Prove that Integration of sqrt sinx from 0 to π/2 is greater than π/3

Question

Prove that $\int_0^{ \pi/2}\sqrt{\sin x}dx>\pi/3$ I can prove it greater than $1$ but this I don't know how to solve

Answer 1

The graph of $\sin(x)$ is concave on $[0,\pi/2]$, so $$\sin(x) \ge \frac{2}{\pi}x \text{ for } x \in [0,\pi/2]$$ Therefore $$\int_0^{\pi/2}\sqrt{\sin(x)}dx \ge \int_0^{\pi/2}\sqrt{\frac{2}{\pi}x} dx = \frac{\pi}{3}$$

Answer 2

Very good question!

The solution is as follow:

$\int_0^{\frac{\pi}{2}} \sqrt{sin(x)}dx \geq \int_0^{\frac{\pi}{2}} \sqrt{\frac{2x}{\pi}}dx=\frac{\pi}{3}$!

It is not a trivial question I agree.

You get the result of integration in the end by by guessing.