show that $F''$ is strictly increasing.

Question

If $f$ is continuous and always positive in $[0,\infty)$ and $$F(x)=\frac{1}{2}\int_0^x(x-t)^2f(t)\,dt$$then show that $F''$ is strictly increasing.

I found that the integrand is continuous and so , $F$ is differentiable and $F'(x)=0$. But to show $F''$ is strictly increasing we have to show that $F'''(x)>0$ in $[0,\infty)$. How it is possible ?

Answer 1

HINT:

$$\begin{align} F'(x)&=\int_0^x 2(x-t)f(t)dt\\\\ F''(x)&=\int_0^x 2f(t)dt\\\\ F'''(x)&=2f(x)>0 \end{align}$$

Answer 2

Expand the square: $$2F(x)=x^2\int_0^xf(t)dt-2x\int_0^xtf(t)dt+\int_0^xt^2f(t)dt$$ Differentiate: $$\begin{align} 2F'(x)&=2x\int_0^xf(t)dt+x^2f(x)-2\int_0^xtf(t)dt-2x^2f(x)+x^2f(x)\\ &=2x\int_0^xf(t)dt-2\int_0^xtf(t)dt \end{align}$$ Divide by two and differentiate again: $$F''(x)=\int_0^xf(t)dt+xf(x)-xf(x)=\int_0^xf(t)dt$$ And again: $$F'''(x)=f(x)>0$$