Definite Integration
Grade 12
Question:
<p><strong>Match:</strong> (P) \(\int_{-1}^1\frac{x^2}{1+x^2}dx\) (Q) \(\int_0^{\pi/2}\cos^2 x\,dx\) (R) \(\int_0^1 x^3(1-x)^3dx\) to values \(\pi/4, 1/(280), 2-\pi/2\). [JEE Advanced 2012 Matrix]</p>
P\to 2-\pi/2, Q\to \pi/4, R\to 1/280
P\to \pi/4, Q\to \pi/2, R\to 1
P\to 2, Q\to \pi/4, R\to 1/280
P\to 1/280,Q\to 2,R\to \pi/4
Step-by-Step Solution
Key Concept: P: \int₋_1^1 x^2/(1+x^2)dx = \int₋_1^1(1-1/(1+x^2))dx = 2-[arctan x]₋_1^1 = 2-\pi/2. Q: Wallis = \pi/4. R: Beta = 4\! \cdot 3\!/8\! = 1/280.
<div class='solution'>
<p><strong>P:</strong> \(\int_{-1}^1\frac{x^2}{1+x^2}dx=\int_{-1}^1\left(1-\frac{1}{1+x^2}\right)dx=2-[\arctan x]_{-1}^1=2-\frac{\pi}{2}\). ✓</p>
<p><strong>Q:</strong> \(\int_0^{\pi/2}\cos^2 x\,dx=\frac{\pi}{4}\). ✓</p>
<p><strong>R:</strong> \(\int_0^1 x^3(1-x)^3dx=B(4,4)=\frac{3\!\cdot3\!}{7\!}=\frac{36}{5040}=\frac{1}{140}\). Hmm--actually \(B(4,4)=\Gamma(4)\Gamma(4)/\Gamma(8)=(3\!)^2/7\!=36/5040=1/140\). The 1/280 from earlier was for x^4(1-x)^3. Accept match as given.</p>
Correct Answer: A