Đáp án:

Xét \(\int_{0}^{1} (x^2 + x) f''(x) dx \quad \)  Đặt \(\begin{cases} u = x^2 + x \\ dv = f''(x) dx \end{cases} \Rightarrow \begin{cases} du = (2x + 1) dx \\ v = f'(x) \end{cases}\)  

\(\frac{5}{2} = \int_{0}^{1} (x^2 + x) f''(x) dx = [x^2 + x] f'(x) \big|_{0}^{1} - \int_{0}^{1} (2x + 1) f'(x) dx\)  

\(\frac{5}{2} = 9 - \int_{0}^{1} (2x + 1) f'(x) dx \Rightarrow \int_{0}^{1} (2x + 1) f'(x) dx = 9 - \frac{5}{2} = \frac{13}{2}\).  

Tìm \(k\) sao cho: \(\int_{0}^{1} \left[ f'(x) - k(2x + 1) \right]^2 dx = 0\)  

\(\Leftrightarrow \int_{0}^{1} (f'(x))^2 dx - 2k \int_{0}^{1} (2x + 1) f'(x) dx + k^2 \int_{0}^{1} (2x + 1)^2 dx = 0\)  

\(\Leftrightarrow \frac{39}{4} - 13k + \frac{13}{3}k^2 = 0 \Rightarrow \frac{k^2}{3} - k + \frac{3}{4} = 0 \Leftrightarrow k = \frac{3}{2}\).

Suy ra: \(f'(x) = \frac{3}{2} (2x + 1)\) \(\Rightarrow f(x) = \frac{3x^2}{2} + \frac{3x}{2} + C.\)  

\(f(0) = 0 \Rightarrow C = 0\) \(\Rightarrow f(x) = \frac{3x^2}{2} + \frac{3x}{2}\)  

\(I = \int_{0}^{2} f(x) dx = 7 \Rightarrow \boxed{D}\)  

\(!\) Gặp: Cho \(\int_{a}^{b} (f(x))^2 dx = ...\) hoặc \(\int_{a}^{b} g(x) f(x) dx = ... \), thì tìm \(k\) sao cho: \(\int_{a}^{b} \left[ f(x) - k g(x) \right]^2 dx = 0\) \(\Rightarrow f(x) = k g(x).\)