Cho hàm số $f\left( x \right)$ liên tục trên \(\...

B

- Hướng dẫn giải

Phương pháp giải:

${\left( {f.g} \right)^\prime } = f'.g + f.g'$.

Giải chi tiết:

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) = \dfrac{1}{2}{e^{2x}} + C$

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

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

Chọn: B