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 given integral is of the form of the Beta function: \[ I(m, n) = \int_0^1 x^{m-1} (1-x)^{n-1} \, dx = B(m, n) \] where \( B(m, n) \) is the Beta function. We are asked to find \( I(9, 14) + I(10, 13) \). 
Step 1: Use the recurrence relation of the Beta function The Beta function has the following recurrence relation: \[ B(m, n) + B(m+1, n-1) = B(m+1, n) \] Substituting the values of \( m = 9 \) and \( n = 14 \) into this recurrence relation, we get: \[ I(9, 14) + I(10, 13) = I(9, 13) \] This is because the integral \( I(9, 14) \) corresponds to \( B(9, 14) \) and \( I(10, 13) \) corresponds to \( B(10, 13) \), and using the recurrence relation we get that their sum is equal to \( I(9, 13) \), which corresponds to \( B(9, 13) \). Thus, the sum of the two integrals is: \[ I(9, 14) + I(10, 13) = I(9, 13) \]