Which Part Should I Integrate?

By integration by parts:

$\begin{aligned} I & = \int_0^1 xf''(2x) \ dx \\ & =x\cdot \frac 12 f'(2x) \bigg|_0^1 - \int_0 ^1 \frac 12 f'(2 x) \ dx \\ & =1\cdot \frac 12 f’(2) - 0 - \frac 14 f(2x)\bigg|_0 ^1 \\ &= \frac 52 - \frac 14(f(2) - f(0)) \\&= \frac 52 - \frac 14(3 - 1)\\&=\boxed {2} \end{aligned}$

Relevant wiki: Integration by Parts - Intermediate

Substitute $2x=t$ such that :- $4I=\int_0^2 tf''(t)\mathrm{d}t$

Now apply integration by parts:-

$\large 4I= \left[tf'(t)\right]_0^2-\overbrace{\displaystyle\int_0^2f'(t)\mathrm{d}t}^{\large \left[f(t)\right]_0^2}$

$\large ~~~=2f'(2)-(f(2)-f(1))=8$

$\Large \implies I=\boxed 2$

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$\color{#3D99F6}{\underline{\textbf{NOTE:-}}}$

$\small{\because \dfrac{\mathrm d(f(t)+C)}{\mathrm dt}=f'(t)~~\therefore \displaystyle\int f'(t)\mathrm{d}t=f(t)+C \\\because \dfrac{\mathrm d(f'(t)+C)}{\mathrm dt}=f''(t)~~\therefore \displaystyle\int f''(t)\mathrm{d}t=f'(t)+C }$ $$ $\small{\color{grey}{C-\text{Arbitrary real constant }}}$

Integrating by parts, let u = x, dv = f""(2x)dx. Then du = dx, v = (1/2)f"(2x).. (1/2)xf"(2x) - (1/2)*integral of f"(2x)dx , all evaluated from o to 1 gives: (1/2)f"2) - (1/4)f(2) + (1/4)f(0) = 5/2 - 3/4 + 1/4 = 2. Ed Gray

do integration by part!

let:

$u = x$ and $dv = f''(2x)$

do it once again with

$u = 1$ and $dv = 1/2 f'(2x)$

if you don't want to do some tedious work, do it in tabular way!