Bài 2 (Trang 112 SGK Toán Giải tích 12):

Tính các tích phân sau:

a) ${\int }_0^2\left|1-\mathrm{x}\right|\mathrm{dx}$;

b) ${\int }_0^{\frac{\mathrm{\pi }}{2}}{\sin }^2\mathrm{xdx}$;

c) ${\int }_0^{\ln 2}\frac{{\mathrm{e}}^{2\mathrm{x}+1}+1}{{\mathrm{e}}^{\mathrm{x}}}\mathrm{dx}$;

d) ${\int }_0^{\mathrm{\pi }}\sin 2{\mathrm{xcos}}^2\mathrm{xdx}$.

Hướng dẫn giải:

a) Ta có: ${\int }_0^2\left|1-\mathrm{x}\right|\mathrm{dx}={\int }_0^1\left|1-\mathrm{x}\right|\mathrm{dx} + {\int }_1^2\left|1-\mathrm{x}\right|\mathrm{dx} ={\int }_0^1(1-\mathrm{x})\mathrm{dx}+{\int }_1^2(\mathrm{x}-1)\mathrm{dx}=(\mathrm{x}-\frac{{\mathrm{x}}^2}{2}){|}_0^1 +(\frac{{\mathrm{x}}^2}{2}-\mathrm{x}){|}_1^2=1$

b) ${\int }_0^{\frac{\mathrm{\pi }}{2}}{\sin }^2\mathrm{xdx} = {\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{1-\cos 2\mathrm{x}}{2}\mathrm{dx} = \frac{1}{2}(\mathrm{x}-\frac{1}{2}\sin 2\mathrm{x}){|}_0^{\frac{\mathrm{\pi }}{2}} = \frac{\mathrm{\pi }}{4}$

c) ${\int }_0^{\ln 2}\frac{{\mathrm{e}}^{2\mathrm{x}+1}+1}{{\mathrm{e}}^{\mathrm{x}}}\mathrm{dx} = {\int }_0^{\ln 2}({\mathrm{e}}^{\mathrm{x}+1}+{\mathrm{e}}^{-\mathrm{x}})\mathrm{dx} = ({\mathrm{e}}^{\mathrm{x}+1}-{\mathrm{e}}^{-\mathrm{x}}){|}_0^{\ln 2} = \mathrm{e}+\frac{1}{2}$

d) ${\int }_0^{\pi }\sin 2{\mathrm{xcos}}^2\mathrm{xdx}=\frac{1}{2}{\int }_0^{\mathrm{\pi }}\sin 2\mathrm{x}(1+\cos 2\mathrm{x})\mathrm{dx}=\frac{1}{2}{\int }_0^{\mathrm{\pi }}\sin 2\mathrm{xdx} + \frac{1}{4}{\int }_0^{\mathrm{\pi }}\sin 4\mathrm{xdx} =(-\frac{1}{4}\cos 2\mathrm{x}-\frac{1}{16}\cos 4\mathrm{x}){|}_0^{\mathrm{\pi }} = 0$