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UPSC NDA Math Model Paper 5

Question : 29 of 120

Marks: +1, -0

What is

\int\limits_{-\frac{\pi}{2}}^{\frac{\pi}{2}} x \sin x d x

equal to?

Solution:

Let f(x) = xsinx, put -x instead of x then we get

f(-x) = (-x) \sin(-x) = x \sin x

Therefore, f(-x) = f(x) implies that f(x) is an even function.

Thus, the integral is given by,

\int\limits_{-\frac{\pi}{2}}^{\frac{\pi}{2}} x \sin x d x

= \int\limits_{0}^{\frac{\pi}{2}} x \sin x d x

=2\left[ x(-\cos x) - \int (-\cos x d x) \right]^{\frac{\pi}{2}}

=2\left[ -x \cos x + \sin x \right]^{\frac{\pi}{\theta}}

=2\left[ \left(-\frac{\pi}{2} \cos \left(\frac{\pi}{2}\right) + \sin \left(\frac{\pi}{2}\right)\right)\right]

- \left(-0 \cos 0 + \sin 0\right)

=2

Hence, the value of the given integral is 2.

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