Question

Tính:

a) \(\int\limits^1_0e^xdx;\)                   b) \(\int\limits^e_1\dfrac{1}{x}dx;\)                     c) \(\int\limits^{\dfrac{\pi}{2}}_0\sin xdx;\)                       d) \(\int\limits^{\dfrac{\pi}{3}}_{\dfrac{\pi}{6}}\dfrac{dx}{\sin^2x}.\)

Answer 1

a) \(\int\limits_0^1 {{e^x}dx}  = {e^x}\left| \begin{array}{l}1\\0\end{array} \right. = {e^1} - {e^0} = e - 1\);

b) \(\int\limits_1^e {\frac{1}{x}dx}  = \ln \left| x \right|\left| \begin{array}{l}e\\1\end{array} \right. = \ln e - \ln 1 = 1\);

c) \(\int\limits_0^{\frac{\pi }{2}} {\sin xdx}  =  - \cos x\left| \begin{array}{l}\frac{\pi }{2}\\0\end{array} \right. =  - \cos \frac{\pi }{2} + \cos 0 = 1\);

d) \(\int\limits_{\frac{\pi }{6}}^{\frac{\pi }{3}} {\frac{{dx}}{{{{\sin }^2}x}}}  =  - \tan x\left| \begin{array}{l}\frac{\pi }{3}\\\frac{\pi }{6}\end{array} \right. =  - \cot \frac{\pi }{3} + \cot \frac{\pi }{6} =  - \frac{{\sqrt 3 }}{3} + \sqrt 3  = \frac{{2\sqrt 3 }}{3}\).