Cho hàm số \(f(x)\) liên tục trên R có đạo hàm thỏa mãn \(f\left( x \right) + 2f\left( x \right) = 1,\forall x \in R\) và \(

* Hướng dẫn giải

Ta có: \(f'\left( x \right) + 2f\left( x \right) = 1 \Leftrightarrow {e^{2x}}f'\left( x \right) + {e^{2x}}.2f\left( x \right) = {e^{2x}} \Leftrightarrow {\left( {{e^{2x}}.f\left( x \right)} \right)^\prime } = {e^{2x}}\) 

\( \Rightarrow {e^{2x}}.f\left( x \right) = \int {{e^{2x}}dx}  \Leftrightarrow {e^{2x}}.f\left( x \right) = \frac{1}{2}{e^{2x}} + C\) 

Mà \(f\left( 0 \right) = 1 \Rightarrow 1 = \frac{1}{2} + C \Rightarrow C = \frac{1}{2} \Rightarrow {e^{2x}}.f\left( x \right) = \frac{1}{2}{e^{2x}} + \frac{1}{2} \Leftrightarrow f\left( x \right) = \frac{{{e^{2x}} + 1}}{{2{e^{2x}}}}\) 

\(\int\limits_0^1 {f\left( x \right)dx}  = \int\limits_0^1 {\frac{{{e^{2x}} + 1}}{{2{e^{2x}}}}dx}  = \int\limits_0^1 {\left( {\frac{1}{2} + \frac{1}{2}{e^{ - 2x}}} \right)dx}  = \left. {\left( {\frac{1}{2}x - \frac{1}{4}{e^{ - 2x}}} \right)} \right|_0^1 = \left( {\frac{1}{2} - \frac{1}{{4{e^2}}}} \right) - \left( { - \frac{1}{4}} \right) = \frac{3}{4} - \frac{1}{{4{e^2}}}\)