Area bounded by the curves \( y=cos^2x\ ,x=0\ ,y=0 \) and \( x=\pi \), is:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( \displaystyle\frac{\pi}{2} \) Square units
The curve is never below the x-axis, and \( \cos^2x=\frac{1+\cos 2x}{2} \), so
\( \displaystyle \int_0^{\pi}\cos^2x\,dx=\frac{1}{2}\left[x+\frac{\sin 2x}{2}\right]_0^{\pi} \)
\( \displaystyle =\frac{1}{2}(\pi+0) \)
\( \displaystyle =\frac{\pi}{2}\ \text{square units} \)