Question

In \( I(m, n) = \int_0^1 x^{m-1} (1-x)^{n-1} \, dx \), where \( m, n > 0 \), then \( I(9, 14) + I(10, 13) \) is:

• A. \( I(9, 1) \)
• B. \( I(19, 27) \)
• C. \( I(1, 13) \)
• D. \( I(9, 13) \)

Hint:

The Beta function has a recurrence relation that simplifies integrals of this type. If you know \( I(m, n) \) and \( I(m+1, n-1) \), you can use the identity: \[ I(m, n) + I(m+1, n-1) = I(m+1, n) \] This identity is very useful when working with sums of Beta integrals.

Answer 1

The Correct Option is D

The provided integral is recognized as the Beta function: \[ I(m, n) = \int_0^1 x^{m-1} (1-x)^{n-1} \, dx = B(m, n) \] where \( B(m, n) \) represents the Beta function. The objective is to compute \( I(9, 14) + I(10, 13) \).
Step 1: Apply the Beta function's recurrence relation The Beta function satisfies the relation: \[ B(m, n) + B(m+1, n-1) = B(m+1, n) \] For \( m = 9 \) and \( n = 14 \), this relation yields: \[ I(9, 14) + I(10, 13) = I(9, 13) \] This simplification is derived from \( I(9, 14) = B(9, 14) \) and \( I(10, 13) = B(10, 13) \), where their sum, by the recurrence relation, equals \( B(9, 13) \), which corresponds to \( I(9, 13) \). Therefore, the sum is: \[ I(9, 14) + I(10, 13) = I(9, 13) \]