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Concept understanding β Beta Function
The Beta Function: A First Look
Think of the Beta function as a tool that measures how two quantities blend together in a certain integral. You've seen the Gamma function Ξ(n)=(nβ1)! as a generalization of factorial. The Beta function is like a "product" of two Gamma functions, appearing in its own right in probabilities, integrals of powers, and combinatorial identities.
Intuition: Why "Beta"?
Given two positive numbers a and b, the Beta function B(a,b) answers: if I integrate taβ1(1βt)bβ1 from t=0 to t=1, what value do I get?
The integrand is a product of two powers β one grows near t=0 (if a>1), the other near t=1 (if b>1). The integral "balances" these influences. This shape appears in statistics (the Beta distribution), physics (phase space integrals), and combinatorics (binomial coefficients).
The name comes from Euler's second integral (the first being Gamma). It's also called the Eulerian integral of the first kind.
Precise Definition
For any two positive reals a>0, b>0:
B(a,b)=β«01βtaβ1(1βt)bβ1dt
The integral converges because near t=0, taβ1 is integrable when a>0, and near t=1, (1βt)bβ1 is integrable when b>0.
Key Properties (Memorize These)
- Symmetry:
B(a,b)=B(b,a)
(Substitute u=1βt in the integral.)
- Relation to Gamma function (the most important formula):
B(a,b)=Ξ(a+b)Ξ(a)Ξ(b)β
For positive integers m,n:
B(m,n)=(m+nβ1)!(mβ1)!(nβ1)!β
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Special values:
- B(1,1)=1
- B(21β,21β)=Ο (since Ξ(1/2)=Οβ, B(1/2,1/2)=Ξ(1)Οββ Οββ=Ο)
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Trigonometric form (substitute t=sin2ΞΈ):
B(a,b)=2β«0Ο/2βsin2aβ1ΞΈcos2bβ1ΞΈdΞΈ
B(a,b)=Ξ(a+b)Ξ(a)Ξ(b)β
How to Use It: A Simple Example
Problem: Evaluate β«01βx3(1βx)2dx.
Solution: This is B(4,3) because aβ1=3βa=4, bβ1=2βb=3.
B(4,3)=Ξ(7)Ξ(4)Ξ(3)β=6!3!β 2!β=7206β 2β=601β
When you see β«01βxp(1βx)qdx, recognize it as B(p+1,q+1) β no integration by parts needed!
Common Mistake to Avoid
The exponents are taβ1 and (1βt)bβ1, not ta and (1βt)b. For β«01βt3(1βt)2dt, the parameters are a=4, b=3, not a=3, b=2. Always add 1 to each exponent.
Why It Matters for Exams
In Indian competitive exams (JEE, GATE, etc.), the Beta function appears as direct evaluation of β«01βxm(1βx)ndx and as a shortcut to compute definite integrals via Gamma functions. Once you express an integral as a Beta, rewrite it as a ratio of Gamma functions and use known values (Ξ(n+1)=n!, Ξ(1/2)=Οβ) for a numeric answer.
Final takeaway: The Beta function is just a compact name for a specific integral that appears repeatedly. Learn its Gamma connection, and you'll solve many integrals in seconds.
The Beta function goes beyond the standard NCERT CBSE Class 11/12 syllabus, but it directly extends the Gamma-function and definite-integral ideas that appear at the edges of the Class 12 Integrals chapter, and it's a genuinely useful shortcut for JEE Advanced and engineering-entrance level integral problems. Students searching 'beta function formula and examples' or 'relation between beta and gamma function' will find this B(a,b) = Ξ(a)Ξ(b)/Ξ(a+b) identity is exactly the result those references point to.
The key idea is to recognise the integral as a Beta function. The Beta function is defined as:
B(p,q)=β«01βxpβ1(1βx)qβ1dx=Ξ(p+q)Ξ(p)Ξ(q)β
Step 1: Compare the given integral β«01βx(1βx)ndx with the standard form. Here, the exponent of x is 1, so pβ1=1βΉp=2. The exponent of (1βx) is n, so qβ1=nβΉq=n+1.
Step 2: Therefore, the integral equals B(2,n+1).
Step 3: Using the Gamma function relation: B(2,n+1)=Ξ(2+n+1)Ξ(2)Ξ(n+1)β=(n+2)!1!β n!β.
Step 4: Simplify: (n+2)!n!β=(n+2)(n+1)n!n!β=(n+1)(n+2)1β.
The value is (n+1)(n+2)1ββ.
Substituting t=1βx turns the integral into β«01β(tnβtn+1)dt, giving β«01βx(1βx)ndx=(n+1)(n+2)1β.
Evaluate β«01βx(1βx)ndx for nβN.
Put t=1βx, so x=1βt and dx=βdt. The limits swap (x=0βt=1, x=1βt=0):
β«01βx(1βx)ndx=β«01β(1βt)tndt=β«01β(tnβtn+1)dt.
Integrate:
=[n+1tn+1ββn+2tn+2β]01β=n+11ββn+21β=(n+1)(n+2)(n+2)β(n+1)β=(n+1)(n+2)1β.
(Check n=1: β«01βx(1βx)dx=21ββ31β=61β=2β 31β.)
β«01βx(1βx)ndx=(n+1)(n+2)1β.
- CBSE 2022Set ANNUAL1 markMCQQ.β«01βx(1βx)10dx=(a) 1321β(b) 1325β(c) 1327β(d) 24445β
βΊReveal solutionSolution
β«01βx(1βx)10dx=12!1!10!β=11β 121β=1321β.
Use the standard result β«01βxm(1βx)ndx=(m+n+1)!m!n!β.
Here m=1,Β n=10: β«01βx(1βx)10dx=12!1!β 10!β=12!10!β=12β 111β=1321β.
(Equivalently, substitute t=1βx and expand.)
βFinal answer(a) 1321β.
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