Question

Make a rough sketch of the graph of y=$si{n}^{2}x$,0<x<x/2 then find the area

enclosed between the curve,x axis and the line

Answer 1

The equation of the curve is $y={\sin }^{2}x,0\le x\le \frac{\pi }{2}$

If $x=0$, then $y=0$, and if $x=\frac{\pi }{2}$, then $y=1$

The curve passes through the points $(0,0)$ and $(\frac{x}{2},1)$ from $\left(1\right)$, $\frac{dy}{dx}=2sinxcosx=\sin 2x$

$\therefore$ $\frac{dy}{dx}>0$ in $(0,\frac{\pi }{2})$ The curve is increasing in $(0,\frac{\pi }{2})$ with these ideas we draw a rough sketch of the curve in figure and shade the required area $OA$, $B$ in which $x$ varies from $0$ to $\frac{\pi }{2}$

Hence the required area $={\int }_0^{\pi /2}ydx={\int }_0^{\pi /2}{\sin }^{2}xdx$

$=\frac{1}{2}{\int }_0^{\pi /2}(1-\cos 2x)dx$

$=\frac{1}{2}{[x-\frac{\sin 2x}{2}]}_0^{\pi /2}=\frac{1}{2}[\{\frac{\pi }{2}-0\}-0]$

$=\frac{\pi }{4}$ square units.