Discussion Forum

 The area bounded by the curves y= cos x and y= sin x between the ordinates x=0 and   $x=\frac{3\pi}{2}$  is

Answer:

Explanation:

 Required area 

$\int_{0}^{\frac{\pi}{4}} (\cos x-\sin x)dx+\int_{\pi/4}^{5\pi/4} (\sin x-\cos x)dx$

                                                                    $\int_{5\pi/4}^{3\pi/2} (\cos x-\sin x)dx$

=$\left[\sin x+\cos x\right]_{0}^{\pi/4}+[-cos x-sinx]^{5\pi/4}_{\pi/4}+[sin x+cos x]^{3\pi/4}_{5x/4}$

=  $(4\sqrt{2}-2)$ sq.units