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Câu hỏi: Cho hàm số $f\left(x \right)$ có đạo hàm xác định trên $\mathbb{R}$. Biết $f\left(1 \right)=2$ và $\int\limits_{0}^{1}{{{x}^{2}}{f}'\left(x \right)}\text{d}x=\int\limits_{1}^{4}{\frac{1+3\sqrt{x}}{2\sqrt{x}}f\left(2-\sqrt{x} \right)}\text{d}x=4$. Giá trị của $\int\limits_{0}^{1}{f\left(x \right)}\text{d}x$ bằng
A. $1$.
B. $\frac{5}{7}$.
C. $\frac{3}{7}$.
D. $\frac{1}{7}$.
A. $1$.
B. $\frac{5}{7}$.
C. $\frac{3}{7}$.
D. $\frac{1}{7}$.
Ta có: $4=\int\limits_{0}^{1}{{{x}^{2}}{f}'\left( x \right)}\text{d}x=\int\limits_{0}^{1}{{{x}^{2}}}\text{d}\left( f\left( x \right) \right)=\left( {{x}^{2}}f\left( x \right) \right)\left| \begin{aligned}
& 1 \\
& 0 \\
\end{aligned} \right.-\int\limits_{0}^{1}{2xf\left( x \right)}\text{d}x$
$\Leftrightarrow 4=f\left( 1 \right)-2\int\limits_{0}^{1}{xf\left( x \right)}\text{d}x\Leftrightarrow 4=2-2\int\limits_{0}^{1}{xf\left( x \right)}\text{d}x$ $\Rightarrow \int\limits_{0}^{1}{xf\left( x \right)}\text{d}x=-1$.
Xét $\int\limits_{1}^{4}{\frac{1+3\sqrt{x}}{2\sqrt{x}}f\left( 2-\sqrt{x} \right)}\text{d}x$.
Đặt $t=2-\sqrt{x}\Rightarrow \text{d}t=-\frac{1}{2\sqrt{x}}\text{d}x$.
Với $x=1\Rightarrow t=1$ và $x=4\Rightarrow t=0$.
Khi đó $4=\int\limits_{1}^{4}{\frac{1+3\sqrt{x}}{2\sqrt{x}}f\left( 2-\sqrt{x} \right)}\text{d}x=-\int\limits_{1}^{0}{\left[ 1+3\left( 2-t \right) \right]f\left( t \right)}\text{d}t$
$\Leftrightarrow 4=\int\limits_{0}^{1}{\left( 7-3t \right)f\left( t \right)}\text{d}t\Leftrightarrow 4=7\int\limits_{0}^{1}{f\left( t \right)}\text{d}t-3\int\limits_{0}^{1}{tf\left( t \right)}\text{d}t$ $\Leftrightarrow 4=7\int\limits_{0}^{1}{f\left( t \right)}\text{d}t-3\left( -1 \right)\Leftrightarrow \int\limits_{0}^{1}{f\left( t \right)}\text{d}t=\frac{1}{7}$.
Vậy $\int\limits_{0}^{1}{f\left( x \right)}\text{d}x=\frac{1}{7}$.
& 1 \\
& 0 \\
\end{aligned} \right.-\int\limits_{0}^{1}{2xf\left( x \right)}\text{d}x$
$\Leftrightarrow 4=f\left( 1 \right)-2\int\limits_{0}^{1}{xf\left( x \right)}\text{d}x\Leftrightarrow 4=2-2\int\limits_{0}^{1}{xf\left( x \right)}\text{d}x$ $\Rightarrow \int\limits_{0}^{1}{xf\left( x \right)}\text{d}x=-1$.
Xét $\int\limits_{1}^{4}{\frac{1+3\sqrt{x}}{2\sqrt{x}}f\left( 2-\sqrt{x} \right)}\text{d}x$.
Đặt $t=2-\sqrt{x}\Rightarrow \text{d}t=-\frac{1}{2\sqrt{x}}\text{d}x$.
Với $x=1\Rightarrow t=1$ và $x=4\Rightarrow t=0$.
Khi đó $4=\int\limits_{1}^{4}{\frac{1+3\sqrt{x}}{2\sqrt{x}}f\left( 2-\sqrt{x} \right)}\text{d}x=-\int\limits_{1}^{0}{\left[ 1+3\left( 2-t \right) \right]f\left( t \right)}\text{d}t$
$\Leftrightarrow 4=\int\limits_{0}^{1}{\left( 7-3t \right)f\left( t \right)}\text{d}t\Leftrightarrow 4=7\int\limits_{0}^{1}{f\left( t \right)}\text{d}t-3\int\limits_{0}^{1}{tf\left( t \right)}\text{d}t$ $\Leftrightarrow 4=7\int\limits_{0}^{1}{f\left( t \right)}\text{d}t-3\left( -1 \right)\Leftrightarrow \int\limits_{0}^{1}{f\left( t \right)}\text{d}t=\frac{1}{7}$.
Vậy $\int\limits_{0}^{1}{f\left( x \right)}\text{d}x=\frac{1}{7}$.
Đáp án D.