Question

Make a rough sketch of the graph y=sin 2x,0$⩽x⩽\pi /2$and find the area enclosed by curve and x axis.

Answer 1

The equation of the curve is $y=sin2x$ $0\le x\le \frac{\pi }{2}$

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

In $0\le x\le \frac{\pi }{2}$, the curve meets the $x$ axis at points $(0,0)$ and from $\left(1\right)$, $\frac{dy}{dx}=2$ and $\cos 2x$ and $\frac{d^{2}y}{{dx}^2}=-4\sin 2x$.

$\frac{dy}{dx}>0$ in $(0,\frac{\pi }{4})$ and $\frac{dy}{dx}<0$ in $(\frac{\pi }{4},\frac{\pi }{2})$

The curve is in creasing in $(0,\frac{\pi }{4})$ and decreased in $(\frac{\pi }{4},\frac{\pi }{2})$.

The point $x=\frac{\pi }{4}$ is the point of local maximum and the minimum value is $1$

A rough sketch of the curve $\left(1\right)$ is drawn is fig and the required area is shaded.

Hence the required are.

$={\int }_0^{\pi /2}ydx={\int }_0^{\pi /2}\sin 2xdx={\{-\frac{\cos 2x}{2}\}}_0^{\pi /2}$

$=\frac{-1}{2}(cosx-\cos 0)=\frac{-1}{2}(-1-1)=1$ sq unit.