Bài 2 trang 112 SGK Giải tích 12

Tính các tích phân sau:

a) \(\int_0^2 {\left| {1 - x} \right|} dx\)

b) \(\int_0^{{\pi  \over 2}} \sin ^2xdx\)

c) \(\displaystyle \int_0^{\ln 2} {{{{e^{2x + 1}} + 1} \over {{e^x}}}} dx\)

d) \(\int_0^\pi  \sin 2x\cos ^2xdx\)

a) Ta có: \(\left| {1 - x} \right| = \left[ \begin{array}{l}1 - x\,\,khi\,\,x \le 1\\x - 1\,\,khi\,\,x > 1\end{array} \right.\)

\(\Rightarrow \int_0^2 {\left| {1 - x} \right|} dx = \int_0^1 {\left| {1 - x} \right|} dx + \int_1^2 {\left| {1 - x} \right|} dx\)

\(=   \int_0^1 {(1 - x)} dx + \int_1^2 {(x - 1)} dx\)

\( = \left. {\left( {x - \dfrac{{{x^2}}}{2}} \right)} \right|_0^1 + \left. {\left( {\dfrac{{{x^2}}}{2} - x} \right)} \right|_1^2 = \dfrac{1}{2} + \dfrac{1}{2} = 1\)

\(\begin{array}{l}b)\,\,\int\limits_0^{\frac{\pi }{2}} {{{\sin }^2}xdx} = \dfrac{1}{2}\int\limits_0^{\frac{\pi }{2}} {\left( {1 - \cos 2x} \right)dx} \\= \dfrac{1}{2}\left. {\left( {x - \dfrac{{\sin 2x}}{2}} \right)} \right|_0^{\frac{\pi }{2}}\\= \dfrac{1}{2}.\dfrac{\pi }{2} = \dfrac{\pi }{4}\end{array}\)

\(\begin{array}{l}c)\,\,\int\limits_0^{\ln 2} {\dfrac{{{e^{2x + 1}} + 1}}{{{e^x}}}dx} = \int\limits_0^{\ln 2} {\left( {{e^{2x + 1 - x}} + {e^{ - x}}} \right)dx} \\= \int\limits_0^{\ln 2} {\left( {{e^{x + 1}} + {e^{ - x}}} \right)dx} \\= \left. {\left( {{e^{x + 1}} - {e^{ - x}}} \right)} \right|_0^{\ln 2}\\= {e^{\ln 2 + 1}} - {e^{ - \ln 2}} - \left( {e - 1} \right)\\= {e^{\ln 2}}.e - \dfrac{1}{2} - e + 1\\= e + \dfrac{1}{2}\end{array}\)

\(\begin{array}{l}d)\,\,\sin 2x\cos 2x = \sin 2x\dfrac{{1 + \cos 2x}}{2}\\\,\,\, = \dfrac{1}{2}\sin 2x + \dfrac{1}{2}\sin 2x\cos 2x = \dfrac{1}{2}\sin 2x + \dfrac{1}{4}\sin 4x\\\Rightarrow \int\limits_0^\pi {\sin 2x\cos 2xdx} = \int\limits_0^\pi {\left( {\dfrac{1}{2}\sin 2x + \dfrac{1}{4}\sin 4x} \right)dx} \\= \left. {\left( { - \dfrac{1}{4}\cos 2x - \dfrac{1}{{16}}\cos 4x} \right)} \right|_0^\pi \\= - \dfrac{1}{4} - \dfrac{1}{{16}} - \left( { - \dfrac{1}{4} - \dfrac{1}{{16}}} \right) = 0\end{array}\)