Definite IntegrationhardFree

∫ {x/2}·sinπx dx from −2n to 2n+1/2 | JEE

JEE Maths question with a full step-by-step solution.

Question
The value of ∫−2n 2n+12{x2}sin⁡πx dx\displaystyle\int_{-2n}^{\,2n+\frac12}\left\{\dfrac{x}{2}\right\}\sin\pi x\,dx, where { ⋅ }\{\,\cdot\,\} denotes the fractional part, is
Anπ\dfrac{n}{\pi}
B−4nπ+12π2\dfrac{-4n\pi+1}{2\pi^{2}}correct
C1−2nππ2\dfrac{1-2n\pi}{\pi^{2}}
D2nπ−1π2\dfrac{2n\pi-1}{\pi^{2}}
Solution
Step 1: The integrand has period 22. Both {x2}\left\{\dfrac{x}{2}\right\} and sin⁡πx\sin\pi x repeat with period 22, so their product does too. Split the interval at x=2nx=2n:
I=∫−2n2n{x2}sin⁡πx dx+∫2n2n+12{x2}sin⁡πx dx.I=\int_{-2n}^{2n}\left\{\dfrac{x}{2}\right\}\sin\pi x\,dx+\int_{2n}^{2n+\frac12}\left\{\dfrac{x}{2}\right\}\sin\pi x\,dx.
 Step 2: Evaluate the periodic bulk. The interval [−2n,2n][-2n,2n] has length 4n=2n4n=2n periods, so its integral is 2n2n times the integral over one period [0,2][0,2], where {x2}=x2\left\{\dfrac{x}{2}\right\}=\dfrac{x}{2}:
∫02x2sin⁡πx dx=12[−xcos⁡πxπ+sin⁡πxπ2]02=12(−2π)=−1π.\int_{0}^{2}\dfrac{x}{2}\sin\pi x\,dx=\dfrac12\left[-\dfrac{x\cos\pi x}{\pi}+\dfrac{\sin\pi x}{\pi^{2}}\right]_{0}^{2}=\dfrac12\left(-\dfrac{2}{\pi}\right)=-\dfrac{1}{\pi}.
 Hence the bulk contributes 2n(−1π)=−2nπ.2n\left(-\dfrac1\pi\right)=-\dfrac{2n}{\pi}. Step 3: Evaluate the short tail. Shifting by the period, ∫2n2n+12{x2}sin⁡πx dx=∫01/2x2sin⁡πx dx\displaystyle\int_{2n}^{2n+\frac12}\left\{\dfrac{x}{2}\right\}\sin\pi x\,dx=\int_{0}^{1/2}\dfrac{x}{2}\sin\pi x\,dx:
12[−xcos⁡πxπ+sin⁡πxπ2]01/2=12(0+1π2)=12π2.\dfrac12\left[-\dfrac{x\cos\pi x}{\pi}+\dfrac{\sin\pi x}{\pi^{2}}\right]_{0}^{1/2}=\dfrac12\left(0+\dfrac{1}{\pi^{2}}\right)=\dfrac{1}{2\pi^{2}}.
 Step 4: Add the two parts:
I=−2nπ+12π2=−4nπ+12π2.I=-\dfrac{2n}{\pi}+\dfrac{1}{2\pi^{2}}=\dfrac{-4n\pi+1}{2\pi^{2}}.
 Correct answer: (2)
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