Definite Integration Ques 118
- If $I(m, n)=\int _0^{1} t^{m}(1+t)^{n} d t$, then the expression for $I(m, n)$ in terms of $I(m+1, n-1)$ is
(2003, 1M)
(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) $\frac{m}{m+1} I(m+1, n-1)$
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Answer:
Correct Answer: 118.(a)
Solution:
- Here, $I(m, n)=\int _0^{1} t^{m}(1+t)^{n} d t$ reduce into $I(m+1, n-1)$ [we apply integration by parts taking $(1+t)^{n}$ as first and $t^{m}$ as second function]
$$ \begin{aligned} \therefore I(m, n) & =(1+t)^{n} \cdot \frac{t^{m+1}}{m+1}{ } _0^{1}-\int _0^{1} n(1+t)^{(n-1)} \cdot \frac{t^{m+1}}{m+1} d t \\ & =\frac{2^{n}}{m+1}-\frac{n}{m+1} \int _0^{1}(1+t)^{(n-1)} \cdot t^{m+1} d t \\ \therefore \quad I(m, n) & =\frac{2^{n}}{m+1}-\frac{n}{m+1} \cdot I(m+1, n-1) \end{aligned} $$
BETA
