Đáp án:

Xét \( \int_0^1 x^3 f(x) dx. \)
\( \begin{cases} 
u = f(x)  \\ 
dv = x^3 dx 
\end{cases} \Rightarrow \begin{cases} du = f'(x)  \\ v = \frac{x^4}{4} \end{cases}\)

\( \frac{1}{2} = \int_0^1 x^3 f(x) dx =  \frac{x^4}{4} f(x) \big|_0^1 - \int_0^1 \frac{x^4}{4} f'(x) dx. \)

\( \Rightarrow \int_0^1 \frac{x^4}{4} f'(x) dx = -1. \)

Tìm \( k \) sao cho:

\( \int_0^1 \left[ f'(x) - kx^4 \right]^2 dx = 0 \Leftrightarrow \int_0^1 \left( f'(x) \right)^2 dx - 2k \int_0^1 x^4 f'(x) dx + k^2 \int_0^1 x^8 dx = 0. \)

\( \Rightarrow 9 + 2k  +   \frac{k^2}{9} = 0 \Leftrightarrow k^2 + 18k + 81 = 0 \Leftrightarrow  k=-9\)

\( \Rightarrow f'(x) = -9x^4 \implies f(x) = -\frac{9x^5}{5} + C. \)

\( f(1) = 1 \Rightarrow C = 1 + \frac{9}{5} = \frac{14}{5}. \)

\( \Rightarrow\int_0^1 f(x) dx = \int_0^1 \left( -\frac{9x^5}{5} + \frac{14}{5} \right) dx = \frac{5}{2}. \)