# If $f'' \ge 0$, $\int_0^2 f(x)dx \ge 2f(1)$

Question:

If $$f'' \ge 0$$ in interval $$[0, 2]$$, prove that $$\int_0^2 f(x)dx \ge 2f(1)$$

The question is graphically trivial I think, but not in mathematically.

I wanted to use the fact that there $$\exists c$$ s.t. $$\int_0^2f(x)dx = 2f(c)$$ and the Jensen's Thm, which is $$f(tx_1+(1-t)x_2)\le tf(x_1)+(1-t)f(x_2)$$

I tried to find the characteristics that $$c$$ can have, but it was still hard for me to get an idea. Could you please give some key points to the problem? Thanks.

By a function form of Jensen's inequality, it holds that $$\frac{1}{2} \int_0^2f(x)dx \geq f(\frac{1}{2}\int_0^2xdx) = f(1)$$. (See https://en.wikipedia.org/wiki/Jensen%27s_inequality)

• +1. Nice proof. – Kavi Rama Murthy Feb 9 at 5:16
• I think I got an idea. Thanks:D – ToBY Feb 9 at 5:18

$$\int_0^2 f(x)dx$$ =

$$lim_{h\rightarrow 0}\ h[f(0)+f(0+h) + ........f(0 + (n-1)h)]\ =\ I$$ where $$nh=2$$.

By jenson inquality-

$$\frac{I}{\sum_{r=0}^{n-1}h}\ =\ \frac{I}{2}\ \ge\$$

$$f(lim_{h\rightarrow 0}\ h.0 + h(0+h) + h(0+2h)...... h(0+(n-1)h)$$ (where nh=2) =

$$f(\int_0^2 xdx)$$ = $$f(1)$$.