The area, enclosed by the curves y= x+ x and y=| x- x|, and the lines x=0, x=2, is:

Question

The area, enclosed by the curves $y=\sin x+\cos x$ and $y=|\cos x-\sin x|$, and the lines $x=0$, $x=\dfrac{\pi}{2}$, is:

Accepted Answer

$$A = \int_0^{{\pi  \over 2}} {\left( {(\sin x + \cos x) - \left| {\cos x - \sin x} \right|} \right)\,dx} $$

$$A = \int_0^{{\pi  \over 2}} {\left( {(\sin x + \cos x) - (\cos x - \sin x)} \right)\,dx}  + \int_{{\pi  \over 4}}^{{\pi  \over 2}} {\left( {(\sin x + \cos x) - (\sin x - \cos x)} \right)\,dx} $$

$$A = 2\int_0^{{\pi  \over 2}} {\sin x\,dx + 2\int_{{\pi  \over 4}}^{{\pi  \over 2}} {\cos x\,dx} } $$

$$A =  - 2\left( {{1 \over {\sqrt 2 }} - 1} \right) + \left( {1 - {1 \over {\sqrt 2 }}} \right)$$

$$A = 4 - 2\sqrt 2  = 2\sqrt 2 (\sqrt 2  - 1)$$

Option (a)

Solutions · 1