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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 $A_1$ and $A_2$. Then $A_1, A_2$ equals
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Verified Answer
The correct answer is:
1: 1
Area,
$\begin{aligned}
A_1 & =\int_0^{\pi / 4} \sin x d x \\
& =-[\cos x]_0^{\pi / 4} \\
& =1-\frac{1}{\sqrt{2}}=\frac{\sqrt{2}-1}{\sqrt{2}}
\end{aligned}$
and area, $A_2=\int_{\pi / 4}^{\pi / 2} \cos x d x$
$\begin{aligned} & =[\sin x]_{\pi / 4}^{\pi / 2}=\left[1-\frac{1}{\sqrt{2}}\right]=\frac{\sqrt{2}-1}{\sqrt{2}} \\ & \therefore \quad A_1: A_2=\frac{\sqrt{2}-1}{\sqrt{2}}: \frac{\sqrt{2}-1}{\sqrt{2}}=1: 1 \\ & \end{aligned}$
$\begin{aligned}
A_1 & =\int_0^{\pi / 4} \sin x d x \\
& =-[\cos x]_0^{\pi / 4} \\
& =1-\frac{1}{\sqrt{2}}=\frac{\sqrt{2}-1}{\sqrt{2}}
\end{aligned}$
and area, $A_2=\int_{\pi / 4}^{\pi / 2} \cos x d x$
$\begin{aligned} & =[\sin x]_{\pi / 4}^{\pi / 2}=\left[1-\frac{1}{\sqrt{2}}\right]=\frac{\sqrt{2}-1}{\sqrt{2}} \\ & \therefore \quad A_1: A_2=\frac{\sqrt{2}-1}{\sqrt{2}}: \frac{\sqrt{2}-1}{\sqrt{2}}=1: 1 \\ & \end{aligned}$
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