Question:

Step 1: Find intersection point To find the intersection point of the curves $y = \sin x$ and $y = \cos x$ in the first quadrant, we set them equal: $$\sin x = \cos x$$ Dividing by $\cos x$ (since $\cos x \neq 0$ in the first quadrant at the intersection): $$\tan x = 1$$ In the first quadrant, this occurs at: $$x = \frac{\pi}{4}$$ Step 2: Calculate $A_1$ The region $A_1$ is bounded by the curves $y = \sin x$, $y = \cos x$, and the Y-axis ($x=0$) in the first quadrant. In the interval $[0, \frac{\pi}{4}]$, $\cos x \ge \sin x$. The area $A_1$ is given by the integral: $$A_1 = \int_0^{\pi/4} (\cos x - \sin x) \, dx$$ Evaluating the integral: $$A_1 = [\sin x + \cos x]_0^{\pi/4}$$ $$A_1 = (\sin(\frac{\pi}{4}) + \cos(\frac{\pi}{4})) - (\sin(0) + \cos(0))$$ $$A_1 = \left(\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}\right) - (0 + 1)$$ $$A_1 = \frac{2}{\sqrt{2}} - 1 = \sqrt{2} - 1$$ Step 3: Calculate $A_2$ The region $A_2$ is bounded by the curves $y = \sin x$, $y = \cos x$, the X-axis, and $x = \frac{\pi}{2}$ in the first quadrant. This region can be interpreted as the sum of the areas under each curve from $x=0$ to $x=\frac{\pi}{2}$, minus the area under the upper envelope of the two curves. The area under $y=\sin x$ from $0$ to $\frac{\pi}{2}$ is: $$\int_0^{\pi/2} \sin x \, dx = [-\cos x]_0^{\pi/2} = -\cos(\frac{\pi}{2}) - (-\cos(0)) = 0 - (-1) = 1$$ The area under $y=\cos x$ from $0$ to $\frac{\pi}{2}$ is: $$\int_0^{\pi/2} \cos x \, dx = [\sin x]_0^{\pi/2} = \sin(\frac{\pi}{2}) - \sin(0) = 1 - 0 = 1$$ The area under the upper envelope of the curves $y=\max(\sin x, \cos x)$ from $0$ to $\frac{\pi}{2}$ is: $$\int_0^{\pi/2} \max(\sin x, \cos x) \, dx = \int_0^{\pi/4} \cos x \, dx + \int_{\pi/4}^{\pi/2} \sin x \, dx$$ $$= [\sin x]_0^{\pi/4} + [-\cos x]_{\pi/4}^{\pi/2}$$ $$= (\sin(\frac{\pi}{4}) - \sin(0)) + (-\cos(\frac{\pi}{2}) - (-\cos(\frac{\pi}{4})))$$ $$= \left(\frac{1}{\sqrt{2}} - 0\right) + \left(0 - \left(-\frac{1}{\sqrt{2}}\right)\right)$$ $$= \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} = \frac{2}{\sqrt{2}} = \sqrt{2}$$ Therefore, $A_2$ is calculated as the sum of the individual areas under each curve minus the area under their maximum envelope: $$A_2 = 1 + 1 - \sqrt{2} = 2 - \sqrt{2}$$ Step 4: Compare $A_1$ and $A_2$ and their sum We have $A_1 = \sqrt{2} - 1$ and $A_2 = 2 - \sqrt{2}$. Comparing the values: $A_1 \approx 1.414 - 1 = 0.414$ $A_2 \approx 2 - 1.414 = 0.586$ Thus, $A_1 \neq A_2$. Now, calculate their sum: $$A_1 + A_2 = (\sqrt{2} - 1) + (2 - \sqrt{2})$$ $$A_1 + A_2 = 1$$