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BITSAT 2005 Solved Question Paper

Section: Mathematics

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Question : 11 of 150

Marks: +1, -0

\int\limits_{0}^{2 n \pi} \left\{ \left| \sin x \right| - \left| \frac{1}{2} \sin x \right| \right\} dx

equals :

$n$
$2n$
$-2n$
none of the above

Solution:

Let

I = \int\limits_{0}^{2 n \pi} \left\{ \left| \sin x \right| - \left| \frac{1}{2} \sin x \right| \right\} dx

= \int\limits_{0}^{2 n \pi} \left\{ \left| \sin x \right| - \frac{1}{2} \left| \sin x \right| \right\} dx

= \int\limits_{0}^{2 n \pi} \frac{1}{2} \left| \sin x \right| dx

= \frac{1}{2} \biggl[ \int\limits_{0}^{2 \pi} \left| \sin x \right| dx + \int\limits_{2 \pi}^{4 \pi} \left| \sin x \right| dx + \cdots

+ \int\limits_{2(n-1) \pi}^{2 n \pi} \left| \sin x \right| dx \biggr]

Now,

I_1 = \int\limits_{0}^{2 \pi} \frac{1}{2} \left| \sin x \right| dx

I_1 = \int\limits_{0}^{\pi} \sin x dx - \int\limits_{0}^{2 \pi} \sin x dx

= \left[ -\cos x \right]_{0}^{\pi} + \left[ \cos x \right]_{\pi}^{2 \pi}

= -[-1-1] + [+1+1]

= 2 + 2

\therefore I = \frac{1}{2} \left[ 4+4+4+ \cdots n \text{ times} \right]

= \frac{1}{2}(4n) = 2n

= 4

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