Giải bài 6 trang 113 – SGK môn Giải tích lớp 12
Tính \(\int\limits_{0}^{1}{x{{\left( 1-x \right)}^{5}}dx}\) bằng hai phương pháp:
a) Đổi biến số \(u=1-x\).
b) Tính tích phân từng phần.
a) Đặt \(u=1-x\Rightarrow \left\{ \begin{aligned} & du=-dx\Rightarrow dx=-du \\ & x=1-u \\ \end{aligned} \right. \)
Đổi cận:
| x | 0 | 1 |
| u | 1 | 0 |
\(\begin{aligned} \int\limits_{0}^{1}{x{{\left( 1-x \right)}^{5}}dx}&=-\int\limits_{1}^{0}{\left( 1-u \right){{u}^{5}}du} \\ & =\int\limits_{0}^{1}{\left( {{u}^{5}}-{{u}^{6}} \right)du} \\ & =\left( \dfrac{{{u}^{6}}}{6}-\dfrac{{{u}^{7}}}{7} \right)\left| _{\begin{smallmatrix} \\ 0 \end{smallmatrix}}^{\begin{smallmatrix} 1 \\ \end{smallmatrix}} \right. \\ & =\dfrac{1}{6}-\dfrac{1}{7} \\ & =\dfrac{1}{42} \\ \end{aligned} \)
b) Đặt \(\left\{ \begin{aligned} & u=x \\ & {{\left( 1-x \right)}^{5}}dx=dv \\ \end{aligned} \right.\Rightarrow \left\{ \begin{aligned} & du=dx \\ & v=-\dfrac{{{\left( 1-x \right)}^{6}}}{6} \\ \end{aligned} \right. \)
\(\begin{aligned} \int\limits_{0}^{1}{x{{\left( 1-x \right)}^{5}}dx}&=-\dfrac{x{{\left( 1-x \right)}^{6}}}{6}\left| _{\begin{smallmatrix} \\ 0 \end{smallmatrix}}^{\begin{smallmatrix} 1 \\ \end{smallmatrix}} \right.+\int\limits_{0}^{1}{\dfrac{{{\left( 1-x \right)}^{6}}}{6}dx} \\ & =-\dfrac{{{\left( 1-x \right)}^{7}}}{42}\left| _{\begin{smallmatrix} \\ 0 \end{smallmatrix}}^{\begin{smallmatrix} 1 \\ \end{smallmatrix}} \right. \\ & =\dfrac{1}{42} \\ \end{aligned} \)