← 2021 Paper 1

UPSC 2021 Maths Optional Paper 1 Q3a-iii — Step-by-Step Solution

8 marks · Section A

Indefinite integrals · Calculus · asked 7× in 13 yrs · Read the full method →

Question

Express ab(xa)m(bx)ndx\int_a^b(x-a)^m(b-x)^n\,dx in terms of the Beta function.

Technique

Substitute x=a+(ba)tx=a+(b-a)t to map [a,b][0,1][a,b]\to[0,1]; the factor (ba)m+n+1(b-a)^{m+n+1} comes from (xa)m,(bx)n,dx(x-a)^m,(b-x)^n,dx.

Solution

Beta function definition: B(p,q)=01tp1(1t)q1dtB(p,q)=\int_0^1 t^{p-1}(1-t)^{q-1}\,dt, p,q>0p,q>0.

Step 1 — Substitute x=a+(ba)tx=a+(b-a)t, t[0,1]t\in[0,1]

dx=(ba)dtdx=(b-a)\,dt.

xa=(ba)tx-a=(b-a)t, so (xa)m=(ba)mtm(x-a)^m=(b-a)^m t^m.

bx=(ba)(1t)b-x=(b-a)(1-t), so (bx)n=(ba)n(1t)n(b-x)^n=(b-a)^n(1-t)^n.

ab(xa)m(bx)ndx=01(ba)mtm(ba)n(1t)n(ba)dt\int_a^b(x-a)^m(b-x)^n\,dx=\int_0^1(b-a)^m t^m\cdot(b-a)^n(1-t)^n\cdot(b-a)\,dt

=(ba)m+n+101tm(1t)ndt=(b-a)^{m+n+1}\int_0^1 t^m(1-t)^n\,dt

=(ba)m+n+1B(m+1,n+1)=(b-a)^{m+n+1}\cdot B(m+1,n+1).

Answer

  ab(xa)m(bx)ndx=(ba)m+n+1B(m+1,n+1).  \boxed{\;\int_a^b(x-a)^m(b-x)^n\,dx=(b-a)^{m+n+1}B(m+1,n+1).\;}
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