Đáp án:

\( \frac{2}{5} = \int_0^1 f(\sqrt{x}) dx = \int_0^1 f(\sqrt{x}) \cdot 2\sqrt{x} \, d(\sqrt{x}) = 2 \int_0^1 t f(t) dt. \)

\( \begin{cases}
u = f(t)  \\
dv = t dt
\end{cases}  \implies \begin{cases} du = f'(t) dt \\ v = \frac{t^2}{2} \end{cases}\)

\( \frac{1}{5} = \int_0^1 t f(t) dt = \frac{t^2}{2} f(t) \big|_0^1 - \int_0^1 \frac{t^2}{2} f'(t) dt \)

\( = \frac{1}{2} - \int_0^1 \frac{t^2}{2} f'(t) dt \implies \int_0^1 t^2 f'(t) dt = 2 \left( \frac{1}{2} - \frac{1}{5} \right) = \frac{3}{5}. \)

\(\implies \int_0^1 x^2 f'(x) dx = \frac{3}{5}\)

Tìm \( k \) sao cho:

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

\( \frac{9}{5} - \frac{6k}{5} + \frac{k^2}{5} = 0 \Leftrightarrow (k-3)^3=0 \Leftrightarrow k = 3. \)

Suy ra: \( f'(x) = 3x^2 \implies f(x) = x^3 + C. \)

\( f(1) = 1 \Rightarrow C = 0. \)

\( f(x) = x^3 \Rightarrow \int_0^1 f(x) dx = \frac{1}{4} \implies \boxed{D} \)