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Câu hỏi: Cho hàm số $f(x)$ có đạo hàm liên tục trên $\mathbb{R}$. Biết $f(3)=1$ và $\int_0^1 x f(3 x) \mathrm{d} x=1$, khi đó $\int_0^3 x^2 f^{\prime}(x) \mathrm{d} x$ bằng
A. -9 .
B. $\dfrac{25}{3}$.
C. 3 .
D. 7 .
A. -9 .
B. $\dfrac{25}{3}$.
C. 3 .
D. 7 .
Đặt $t=3 x \Rightarrow \mathrm{d} t=3 \mathrm{~d} x \Rightarrow \mathrm{d} x=\dfrac{1}{3} \mathrm{~d} t$.
Suy ra $1=\int_0^1 x f(3 x) \mathrm{d} x=\dfrac{1}{9} \int_0^3 t f(t) \mathrm{d} t \Leftrightarrow \int_0^3 t f(t) d t=9$.
Đặt $\left\{\begin{array}{l}u=f(t) \\ \mathrm{d} v=t \mathrm{~d} t\end{array} \Rightarrow\left\{\begin{array}{l}\mathrm{d} u=f^{\prime}(t) \mathrm{d} t \\ v=\dfrac{t^2}{2}\end{array}\right.\right.$.
$\Rightarrow \int_0^3 t f(t) \mathrm{d} t=\left.\dfrac{t^2}{2} f(t)\right|_0 ^3-\int_0^3 \dfrac{t^2}{2} f^{\prime}(t) \mathrm{d} t=\dfrac{9}{2} f(3)-\dfrac{1}{2} \int_0^3 t^2 f^{\prime}(t) \mathrm{d} t$
$\Leftrightarrow 9=\dfrac{9}{2}-\dfrac{1}{2} \int_0^3 t^2 f^{\prime}(t) \mathrm{d} t \Leftrightarrow \int_0^3 t^2 f^{\prime}(t) \mathrm{d} t=-9$.
Vậy $\int_0^3 x^2 f^{\prime}(x) \mathrm{d} x=-9$.
Suy ra $1=\int_0^1 x f(3 x) \mathrm{d} x=\dfrac{1}{9} \int_0^3 t f(t) \mathrm{d} t \Leftrightarrow \int_0^3 t f(t) d t=9$.
Đặt $\left\{\begin{array}{l}u=f(t) \\ \mathrm{d} v=t \mathrm{~d} t\end{array} \Rightarrow\left\{\begin{array}{l}\mathrm{d} u=f^{\prime}(t) \mathrm{d} t \\ v=\dfrac{t^2}{2}\end{array}\right.\right.$.
$\Rightarrow \int_0^3 t f(t) \mathrm{d} t=\left.\dfrac{t^2}{2} f(t)\right|_0 ^3-\int_0^3 \dfrac{t^2}{2} f^{\prime}(t) \mathrm{d} t=\dfrac{9}{2} f(3)-\dfrac{1}{2} \int_0^3 t^2 f^{\prime}(t) \mathrm{d} t$
$\Leftrightarrow 9=\dfrac{9}{2}-\dfrac{1}{2} \int_0^3 t^2 f^{\prime}(t) \mathrm{d} t \Leftrightarrow \int_0^3 t^2 f^{\prime}(t) \mathrm{d} t=-9$.
Vậy $\int_0^3 x^2 f^{\prime}(x) \mathrm{d} x=-9$.
Đáp án A.