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Câu hỏi: Cho hàm số $f(x)$ có đạo hàm liên tục trên $\mathbb{R}$. Biết $f(5)=1$ và $\int\limits_{0}^{1}{xf(5\text{x})d\text{x}=1}$, khi đó $\int\limits_{0}^{5}{{{x}^{2}}{{f}^{'}}(\text{x})d\text{x}}$ bằng:
A. 15.
B. 23.
C. $\dfrac{123}{5}$
D. -25.
A. 15.
B. 23.
C. $\dfrac{123}{5}$
D. -25.
Cách 1:
$
\int_{0}^{5} x^{2} f^{\prime}(x) \mathrm{d} x=\left.x^{2} f(x)\right|_{0} ^{5}-\int_{0}^{5} 2 x f(x) \mathrm{d} x=25.1-2 \int_{0}^{1} 5 t f(5 t) \mathrm{d}(5 t)=25-50.1=-25 .
$
Cách 2:
Ta có: $1=\int_{0}^{1} x f(5 x) \mathrm{d} x$
Đặt $t=5 x \Rightarrow \mathrm{d} t=5 \mathrm{~d} x \Rightarrow \dfrac{1}{5} \mathrm{~d} t=\mathrm{d} x$
$\Rightarrow 1=\int_{0}^{5} \dfrac{1}{5} t \cdot f(t) \cdot \dfrac{1}{5} \mathrm{~d} t \Leftrightarrow 1=\dfrac{1}{25} \int_{0}^{5} t \cdot f(t) \mathrm{d} t \Leftrightarrow \int_{0}^{5} t \cdot f(t) \mathrm{d} t=25 \Rightarrow \int_{0}^{5} x \cdot f(x) \mathrm{d} x=25$
Đặt $I=\int_{0}^{5} x^{2} \cdot f^{\prime}(x) \mathrm{d} x$
Đặt: $\left\{\begin{array}{l}u=x^{2} \\ \mathrm{~d} v=f^{\prime}(x) \mathrm{d} x\end{array} \Rightarrow\left\{\begin{array}{l}\mathrm{d} u=2 x \mathrm{~d} x \\ v=f(x)\end{array}\right.\right.$
$\Rightarrow I=\left.x^{2} \cdot f(x)\right|_{0} ^{5}-2 \int_{0}^{5} x f(x) \mathrm{d} x=25 \cdot f(5)-2 \cdot 25=-25$
$
\int_{0}^{5} x^{2} f^{\prime}(x) \mathrm{d} x=\left.x^{2} f(x)\right|_{0} ^{5}-\int_{0}^{5} 2 x f(x) \mathrm{d} x=25.1-2 \int_{0}^{1} 5 t f(5 t) \mathrm{d}(5 t)=25-50.1=-25 .
$
Cách 2:
Ta có: $1=\int_{0}^{1} x f(5 x) \mathrm{d} x$
Đặt $t=5 x \Rightarrow \mathrm{d} t=5 \mathrm{~d} x \Rightarrow \dfrac{1}{5} \mathrm{~d} t=\mathrm{d} x$
$\Rightarrow 1=\int_{0}^{5} \dfrac{1}{5} t \cdot f(t) \cdot \dfrac{1}{5} \mathrm{~d} t \Leftrightarrow 1=\dfrac{1}{25} \int_{0}^{5} t \cdot f(t) \mathrm{d} t \Leftrightarrow \int_{0}^{5} t \cdot f(t) \mathrm{d} t=25 \Rightarrow \int_{0}^{5} x \cdot f(x) \mathrm{d} x=25$
Đặt $I=\int_{0}^{5} x^{2} \cdot f^{\prime}(x) \mathrm{d} x$
Đặt: $\left\{\begin{array}{l}u=x^{2} \\ \mathrm{~d} v=f^{\prime}(x) \mathrm{d} x\end{array} \Rightarrow\left\{\begin{array}{l}\mathrm{d} u=2 x \mathrm{~d} x \\ v=f(x)\end{array}\right.\right.$
$\Rightarrow I=\left.x^{2} \cdot f(x)\right|_{0} ^{5}-2 \int_{0}^{5} x f(x) \mathrm{d} x=25 \cdot f(5)-2 \cdot 25=-25$
Đáp án D.