n - 1)$
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) $\frac{m}{n + 1}\ I(m + 1,\ n - 1)$
Explanation
(a)
We have, $I(m,\ n) = I = \int_{0}^{1}{t^{m}(1 + t)^{n}dt}$
$\Rightarrow \ I(m,\ n) = \left\lbrack (1 + t)^{n} \bullet \ \frac{t^{m + 1}}{m + 1} \right\rbrack_{0}^{1} - \frac{n}{m + 1}\int_{0}^{1}{(1 + t)^{n - 1}t^{m + 1}dt}$
$\Rightarrow \ I(m,\ n) = \frac{2^{n}}{m + 1} - \frac{n}{m + 1}I(m + 1,\ n - 1)$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.