Đáp án:
$\begin{array}{l}
b)J = \int\limits_0^{\pi /6} {\sin \left( {2x + \dfrac{\pi }{6}} \right)dx} \\
= \dfrac{1}{2}.\int\limits_0^{\pi /6} {\sin \left( {2x + \dfrac{\pi }{6}} \right)2dx} \\
= \dfrac{1}{2}\int\limits_0^{\pi /6} {\sin \left( {2x + \dfrac{\pi }{6}} \right)d\left( {2x + \dfrac{\pi }{6}} \right)} \\
= \dfrac{1}{2}.\left( { - cos\left( {2x + \dfrac{\pi }{6}} \right)} \right)\left| {_0^{\dfrac{\pi }{6}}} \right.\\
= \dfrac{1}{2}.\dfrac{{\sqrt 3 }}{2} = \dfrac{{\sqrt 3 }}{4}\\
c)K = \int\limits_0^1 {{e^{2 - 5x}}dx} \\
= - \dfrac{1}{5}.\int\limits_0^1 {{e^{2 - 5x}}.\left( { - 5dx} \right)} \\
= - \dfrac{1}{5}\int\limits_0^1 {{e^{2 - 5x}}.d\left( {2 - 5x} \right)} \\
= \left( { - \dfrac{1}{5}.{e^{2 - 5x}}} \right)\left| {_0^1} \right.\\
= - \dfrac{1}{5}.{e^{2 - 5}} + \dfrac{1}{5}.{e^{2 - 5.0}}\\
= - \dfrac{1}{5}{e^3} + \dfrac{1}{5}{e^2}
\end{array}$
