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Official PYQ · 2014Set 3 · Q47Maxima and Minima, Mean Value Theorem, IntegrationMCQ1 mark

GATE CS 2014 Set 3 Q47 — Maxima and Minima, Mean Value Theorem, Integration

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The value of the integral given below is
∫0π x2 cos⁡x dx\int \limits_0^{\pi} \: x^2 \: \cos x\:dx

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Answer: A

We need to evaluate the definite integral:

I=∫0πx2cos⁡x dxI = \int_{0}^{\pi} x^2 \cos x \, dx

We can solve this using integration by parts, or more quickly using the tabular integration (DI method):

Sign Differentiate (uu) Integrate (dvdv)
++ x2x^2 cos⁡x\cos x
−- 2x2x sin⁡x\sin x
++ 22 −cos⁡x-\cos x
−- 00 −sin⁡x-\sin x

Multiplying along the diagonals, we get the antiderivative:

∫x2cos⁡x dx=x2sin⁡x+2xcos⁡x−2sin⁡x\int x^2 \cos x \, dx = x^2 \sin x + 2x \cos x - 2 \sin x

Now, applying the limits from 00 to π\pi:

I=[x2sin⁡x+2xcos⁡x−2sin⁡x]0πI = \Big[ x^2 \sin x + 2x \cos x - 2 \sin x \Big]_{0}^{\pi}

Evaluating at the upper limit x=πx = \pi:

π2sin⁡π+2πcos⁡π−2sin⁡π=0+2π(−1)−0=−2π\pi^2 \sin \pi + 2\pi \cos \pi - 2 \sin \pi = 0 + 2\pi(-1) - 0 = -2\pi

Evaluating at the lower limit x=0x = 0:

02sin⁡0+2(0)cos⁡0−2sin⁡0=00^2 \sin 0 + 2(0) \cos 0 - 2 \sin 0 = 0

Subtracting the lower limit value from the upper limit value:

I=−2π−0=−2πI = -2\pi - 0 = -2\pi

Therefore, the correct option is A.

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