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Câu hỏi: Cho hàm số $f\left( x \right)$ liên tục trên $R$ và thỏa $\int\limits_{0}^{3}{f\left( \sqrt{{{x}^{2}}+16}+x \right)\text{d}x=2019,}$ $\int\limits_{4}^{8}{\dfrac{f\left( x \right)}{{{x}^{2}}}\text{d}x=1.}$
Tính $\int\limits_{4}^{8}{f\left( x \right)\text{d}x}.$
A. 2019
B. 4022.
C. 2020.
D. 4038.
Tính $\int\limits_{4}^{8}{f\left( x \right)\text{d}x}.$
A. 2019
B. 4022.
C. 2020.
D. 4038.
Xét ${{I}_{1}}=\int\limits_{0}^{3}{f\left( \sqrt{{{x}^{2}}+16}+x \right)\text{d}x=2019}$.
Đặt $u=\sqrt{{{x}^{2}}+16}+x\Leftrightarrow u-x=\sqrt{{{x}^{2}}+16}\Rightarrow x=\dfrac{{{u}^{2}}-16}{2u}\Rightarrow \ \text{d}x=\dfrac{{{u}^{2}}+16}{2{{u}^{2}}}\text{d}u.$
Khi $x=0\Rightarrow u=4.$
Khi $x=3\Rightarrow u=8.$
$\Rightarrow $ ${{I}_{1}}=\dfrac{1}{2}\int\limits_{4}^{8}{\dfrac{{{u}^{2}}+16}{{{u}^{2}}}f\left( u \right)\text{d}u=2019}\Rightarrow \int\limits_{4}^{8}{\dfrac{{{x}^{2}}+16}{{{x}^{2}}}f\left( x \right)\text{d}x=\int\limits_{4}^{8}{\dfrac{{{u}^{2}}+16}{{{u}^{2}}}f\left( u \right)\text{d}u=4038.}}$
$\Rightarrow $ $\int\limits_{4}^{8}{\dfrac{{{x}^{2}}+16}{{{x}^{2}}}f\left( x \right)\text{d}x}=4038\Leftrightarrow \int\limits_{4}^{8}{f\left( x \right)\text{d}x}+16\int\limits_{4}^{8}{\dfrac{f\left( x \right)}{{{x}^{2}}}\text{d}x=}4038\Leftrightarrow \int\limits_{4}^{8}{f\left( x \right)\text{d}x}=4038-16=4022.$
Do $\int\limits_{4}^{8}{\dfrac{f\left( x \right)}{{{x}^{2}}}\text{d}x=1}.$
Kết luận: $\int\limits_{4}^{8}{f\left( x \right)\text{d}x}=4022.$
Đặt $u=\sqrt{{{x}^{2}}+16}+x\Leftrightarrow u-x=\sqrt{{{x}^{2}}+16}\Rightarrow x=\dfrac{{{u}^{2}}-16}{2u}\Rightarrow \ \text{d}x=\dfrac{{{u}^{2}}+16}{2{{u}^{2}}}\text{d}u.$
Khi $x=0\Rightarrow u=4.$
Khi $x=3\Rightarrow u=8.$
$\Rightarrow $ ${{I}_{1}}=\dfrac{1}{2}\int\limits_{4}^{8}{\dfrac{{{u}^{2}}+16}{{{u}^{2}}}f\left( u \right)\text{d}u=2019}\Rightarrow \int\limits_{4}^{8}{\dfrac{{{x}^{2}}+16}{{{x}^{2}}}f\left( x \right)\text{d}x=\int\limits_{4}^{8}{\dfrac{{{u}^{2}}+16}{{{u}^{2}}}f\left( u \right)\text{d}u=4038.}}$
$\Rightarrow $ $\int\limits_{4}^{8}{\dfrac{{{x}^{2}}+16}{{{x}^{2}}}f\left( x \right)\text{d}x}=4038\Leftrightarrow \int\limits_{4}^{8}{f\left( x \right)\text{d}x}+16\int\limits_{4}^{8}{\dfrac{f\left( x \right)}{{{x}^{2}}}\text{d}x=}4038\Leftrightarrow \int\limits_{4}^{8}{f\left( x \right)\text{d}x}=4038-16=4022.$
Do $\int\limits_{4}^{8}{\dfrac{f\left( x \right)}{{{x}^{2}}}\text{d}x=1}.$
Kết luận: $\int\limits_{4}^{8}{f\left( x \right)\text{d}x}=4022.$
Đáp án B.