Giải bài 1 trang 112 – SGK môn Giải tích lớp 12
Tính các tích phân sau:
a) \(\int\limits_{-\frac{1}{2}}^{\frac{1}{2}}{\sqrt[3]{{{\left( 1-x \right)}^{2}}}}dx\);
b) \(\int\limits_{0}^{\frac{\pi }{2}}{\sin \left( \dfrac{\pi }{4}-x \right)dx}\);
c) \(\int\limits_{\frac{1}{2}}^{2}{\dfrac{1}{x\left( x+1 \right)}dx}\);
d) \(\int\limits_{0}^{2}{x{{\left( x+1 \right)}^{2}}dx}\);
e) \(\int\limits_{\frac{1}{2}}^{2}{\dfrac{1-3x}{{{\left( x+1 \right)}^{2}}}dx}\);
g) \(\int\limits_{-\frac{\pi }{2}}^{\frac{\pi }{2}}{\sin 3x\cos 5xdx}\).
a) \(\int\limits_{-\frac{1}{2}}^{\frac{1}{2}}{\sqrt[3]{{{\left( 1-x \right)}^{2}}}}dx\);
Đặt \(1-x=u\Rightarrow du=-dx\Rightarrow dx=-du \)
Đổi cận
| x | \(-\dfrac{1}{2}\) | \(\dfrac{1}{2}\) |
| t | \(\dfrac{3}{2}\) | \(\dfrac{1}{2}\) |
\(\begin{align} & \int\limits_{-\frac{1}{2}}^{\frac{1}{2}} {\sqrt[3] {{{\left( 1-x \right)}^{2}}}}dx= -\int\limits_ {\frac{3}{2}}^{\frac{1}{2}}{{{u} ^{\frac{2}{3}}}du} \\ & =\int\limits_{\frac{1}{2}} ^{\frac{3}{2}}{{{u} ^{\frac{2}{3}}}du} =\dfrac{3}{5}{{u}^{\frac{5}{3}}}\left| _{\frac{1}{2}}^{\frac{3}{2}} \right.\\& =\dfrac{3}{5}\left[ {{\left( \dfrac{3}{2} \right)}^ {\frac{5}{3}}}-\dfrac{1}{{{2}^{\frac{5}{3}}}} \right] \\ & =\dfrac{3}{5}.\dfrac{3\sqrt[3]{9}-1}{2\sqrt[3]{4}} \\ \end{align} \)
\(\begin{align} b)\,\int\limits_{0}^{\frac{\pi }{2}}{\sin \left( \dfrac{\pi }{4}-x \right)dx}&=\cos \left( \dfrac{\pi }{4}-x \right)\left| _{\begin{smallmatrix} \\ 0 \end{smallmatrix}}^{\frac{\pi }{2}} \right. \\ & =\cos \left( -\dfrac{\pi }{4} \right)-\cos \left( \dfrac{\pi }{4} \right)=0 \\ \end{align} \)
\(\begin{align} &c)\, \int\limits_{\frac{1}{2}}^{2}{\dfrac{1}{x\left( x+1 \right)}dx}=\int\limits_{\frac{1}{2}}^{2}{\left( \dfrac{1}{x}-\dfrac{1}{x+1} \right)dx} \\ & =\ln \left| \dfrac{x}{x+1} \right|\left| _{\frac{1}{2}}^{\begin{smallmatrix} 2\\\\\\ \\ \end{smallmatrix}} \right.=\ln \dfrac{2}{3}-\ln \dfrac{1}{3}=\ln 2 \\ \end{align} \)
\(\begin{align}d)\, \int\limits_{0}^{2}{x{{\left( x+1 \right)}^{2}}dx}&=\int\limits_{0}^{2}{\left( {{x}^{3}}+2{{x}^{2}}+x \right)dx} \\ & =\left( \dfrac{{{x}^{4}}}{4}+\dfrac{2{{x}^{3}}}{3}+\dfrac{{{x}^{2}}}{2} \right) \left| _{\begin{smallmatrix} \\ \\ \\ 0 \end{smallmatrix}}^{\begin{smallmatrix} 2 \\\\\\ \end{smallmatrix}} \right. \\ & =4+\dfrac{16}{3}+2 \\ & =\dfrac{34}{3} \\ \end{align} \)
e) \(\int\limits_{\frac{1}{2}}^{2}{\dfrac{1-3x}{{{\left( x+1 \right)}^{2}}}dx}\);
Đặt \(x+1=u\Rightarrow \left\{ \begin{align} & du=dx \\ & x=u-1 \\ \end{align} \right. \)
Đổi cận
| x | \(\dfrac{1}{2}\) | 2 |
| u | \(\dfrac{3}{2}\) | 3 |
\(\begin{align} & \int\limits_{\frac{1}{2}}^{2}{\dfrac{1-3x}{{{\left( x+1 \right)}^{2}}}dx}=\int\limits_{\frac{3}{2}}^{3}{\dfrac{1-3\left( u-1 \right)}{{{u}^{2}}}du} \\ & =\int\limits_{\frac{3}{2}}^{3}{\left( \dfrac{4}{{{u}^{2}}}-\dfrac{3}{u} \right)du=\left( -\dfrac{4}{u}-3\ln \left| u \right| \right)\left| _{\frac{3}{2}}^{\begin{smallmatrix} 3\\\\ \\ \end{smallmatrix}} \right.} \\ & =-\dfrac{4}{3}-3\ln 3+\dfrac{8}{3}+3\ln \dfrac{3}{2} \\ & =\dfrac{4}{3}-3\ln 2 \\ \end{align} \)
\(\begin{align} g)\,\int\limits_{-\frac{\pi }{2}}^{\frac{\pi }{2}}{\sin 3x\cos 5xdx} & =\int\limits_{-\frac{\pi }{2}}^{\frac{\pi }{2}}{\dfrac{1}{2}\left( \sin 8x-\sin 2x \right)dx} \\ & =\left( \dfrac{\cos 8x}{16}-\dfrac{\cos 2x}{4} \right)\left| _{-\frac{\pi }{2}}^{\frac{\pi }{2}} \\ =\dfrac{1}{16}+\dfrac{1}{4}-\dfrac{1}{16}+\dfrac{1}{4}=\dfrac{1}{2} \right. \\ \end{align} \)