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The line $x=\frac{\pi}{4}$ divides the area of the region bounded by $y=\sin x, y=\cos x$ and $x$-axis $\left(0 \leq x \leq \frac{\pi}{2}\right)$ into two regions of areas $\mathrm{A}_{1}$ and $\mathrm{A}_{2}$. Then $\mathrm{A}_{1}: \mathrm{A}_{2}$ equals
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Verified Answer
The correct answer is:
$1: 1$
$$
\text { Area, } \begin{aligned}
A_{1} &=\int_{0}^{\pi / 4} \sin x d x \\
&=-[\cos x]_{0}^{\pi / 4}=1-\frac{1}{\sqrt{2}} \\
&=\frac{\sqrt{2}-1}{\sqrt{2}}
\end{aligned}
$$
$$
\begin{array}{l}
\text { and area, } A_{2}=\int_{\pi / 4}^{\pi / 2} \cos x d x \\
\quad=[\sin x]_{\pi / 4}^{\pi / 2}=\left[1-\frac{1}{\sqrt{2}}\right]=\frac{\sqrt{2}-1}{\sqrt{2}} \\
\therefore A_{1}: A_{2}=\frac{\sqrt{2}-1}{\sqrt{2}}: \frac{\sqrt{2}-1}{\sqrt{2}}=1: 1
\end{array}
$$
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