(Đề tham khảo 2018 câu 50)

Đáp án:

Xét: \( \int_0^1 x^2 f(x) dx. \)

Đặt:\(
\begin{cases}
u = f(x) \\
dv = x^2 dx
\end{cases} \implies \begin{cases} du = f'(x) dx \\  v = \frac{x^3}{3} \end{cases}\)

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

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

Tìm \( k \) sao cho: 
\( \int_0^1 \left[ f'(x) - kx^3 \right]^2 dx = 0 \Leftrightarrow \int_0^1 \left( f'(x) \right)^2 dx - 2k \int_0^1 x^3 f'(x) dx + k^2 \int_0^1 x^6 dx = 0. \)

\( \Leftrightarrow 7 + 2k + \frac{k^2}{7} = 0 \Leftrightarrow k^2 + 14k + 49 = 0 \Leftrightarrow k = -7. \)

Suy ra: \( f'(x) = kx^3 = -7x^3 \implies f(x) = -\frac{7x^4}{4} + C. \)

\( f(1) = 0 \implies C = \frac{7}{4} \)

\( f(x) = -\frac{7}{4}x^4 + \frac{7}{4} \implies \int_0^1 f(x) dx = \frac{7}{5} \implies \boxed{A} \)