IIT-JEE 1982Question 1
Find the value of $\displaystyle\int_{-1}^{3/2}|x\sin\pi x|\,dx$.
Definite Integration
Answer$\dfrac{3}{\pi}+\dfrac{1}{\pi^{2}}$
Solution
Concept. $x\sin\pi x\ge0$ on $[-1,1]$ (both factors have the same sign) and $\le0$ on $[1,\frac32]$. Split there and use one antiderivative. Formulas used. (1) $\int x\sin\pi x\,dx=-\frac{x\cos\pi x}{\pi}+\frac{\sin\pi x}{\pi^2}=G(x)$Step 1 - Sign.Step 2 - $[-1,1]$. By (1), $G(1)=\frac1\pi$ and $G(-1)=-\frac1\pi$, so $\int_{-1}^1x\sin\pi x\,dx=\frac2\pi$. Step 3 - $[1,\frac32]$. $G\big(\frac32\big)=0-\frac1{\pi^2}$, so $\int_1^{3/2}x\sin\pi x\,dx=-\frac1{\pi^2}-\frac1\pi$; its absolute value is $\frac1\pi+\frac1{\pi^2}$. Step 4 - Add. $\frac2\pi+\frac1\pi+\frac1{\pi^2}=\frac3\pi+\frac1{\pi^2}$. Answer: $\dfrac{3}{\pi}+\dfrac{1}{\pi^{2}}$