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AIEEE 2003

JEE Main 2003 Mathematics question on Definite Integrals and Areas

If I(m,n)=∫01tm(1+t)n dtI(m,n) = \int_0^1 t^m(1+t)^n\, dt, then the expression for I(m,n)I(m,n) in terms of I(m+1,n−1)I(m+1, n-1) is
  1. A.
    2nm+1−nm+1I(m+1,n−1)\dfrac{2^n}{m+1} - \dfrac{n}{m+1}I(m+1, n-1)
  2. B.
    nm+1I(m+1,n−1)\dfrac{n}{m+1}I(m+1, n-1)
  3. C.
    2nm+1+nm+1I(m+1,n−1)\dfrac{2^n}{m+1} + \dfrac{n}{m+1}I(m+1, n-1)
  4. D.
    mm+1I(m+1,n−1)\dfrac{m}{m+1}I(m+1, n-1)
Show answer and solution

Correct option: A

Solution

Integrate by parts: let u = (1+t)n, dv = tm dt.

Then du = n(1+t)^(n-1) dt, v = t^(m+1)/(m+1).

I(m,n) = [t^(m+1)(1+t)n/(m+1)] from 0 to 1 − (n/(m+1)) ∫₀¹ t^(m+1)(1+t)^(n-1) dt

= 2n/(m+1) − (n/(m+1)) · I(m+1, n-1).

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