Integrating a second derivative

Question

Admit that $f$ has a second derivative find the integer $m$.

$$m\int_{0}^{1}xf''(2x)dx = \int_0^2xf''(x)dx$$

So I took $2x=u$ where $du/dx=2$ and I plugged in the integral getting

$$\frac{m}{4}\int_{0}^{2}uf''(u)du =\frac{1}{4} \int_{0}^{2} uf''(\frac{u}{2})du$$

How do I proceed from here?

Answer 1

So as you observed, substituting $2x \to x$ in the LHS we have

$$m\int_0^1 x f''(2x) \, \mathrm d x = \frac m 4 \int_0^2 x f''(x) \, \mathrm d x$$

Thus $m=4$.

Answer 2

I have no idea where your second integral comes from. Your first integral in the second equality is equal to $m\int_0^1 xf''(2x)\,dx$. For this to be $\int_0^2 x f''(x)dx$, you need $m/4=1$, i.e. $m=4$.