Indian Statistical Institute B.Math & B.Stat: Integration and Continuity

If $c \int_{0}^{1} xf(2x) dx = \int_{0}^{2} tf(t) dt,$ where $f$ is a positive continuous functions, then find then value of $c$. In the R.H.S put $2x=t$, this gives $c \int_{0}^{1} xf(2x) dx = 4\int_{0}^{1} xf(2x) dx = (c-4) \int_{0}^{1} xf(2x) dx =0 \implies c =4$ Since $f$ is given to be a positive continuous function, $xf(x)$ is continuous and $>0 \in (0,1)$. Therefore $\int_{0}^{1} xf(2x) dx > 0$. This follows from the fact that if a continuous functions is positive at a point in its domain, the $\exists$ a open neighborhood containing the point and contained in the domain throughout which the function is positive.