SATHEE: Indefinite Integration 3 Question 8

Indefinite Integration 3 Question 8

8. 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) $\frac{n}{m + 1} I (m + 1, n - 1)$

(d) $\frac{m}{m + 1} I (m + 1, n - 1)$

Show Answer

Answer:

Correct Answer: 8. (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]

$\begin{aligned} ∴ I (m, n) & = (1 + t)^{n} \cdot \frac{t^{m + 1}}{m + 1}_{0}^{1} - \int_{0}^{1} n (1 + t)^{(n - 1)} \cdot \frac{t^{m + 1}}{m + 1} d t \\ = \frac{2^{n}}{m + 1} - \frac{n}{m + 1} \int_{0}^{1} (1 + t)^{(n - 1)} \cdot t^{m + 1} d t \\ ∴ I (m, n) & = \frac{2^{n}}{m + 1} - \frac{n}{m + 1} \cdot I (m + 1, n - 1) \end{aligned}$

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Indefinite Integration 3 Question 8

8. If I(m,n)=∫01tm(1+t)ndt, then the expression for I(m,n) in terms of I(m+1,n−1) is

Answer:

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8. 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