Đáp án:
\[I = \dfrac{1}{4}{e^2} + \dfrac{1}{4}\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
I = \int\limits_0^1 {x.{e^{2x}}dx} \\
\left\{ \begin{array}{l}
u = x\\
v' = {e^{2x}}
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
u' = 1\\
v = \dfrac{1}{2}{e^{2x}}
\end{array} \right.\\
\Rightarrow I = \mathop {\left. {\left( {\dfrac{1}{2}x.{e^{2x}}} \right)} \right|}\nolimits_0^1 - \int\limits_0^1 {1.\dfrac{1}{2}{e^{2x}}dx} \\
= \left( {\dfrac{1}{2}.1.{e^{2.1}} - \dfrac{1}{2}.0.{e^{2.0}}} \right) - \int\limits_0^1 {\dfrac{1}{2}{e^{2x}}dx} \\
= \dfrac{{{e^2}}}{2} - \mathop {\left. {\dfrac{1}{4}{e^{2x}}} \right|}\nolimits_0^1 \\
= \dfrac{{{e^2}}}{2} - \left( {\dfrac{1}{4}{e^2} - \dfrac{1}{4}{e^0}} \right)\\
= \dfrac{1}{4}{e^2} + \dfrac{1}{4}
\end{array}\)
