Let A_1 be the area of the region bounded by the curves y= x, y= x and the y-axis in the f…

Question

Let $A_1$ be the area of the region bounded by the curves $y=\sin x$, $y=\cos x$ and the $y$-axis in the first quadrant. Also, let $A_2$ be the area of the region bounded by the curves $y=\sin x$, $y=\cos x$, the $x$-axis, and $x=\dfrac{\pi}{2}$ in the first quadrant. Then:

Accepted Answer

$\sin x=\cos x$ at $x=\pi/4$, giving point $B\left(\dfrac{\pi}{4},\dfrac{1}{\sqrt2}\right)$.

$$A_1 = \int_0^{\pi/4}(\cos x-\sin x)\,dx = [\sin x+\cos x]_0^{\pi/4} = \sqrt2-1$$

$$A_2 = \int_0^{\pi/4}\sin x\,dx + \int_{\pi/4}^{\pi/2}\cos x\,dx = [-\cos x]_0^{\pi/4}+[\sin x]_{\pi/4}^{\pi/2} = \sqrt2(\sqrt2-1)$$

So $A_1 : A_2 = (\sqrt2-1) : \sqrt2(\sqrt2-1) = 1:\sqrt2$, and $A_1+A_2 = (\sqrt2-1)(1+\sqrt2) = 2-1 = 1$.

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