Find an integer to make a given integral formula true

Find a value for $n \in \mathbb{Z}$ such that
$n \int_0^1 x f''(2x) \, dx = \int_0^2 t f''(t) \, dt.$
Given that
$f(0) = 1, \qquad f(2) = 3, \qquad f'(2) = 5$

compute the integral

$\int_0^1 xf''(2x) \, dx.$

To do this we use the expansion/contraction property of the integral (which we proved here). We have
$\begin{align*} \int_0^2 t f''(t) \, dt &= n \int_0^1 x f''(2x) \, dx \\ &= \frac{n}{2} \int_0^2 \frac{x}{2} f''(x) \, dx \\ &= \frac{n}{4} \int_0^2 x f''(x) \, dx \\ &= \frac{n}{4} \int_0^2 t f''(t) \, dt. \end{align*}$

For this equation to be true we must have $n =4$ .
To compute this we use the formula we established in part (a),
$\int_0^1 x f''(2x) \, dx = \frac{1}{4} \int_0^2 x f''(x) \, dx.$

Then, we use integration by parts with

$\begin{align*} u &= x & \implies && du &= dx \\ dv &= f''(x) \, dx & \implies && v &= f'(x) \end{align*}$

Therefore,

$\begin{align*} \frac{1}{4} \int_0^2 xf''(x) \, dx &= \frac{1}{4} \left( xf'(x) \Big \rvert_0^2 - \int_0^2 f'(x) \, dx \right) \\ &= \frac{1}{4} \left( 2f'(2) - f(2) + f(0) \right) \\ &= \frac{1}{4} (10 - 3 + 1) \\ &= 2. \end{align*}$