The line x=frac{pi}{4} divides the area of th | vedclass

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(D) The region is bounded by $y=\sin x$,$y=\cos x$ and the $x$-axis for $0 \leq x \leq \frac{\pi}{2}$.For $0 \leq x \leq \frac{\pi}{4}$,the region is

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$1$

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(D) The region is bounded by $y=\sin x$,$y=\cos x$ and the $x$-axis for $0 \leq x \leq \frac{\pi}{2}$.For $0 \leq x \leq \frac{\pi}{4}$,the region is bounded by $y=\sin x$ and the $x$-axis. Thus,the area $A_1$ is given by:$A_1 = \int_0^{\pi/4} \sin x \, dx = [-\cos x]_0^{\pi/4} = -(\cos \frac{\pi}{4} - \cos 0) = -(\frac{1}{\sqrt{2}} - 1) = 1 - \frac{1}{\sqrt{2}} = \frac{\sqrt{2}-1}{\sqrt{2}}$.For $\frac{\pi}{4} \leq x \leq \frac{\pi}{2}$,the region is bounded by $y=\cos x$ and the $x$-axis. Thus,the area $A_2$ is given by:$A_2 = \int_{\pi/4}^{\pi/2} \cos x \, dx = [\sin x]_{\pi/4}^{\pi/2} = \sin \frac{\pi}{2} - \sin \frac{\pi}{4} = 1 - \frac{1}{\sqrt{2}} = \frac{\sqrt{2}-1}{\sqrt{2}}$.Comparing the two areas,we get $A_1 : A_2 = \frac{\sqrt{2}-1}{\sqrt{2}} : \frac{\sqrt{2}-1}{\sqrt{2}} = 1 : 1$.

The line $x=\frac{\pi}{4}$ divides the area of the region bounded by $y=\sin x$,$y=\cos x$ and the $x$-axis $(0 \leq x \leq \frac{\pi}{2})$ into two regions of areas $A_1$ and $A_2$. Then $A_1 : A_2$ equals (in $: 1$)

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