Giải bài 2 trang 112 – SGK môn Giải tích lớp 12

Tính các tích phân sau:

a) \(\int\limits_{0}^{2}{\left| 1-x \right|dx}\);

b) \(\int\limits_{0}^{\frac{\pi }{2}}{{{\sin }^{2}}xdx}\);

c) \(\int\limits_{0}^{\ln 2}{\dfrac{{{e}^{2x+1}}+1}{{{e}^{x}}}dx}\);

d) \(\int\limits_{0}^{\pi }{\sin 2x{{\cos }^{2}}xdx}\).

Lời giải:

\(\begin{align} & a)\,\int\limits_{0}^{2}{\left| 1-x \right|dx}=\int\limits_{0}^{1}{\left( 1-x \right)dx}+\int\limits_{1}^{2}{\left( x-1 \right)dx} \\ & =\left( x-\dfrac{{{x}^{2}}}{2} \right)\left| _{\begin{smallmatrix} \\ 0 \end{smallmatrix}}^{\begin{smallmatrix} 1 \\ \end{smallmatrix}} \right.+\left( \dfrac{{{x}^{2}}}{2}-x \right)\left| _{\begin{smallmatrix} \\ 1 \end{smallmatrix}}^{\begin{smallmatrix} 2 \\ \end{smallmatrix}} \right. \\ & =1-\dfrac{1}{2}+2-2-\dfrac{1}{2}+1=1 \\ \end{align} \)

\(\begin{align} b)\, \int\limits_{0}^{\frac{\pi }{2}}{{{\sin }^{2}}xdx}& =\int\limits_{0}^{\frac{\pi }{2}}{\dfrac{1-\cos 2x}{2}dx} \\ & =\left( \dfrac{x}{2}-\dfrac{\sin 2x}{4} \right)\left| _{\begin{smallmatrix} \\ 0 \end{smallmatrix}}^{\frac{\pi }{2}} \right. \\ & =\dfrac{\pi }{4} \\ \end{align} \)

\( \begin{align} c)\,\int\limits_{0}^{\ln 2}{\dfrac{{{e}^{2x+1}}+1}{{{e}^{x}}}dx}&=\int\limits_{0}^{\ln 2}{\left( {{e}^{x+1}}+\dfrac{1}{{{e}^{x}}} \right)dx} \\ & =\left( {{e}^{x+1}}-\dfrac{1}{{{e}^{x}}} \right)\left| _{\begin{smallmatrix} \\ 0 \end{smallmatrix}}^{\begin{smallmatrix} \ln 2 \\ \end{smallmatrix}} \right. \\ & =2e-\dfrac{1}{2}-e+1 \\ & =e+\dfrac{1}{2} \\ \end{align} \)

\(\begin{align} d)\,\int\limits_{0}^{\pi }{\sin 2x{{\cos }^{2}}xdx}&=\int\limits_{0}^{\pi }{\sin 2x\dfrac{1+\cos 2x}{2}dx} \\ & =\int\limits_{0}^{\pi }{\left( \dfrac{\sin 2x}{2}+\dfrac{\sin 4x}{4} \right)dx} \\ & =-\left( \dfrac{\cos 2x}{4}+\dfrac{\cos 4x}{16} \right)\left| _{\begin{smallmatrix} \\\\\\ 0 \end{smallmatrix}}^{\begin{smallmatrix} \pi \\\\\\\\\\ \end{smallmatrix}} \right. \\ & =-\left( \dfrac{1}{4}+\dfrac{1}{16}-\dfrac{1}{4}-\dfrac{1}{16} \right) \\ & =0 \\ \end{align} \)