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Câu hỏi: Giả sử hàm số $f(x)$ có đạo hàm cấp 2 trên $\mathbb{R}$ thỏa mãn $f(1)=f^{\prime}(1)=1$ và $f(1-x)+x^2 \cdot f^{\prime \prime}(x)=$ $2 x$ với mọi $x \in \mathbb{R}$. Tính tích phân $I=\int_0^1 x f^{\prime}(x) \mathrm{d} x$.
A. $I=\dfrac{1}{3}$.
B. $I=\dfrac{2}{3}$.
C. $I=1$.
D. $I=2$.
A. $I=\dfrac{1}{3}$.
B. $I=\dfrac{2}{3}$.
C. $I=1$.
D. $I=2$.
Đặt $\left\{\begin{array}{l}u=f^{\prime}(x) \\ d v=x \mathrm{~d} x\end{array} \Rightarrow\left\{\begin{array}{l}\mathrm{d} u=f^{\prime \prime}(x) \mathrm{d} x \\ v=\dfrac{x^2}{2}\end{array}\right.\right.$.
Suy ra $I=\int_0^1 x f^{\prime}(x) \mathrm{d} x=\left.\dfrac{x^2}{2} f^{\prime}(x)\right|_0 ^1-\int_0^1 \dfrac{x^2}{2} f^{\prime \prime}(x) \mathrm{d} x=\dfrac{1}{2}-\int_0^1 \dfrac{x^2}{2} f^{\prime \prime}(x) \mathrm{d} x$.
Do $f(1-x)+x^2 \cdot f^{\prime \prime}(x)=2 x \Rightarrow \dfrac{x^2}{2} \cdot f^{\prime \prime}(x)=x-\dfrac{1}{2} f(1-x)$.
Vậy $I=\dfrac{1}{2}-\int_0^1\left[x-\dfrac{1}{2} f(1-x)\right] \mathrm{d} x=\dfrac{1}{2} \int_0^1 f(1-x) \mathrm{d} x$.
Đặt $t=1-x$ suy ra $I=-\dfrac{1}{2} \int_1^0 f(t) \mathrm{d} t=\dfrac{1}{2} \int_0^1 f(t) \mathrm{d} t=\dfrac{1}{2} \int_0^1 f(x) \mathrm{d} x$.
Đặt $\left\{\begin{array}{l}u=f(x) \\ d v=\mathrm{d} x\end{array} \Rightarrow\left\{\begin{array}{l}\mathrm{d} u=f^{\prime}(x) \mathrm{d} x \\ v=x\end{array}\right.\right.$
Suy ra $I=\dfrac{1}{2}\left[\left.x f(x)\right|_0 ^1-\int_0^1 x f^{\prime}(x) d x\right] \Leftrightarrow I=\dfrac{1}{2}(1-I) \Leftrightarrow I=\dfrac{1}{3}$.
Suy ra $I=\int_0^1 x f^{\prime}(x) \mathrm{d} x=\left.\dfrac{x^2}{2} f^{\prime}(x)\right|_0 ^1-\int_0^1 \dfrac{x^2}{2} f^{\prime \prime}(x) \mathrm{d} x=\dfrac{1}{2}-\int_0^1 \dfrac{x^2}{2} f^{\prime \prime}(x) \mathrm{d} x$.
Do $f(1-x)+x^2 \cdot f^{\prime \prime}(x)=2 x \Rightarrow \dfrac{x^2}{2} \cdot f^{\prime \prime}(x)=x-\dfrac{1}{2} f(1-x)$.
Vậy $I=\dfrac{1}{2}-\int_0^1\left[x-\dfrac{1}{2} f(1-x)\right] \mathrm{d} x=\dfrac{1}{2} \int_0^1 f(1-x) \mathrm{d} x$.
Đặt $t=1-x$ suy ra $I=-\dfrac{1}{2} \int_1^0 f(t) \mathrm{d} t=\dfrac{1}{2} \int_0^1 f(t) \mathrm{d} t=\dfrac{1}{2} \int_0^1 f(x) \mathrm{d} x$.
Đặt $\left\{\begin{array}{l}u=f(x) \\ d v=\mathrm{d} x\end{array} \Rightarrow\left\{\begin{array}{l}\mathrm{d} u=f^{\prime}(x) \mathrm{d} x \\ v=x\end{array}\right.\right.$
Suy ra $I=\dfrac{1}{2}\left[\left.x f(x)\right|_0 ^1-\int_0^1 x f^{\prime}(x) d x\right] \Leftrightarrow I=\dfrac{1}{2}(1-I) \Leftrightarrow I=\dfrac{1}{3}$.
Đáp án A.