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Q.
If $I(m, n)=\int\limits_{0}^{1} t^{m}(1+t)^{n} d t$, then the expression for $I(m, n)$ in terms of $I(m+1, n-1)$ is
ManipalManipal 2012
Solution:
Here, $I(m, n)=\int\limits_{0}^{1} t^{m}(1+t)^{n} d t$
[We apply integration by parts, taking $(1+t)^{n}$ as first and $t^{m}$ as second function]
$I(m, n)=\left[(1 +t)^{n} \cdot \frac{t^{m+1}}{m+1}\right]_{0}^{1}$
$-\int\limits_{0}^{1} n(1+t)^{n-1} \cdot \frac{t^{m+1}}{m+1} d t$
$=\frac{2^{n}}{m+1}-\frac{n}{m+1} \int\limits_{0}^{1}(1+t)^{n-1} \cdot t^{m+1} d t$
$\therefore I(m,\, n)=\frac{2^{n}}{m+1}-\frac{n}{m+1} \cdot I(m+1, n-1)$