If l m - n -01 t m 1- t n dt- then the expression for l m - n in terms of l m -1- n -1 isA- 2n-m-1-n-m-1 1m-1- n-1B- n-m-1 1m-1- n-1C- 2n-m-1-n-m-1 lm-1- n-1D- m-m-1 lm-1- n-1

The correct option is A 2^n/m+1 n/m+1l(m+1,n 1)We have I_m,n=∫_0^1t^m (1+t)^n dt Intergarting by parts considering (1+t)^n as first function, we get I_m,n= [ t ...

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Standard XII

Mathematics

First Fundamental Theorem of Calculus

If l m , n =∫...

Question

If $l (m, n) = \int_{0}^{1} t^{m} (1 + t)^{n} d t$ , then the expression for $l (m, n)$ in terms of $l (m + 1, n - 1)$ is

\frac{2^{n}}{m + 1} - \frac{n}{m + 1} l (m + 1, n - 1)

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\frac{n}{m + 1} l (m + 1, n - 1)

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\frac{2^{n}}{m + 1} + \frac{n}{m + 1} l (m + 1, n - 1)

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\frac{m}{m + 1} l (m + 1, n - 1)

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Solution

The correct option is A $\frac{2^{n}}{m + 1} - \frac{n}{m + 1} l (m + 1, n - 1)$
We have $I_{m, n} = \int_{0}^{1} t^{m} (1 + t)^{n} d t$
Intergarting by parts considering $(1 + t)^{n}$ as first function, we get
$I_{m, n} = {[\frac{t^{m + 1}}{m + 1} (1 + t)^{n}]}_{0}^{n} - \frac{n}{m + 1} \int_{0}^{1} t^{m + 1} (1 + t)^{n - 1} d t I_{m, n} = \frac{2^{n}}{m + 1} - \frac{n}{m + 1} l_{m + 1, n - 1}$

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If l(m,n)=∫10tm(1+t)ndt, then the expression for l(m,n) in terms of l(m+1,n−1) is

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