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

\displaystyle\int_{-2n}^{\,2n+\frac12}\left\{\dfrac{x}{2}\right\}\sin\pi x\,dx

, where

\{\,\cdot\,\}

denotes the fractional part, is

\dfrac{n}{\pi}

\dfrac{-4n\pi+1}{2\pi^{2}}

correct

\dfrac{1-2n\pi}{\pi^{2}}

\dfrac{2n\pi-1}{\pi^{2}}

Solution

Step 1: The integrand has period

2

. Both

\left\{\dfrac{x}{2}\right\}

and

\sin\pi x

repeat with period

2

, so their product does too. Split the interval at

x=2n

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]

has length

4n=2n

periods, so its integral is

2n

times the integral over one period

[0,2]

, where

\left\{\dfrac{x}{2}\right\}=\dfrac{x}{2}

\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\left(-\dfrac1\pi\right)=-\dfrac{2n}{\pi}.

Step 3: Evaluate the short tail. Shifting by the period,

\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

\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=-\dfrac{2n}{\pi}+\dfrac{1}{2\pi^{2}}=\dfrac{-4n\pi+1}{2\pi^{2}}.

Correct answer: (2)

Still stuck on this question?Ask your doubt on WhatsApp

Solve more, learn faster

Sign up free to solve more JEE Maths questions and explore doMath — timed drills, mastery sprints, bookmarks, and chapter-wise progress tracking.