$h:[0,1]\rightarrow\textbf{R}$ a continuous function such that $h(0) =h(1) =0$, then $\forall c \in (0,1)$ is it true? $\int_{0}^{c}h\, dx=(1-c)\int_{0}^{1}h\, dx$??
I had a problem in R.I exercise, and while solving it, I got this equality, but then stuck here. If I substitute $x=ct$ then $\int_{0}^{c}h\, dxdx=c\int_{0}^{1}h(cx)\, dx$, doing little calculation gives $(1-c)\int_{c}^{1}h\,dx=c\int_{0}^{c}h\,dx$. But nothing is helping much. Can anyone prove or disprove(please give an counter-example) the above will be helpful
This is not true. Counter-example: Let $g(x)=x^{2}(1-x)^{2}$ and $h(x)=g'(x)=2x(1-x)^{2}-2x^{2}(1-x)$. Then $h$ is continuous, $h(0)=h(1)=0$ and $\int_0^{1}h(t)dt=g(1)-g(0)=0$. But $\int_0^{c} h(t) dt =g(c) > 0$ for $0<c<1$.