Question

Find the area of the figure bounded by the following curves
$y=2+sinx,y=1.+co{s}^{2}x,x=0.x?$

Answer 1

$y=2+sinx$

$y=1+{\cos }^{2}x$,

$x=0$

$2+sinx=1+{\cos }^{2}x$

$1+sinx=1-{\sin }^{2}x$

$sinx+{\sin }^{2}x=0$

$sinx(1+sinx)=0$

$sinx=0$ , $sinx=-1$

$x=0$ or $x=-\frac{\pi }{2}$

${\int }_{-\frac{\pi }{2}}^0(2+sinx)-(1+{\cos }^{2}x)dx$

$={\int }_{-\frac{\pi }{2}}^{0}1+sinx-{\cos }^{2}xdx$

$={\int }_{-\frac{\pi }{2}}^0(1+sinx-\frac{1+\cos 2x}{2})dx$

$={\int }_{\frac{-\pi }{2}}^0\frac{1}{2}+sinx-\frac{\cos 2x}{2}dx$

$=\frac{x}{2}-cosx+\frac{\sin 2x}{4}{]}_{\frac{-\pi }{2}}^0$

$=\frac{0}{2}-\cos 0+0+\frac{\pi }{2}+\cos \frac{\pi }{2}-\frac{\sin \pi }{4}$

$=-1+\frac{\pi }{2}$ Sq units