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}{\frac{f\left( x \right)}{{{x...

* Hướng dẫn giải

Xét \({{I}_{1}}=\int\limits_{0}^{3}{f\left( \sqrt{{{x}^{2}}+16}+x \right)\text{d}x=2019}\)

Đặt \(u=\sqrt{{{x}^{2}}+16}+x\Leftrightarrow u-x=\sqrt{{{x}^{2}}+16}\Rightarrow x=\frac{{{u}^{2}}-16}{2u}\Rightarrow \ \text{d}x=\frac{{{u}^{2}}+16}{2{{u}^{2}}}\text{d}u.\)

Khi \(x=0\Rightarrow u=4.\)

Khi \(x=3\Rightarrow u=8.\)

\(\Rightarrow {{I}_{1}}=\frac{1}{2}\int\limits_{4}^{8}{\frac{{{u}^{2}}+16}{{{u}^{2}}}f\left( u \right)\text{d}u=2019}\Rightarrow \int\limits_{4}^{8}{\frac{{{x}^{2}}+16}{{{x}^{2}}}f\left( x \right)\text{d}x=\int\limits_{4}^{8}{\frac{{{u}^{2}}+16}{{{u}^{2}}}f\left( u \right)\text{d}u=4038.}}\)

\(\Rightarrow \int\limits_{4}^{8}{\frac{{{x}^{2}}+16}{{{x}^{2}}}f\left( x \right)\text{d}x}=4038\Leftrightarrow \int\limits_{4}^{8}{f\left( x \right)\text{d}x}+16\int\limits_{4}^{8}{\frac{f\left( x \right)}{{{x}^{2}}}\text{d}x=}4038\Leftrightarrow \int\limits_{4}^{8}{f\left( x \right)\text{d}x}=4038-16=4022.\)

Do \(\int\limits_{4}^{8}{\frac{f\left( x \right)}{{{x}^{2}}}\text{d}x=1}.\)

Kết luận: \(\int\limits_{4}^{8}{f\left( x \right)\text{d}x}=4022.\)