Consider the region R bounded by $y = \sin x $ and $y$ =$\frac{4x^2}{\pi^2}$. Find the area of R.

Question

Consider the region R bounded by $y = \sin x $ and $y$ =$\frac{4x^2}{\pi^2}$. Find the area of R.

My attempt

$$\sin x=\frac{4x^2}{\pi^2}$$

$$\pi^2\sin x=4x^2$$

$$4x^2-\pi^2\sin x=0$$

But, I am struck here. Can anyone please help to find the area

Answer 1

The region bounded by the graphs of these functions has an area equal to$$\int_0^{\frac\pi2}\sin(x)-\frac{4x^2}{\pi^2}\,\mathrm dx.$$Can you take it from here?

Answer 2

Hint: The Solutions are given by $x=0$ or $x=\frac{\pi}{2}$