The area of region bounded by the x-axis, $y=cos\, x$ and $y=sin\, x. 0≤x≤\frac{\pi}{2}$
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → $2- \sqrt{2}$ sq.units
The curves $y = \sin x$ and $y = \cos x$ intersect at $x = \frac{\pi}{4}$.
- From $0$ to $\frac{\pi}{4}$, $\sin x < \cos x$, so the region is bounded by the x-axis and $\sin x$.
- From $\frac{\pi}{4}$ to $\frac{\pi}{2}$, $\cos x < \sin x$, so the region is bounded by the x-axis and $\cos x$.
Hence, total area:
$\text{Area} = \int_{0}^{\pi/4} \sin x \, dx + \int_{\pi/4}^{\pi/2} \cos x \, dx$
Evaluating:
- $\int_{0}^{\pi/4} \sin x \, dx = 1 - \frac{\sqrt{2}}{2}$
- $\int_{\pi/4}^{\pi/2} \cos x \, dx = 1 - \frac{\sqrt{2}}{2}$
Adding both:
$\text{Area} = \left( 1 - \frac{\sqrt{2}}{2} \right) + \left( 1 - \frac{\sqrt{2}}{2} \right) = 2 - \sqrt{2}$