CUET Preparation Today
The area bounded by the y-axis, y=cosx and y=sinx when $0≤x≤\frac{\pi }{2} $ is : |
$\sqrt{2}\, sq.units $ $(2\sqrt{2} + 1) sq.units $ $(\sqrt{2} + 1) sq.units $ $(\sqrt{2} - 1) sq.units $ |
$(\sqrt{2} - 1) sq.units $ |
The correct answer is Option (4) → $(\sqrt{2} - 1) sq.units$
Required area = $\int\limits_0^{\pi/4}\cos x-\sin x\,dx$ $=\left[\sin x+\cos x\right]_0^{\pi/4}$ $=\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-1-0$ $=\sqrt{2}-1$ |
