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MCQMediumJEE 2025Definite Integrals

JEE Mathematics 2025 Question with Solution

The integral ∫−132(π2xsin⁡(πx))dx\int_{-1}^{\frac{3}{2}} \left( \pi^2 x \sin(\pi x) \right) dx is equal to:

  • A

    2+3π2 + 3\pi

  • B

    3+2π3 + 2\pi

  • C

    1+3π1 + 3\pi

  • D

    4+π4 + \pi

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given:

I=∫−132∣π2xsin⁡(πx)∣dxI=\int_{-1}^{\frac{3}{2}} \left|\pi^2 x\sin(\pi x)\right|dx

Find: The value of the integral and the correct option.

Use the sign of xsin⁡(πx)x\sin(\pi x) on the intervals [−1,0][-1,0], [0,1][0,1] and [1,32]\left[1,\frac{3}{2}\right].

  • On [−1,0][-1,0], both xx and sin⁡(πx)\sin(\pi x) are negative, so the product is positive.
  • On [0,1][0,1], both are positive, so the product is positive.
  • On [1,32]\left[1,\frac{3}{2}\right], x>0x>0 and sin⁡(πx)<0\sin(\pi x)<0, so the product is negative.

Therefore,

I=π2∫−10xsin⁡(πx) dx+π2∫01xsin⁡(πx) dx−π2∫132xsin⁡(πx) dxI=\pi^2\int_{-1}^{0}x\sin(\pi x)\,dx+\pi^2\int_{0}^{1}x\sin(\pi x)\,dx-\pi^2\int_{1}^{\frac{3}{2}}x\sin(\pi x)\,dx

Now evaluate

∫xsin⁡(πx) dx\int x\sin(\pi x)\,dx

by integration by parts. Take u=xu=x and dv=sin⁡(πx) dxdv=\sin(\pi x)\,dx. Then du=dxdu=dx and v=−cos⁡(πx)πv=-\frac{\cos(\pi x)}{\pi}. So,

∫xsin⁡(πx) dx=−xcos⁡(πx)π+sin⁡(πx)π2+C\int x\sin(\pi x)\,dx=-\frac{x\cos(\pi x)}{\pi}+\frac{\sin(\pi x)}{\pi^2}+C

For [−1,0][-1,0],

∫−10π2xsin⁡(πx) dx=π2[−xcos⁡(πx)π+sin⁡(πx)π2]−10=π\int_{-1}^{0}\pi^2 x\sin(\pi x)\,dx =\pi^2\left[-\frac{x\cos(\pi x)}{\pi}+\frac{\sin(\pi x)}{\pi^2}\right]_{-1}^{0} =\pi

For [0,1][0,1],

∫01π2xsin⁡(πx) dx=π2[−xcos⁡(πx)π+sin⁡(πx)π2]01=π\int_{0}^{1}\pi^2 x\sin(\pi x)\,dx =\pi^2\left[-\frac{x\cos(\pi x)}{\pi}+\frac{\sin(\pi x)}{\pi^2}\right]_{0}^{1} =\pi

For [1,32]\left[1,\frac{3}{2}\right],

∫132−π2xsin⁡(πx) dx=−π2[−xcos⁡(πx)π+sin⁡(πx)π2]132=1+π\int_{1}^{\frac{3}{2}}-\pi^2 x\sin(\pi x)\,dx =-\pi^2\left[-\frac{x\cos(\pi x)}{\pi}+\frac{\sin(\pi x)}{\pi^2}\right]_{1}^{\frac{3}{2}} =1+\pi

Adding all three parts,

I=π+π+(1+π)=1+3πI=\pi+\pi+(1+\pi)=1+3\pi

Therefore, the integral is 1+3π1+3\pi and the correct option is C.

Interval-wise evaluation

The absolute value is the key point. First identify where sin⁡(πx)\sin(\pi x) changes sign inside [−1,32]\left[-1,\frac{3}{2}\right], namely at x=−1,0,1x=-1,0,1.

Hence,

∣π2xsin⁡(πx)∣={π2xsin⁡(πx),−1≤x≤0π2xsin⁡(πx),0≤x≤1−π2xsin⁡(πx),1≤x≤32\left|\pi^2 x\sin(\pi x)\right|= \begin{cases} \pi^2 x\sin(\pi x), & -1\le x\le 0 \\ \pi^2 x\sin(\pi x), & 0\le x\le 1 \\ -\pi^2 x\sin(\pi x), & 1\le x\le \frac{3}{2} \end{cases}

Using

F(x)=−xcos⁡(πx)π+sin⁡(πx)π2F(x)=-\frac{x\cos(\pi x)}{\pi}+\frac{\sin(\pi x)}{\pi^2}

we get

π2(F(0)−F(−1))=π,π2(F(1)−F(0))=π,\pi^2\big(F(0)-F(-1)\big)=\pi, \quad \pi^2\big(F(1)-F(0)\big)=\pi,

and

−π2(F(32)−F(1))=1+π.-\pi^2\big(F(\tfrac{3}{2})-F(1)\big)=1+\pi.

Therefore,

I=π+π+(1+π)=1+3πI=\pi+\pi+(1+\pi)=1+3\pi

So the correct option remains C.

Common mistakes

  • Ignoring the absolute value and integrating π2xsin⁡(πx)\pi^2 x\sin(\pi x) directly over the whole interval is wrong because the sign changes at x=1x=1. Split the interval according to the sign before integrating.

  • Using integration by parts incorrectly for ∫xsin⁡(πx) dx\int x\sin(\pi x)\,dx is a common error. The antiderivative is not just −1πcos⁡(πx)-\frac{1}{\pi}\cos(\pi x); the factor xx must also be handled, giving −xcos⁡(πx)π+sin⁡(πx)π2-\frac{x\cos(\pi x)}{\pi}+\frac{\sin(\pi x)}{\pi^2}.

  • Making sign mistakes with sin⁡(πx)\sin(\pi x) on [1,32]\left[1,\frac{3}{2}\right] leads to a wrong last term. In this interval sin⁡(πx)<0\sin(\pi x)<0, so the absolute value contributes an extra minus sign.

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