If I(m, n)=∫₀¹ xᵐ⁻¹(1-x)ⁿ⁻¹ d x, m, n>0, then I(9,14)+I(10,13) is

Solution

Maths sir · 4 steps

Write out both integrands and pull out the common factor; the leftover bracket is just 1.

1. Read the powers. In I(m,n) the power of x is m-1 and the power of (1-x) is n-1. So I(9,14) has x⁸(1-x)¹³, and I(10,13) has x⁹(1-x)¹².
2. Combine the integrals. Both integrals run from 0 to 1, so add them under one integral sign. The smaller powers x⁸ and (1-x)¹² are common to both terms.

   ∫₀¹ [x⁸(1-x)¹³+x⁹(1-x)¹²]dx=∫₀¹ x⁸(1-x)¹²[(1-x)+x]dx

3. Bracket is 1. (1-x)+x=1, so the sum becomes ∫₀¹ x⁸(1-x)¹² dx. Here m-1=8 and n-1=12, so m=9 and n=13.
4. So. I(9,14)+I(10,13)=I(9,13), which is option (d).

The idea to keep. In general I(m,n+1)+I(m+1,n)=I(m,n). Take out the lowest powers of x and (1-x) as a common factor, and the bracket x+(1-x) collapses to 1.

Answer from the NTA final answer key. Solution by PrepWiser.