Let A_{1} be the area of the region bounded by the curves

Question:

Let $A_{1}$ be the area of the region bounded by the curves $y=\sin x, y=\cos x$ and $y$-axis in the first quadrant. Also, let $A_{2}$ be the area of the region bounded by the curves $y=\sin x, y=\cos x, x$-axis and $x=\frac{\pi}{2}$ in the first quadrant. Then,

1. (1) $\mathrm{A}_{1}=\mathrm{A}_{2}$ and $\mathrm{A}_{1}+\mathrm{A}_{2}=\sqrt{2}$

2. (2) $\mathrm{A}_{1}: \mathrm{A}_{2}=1: 2$ and $\mathrm{A}_{1}+\mathrm{A}_{2}=1$

3. (3) $2 \mathrm{~A}_{1}=\mathrm{A}_{2}$ and $\mathrm{A}_{1}+\mathrm{A}_{2}=1+\sqrt{2}$

4. (4) $\mathrm{A}_{1}: \mathrm{A}_{2}=1: \sqrt{2}$ and $\mathrm{A}_{1}+\mathrm{A}_{2}=1$

Correct Option: , 4

Solution:

$A_{1}+A_{2}=\int_{0}^{\pi / 2} \cos x \cdot d x=\left.\sin x\right|_{0} ^{\pi / 2}=1$

$A_{1}=\int_{0}^{\pi / 4}(\cos x-\sin x) d x=\left.(\sin x+\cos x)\right|_{0} ^{\pi / 4}=\sqrt{2}-1$

$\therefore A_{2}=1-(\sqrt{2}-1)=2-\sqrt{2}$

$\therefore \frac{A_{1}}{A_{2}}=\frac{\sqrt{2}-1}{\sqrt{2}(\sqrt{2}-1)}=\frac{1}{\sqrt{2}}$