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Official PYQ · 2014Set 3 · Q6Maxima and Minima, Mean Value Theorem, IntegrationNAT1 mark

GATE CS 2014 Set 3 Q6 — Maxima and Minima, Mean Value Theorem, Integration

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If ∫02π∣x sin⁡x∣dx=kπ\int \limits_0^{2 \pi} |x \: \sin x| dx=k\pi, then the value of kk is equal to ______.
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Answer: 4

We need to evaluate the integral:

I=∫02π∣xsin⁡x∣ dxI = \int_{0}^{2\pi} |x \sin x| \, dx

Since x≥0x \ge 0 in the interval [0,2π][0, 2\pi], we can write ∣xsin⁡x∣=x∣sin⁡x∣|x \sin x| = x |\sin x|. Thus:

I=∫02πx∣sin⁡x∣ dxI = \int_{0}^{2\pi} x |\sin x| \, dx

Using the definite integral property ∫abf(x) dx=∫abf(a+b−x) dx\int_{a}^{b} f(x) \, dx = \int_{a}^{b} f(a+b-x) \, dx, we substitute xx with 2π−x2\pi - x:

I=∫02π(2π−x)∣sin⁡(2π−x)∣ dx=∫02π(2π−x)∣sin⁡x∣ dxI = \int_{0}^{2\pi} (2\pi - x) |\sin(2\pi - x)| \, dx = \int_{0}^{2\pi} (2\pi - x) |\sin x| \, dx

Adding these two expressions for II gives:

2I=∫02π(x+2π−x)∣sin⁡x∣ dx=2π∫02π∣sin⁡x∣ dx2I = \int_{0}^{2\pi} (x + 2\pi - x) |\sin x| \, dx = 2\pi \int_{0}^{2\pi} |\sin x| \, dx I=π∫02π∣sin⁡x∣ dxI = \pi \int_{0}^{2\pi} |\sin x| \, dx

Since ∣sin⁡x∣|\sin x| has a period of π\pi, we can simplify the integral:

∫02π∣sin⁡x∣ dx=2∫0πsin⁡x dx=2[−cos⁡x]0π=2(1−(−1))=4\int_{0}^{2\pi} |\sin x| \, dx = 2 \int_{0}^{\pi} \sin x \, dx = 2 \left[ -\cos x \right]_{0}^{\pi} = 2(1 - (-1)) = 4

Substituting this back into the equation for II:

I=4πI = 4\pi

Comparing this with I=kπI = k\pi, we get k=4k = 4.

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