Discussion Forum

The area enclosed by the curves y= sin x+cos x and y=|cosx-sinx| over the interval  $\left[0,\frac{\pi}{2}\right]$ is

Answer:

Explanation:

Concept involved

 T o find the bounded area between y=f(x)  and y= g(x)  between  x=a to x= b

 $\therefore$ Area bounded

     $=\int_{a}^{c} [g(x) -f(x)]dx+\int_{c}^{b} [f(x)-g(x)]dx$

  $\Rightarrow \int_{a}^{b} |f(x)-g(x)| dx$

 Here , f(x)=y=sinx+cos x , when

              $0\leq x \leq\frac{\pi}{2}$

 and g(x) = y=|cos x-sinx |

   $=\begin{cases}\cos x- \sin x, & 0\leq  x\leq\frac{\pi}{4} \\\sin x-\cos x, & \frac{\pi}{4}\leq x \leq \frac{\pi}{2}\end{cases}$

Could  be shown as

$\therefore$ Area bounded

$=\int_{0}^{\frac{\pi}{4}} \left\{(sinx+cos x)-(cos x-sinx)\right\} dx$

              $+\int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \left\{(sinx+cos x)-(sin x-cosx)\right\} dx$

  $=\int_{0}^{\frac{\pi}{4}} 2 sin x dx+\int_{\frac{\pi}{4}}^{\frac{\pi}{2}} 2cos x dx$

  $=-2(cos x)^{\frac{\pi}{4}}_{0}+2(sin x.n)^{\frac{\pi}{2}}_{\frac{\pi}{4}}$

    $=4-2\sqrt{2}=2\sqrt{2}(\sqrt{2}-1)$