Q. The area enclosed between the curves $y = \sin^{2}x$ and $y = \cos^{2}x$ in the interval $0 \leq x \leq \pi$ is
A) 2 sq unit
B) $\frac{1}{2}$ sq unit
C) 1 sq unit
D) None of these
Explanation
(c)
Required area $= \int_{\pi/4}^{3\pi/4}{(\sin^{2}{x - \cos^{2}{x)dx}}}$
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$= - \int_{\pi/4}^{3\pi/4}{\cos{(2x)dx = - \left\lbrack \frac{\sin{2x}}{2} \right\rbrack_{\pi/4}^{3\pi/4}}}$
$= - \frac{1}{2}\left( \sin{\frac{3\pi}{2} - \sin\frac{\pi}{2}} \right) = 1\ sq\ unit$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.