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,

• A. $A _{1}: A _{2}=1: \sqrt{2}$ and $A _{1}+ A _{2}=1$
• B. $A _{1}= A _{2}$ and $A _{1}+ A _{2}=\sqrt{2}$
• C. $2 A _{1}= A _{2}$ and $A _{1}+ A _{2}=1+\sqrt{2}$
• D. $A _{1}: A _{2}=1: 2$ and $A _{1}+ A _{2}=1$

Answer 1

Answer: (A)

$A_{1}=\int\limits_{0}^{\pi / 4}(\cos x-\sin x) d x$
$A_{1}=(\sin x+\cos x)_{0}^{\pi / 4}=\sqrt{2}-1$
$A_{2}=\int\limits_{0}^{\pi / 4} \sin x d x+\int\limits_{\pi / 4}^{\pi / 2} \cos x d x$
$=(-\cos x)_{0}^{\pi / 4}+(\sin x)_{\pi / 4}^{\pi / 2}$
$A_{2}=\sqrt{2}(\sqrt{2}-1)$
$A_{1}: A_{2}=1: \sqrt{2}, A_{1}+A_{2}=1$