# Bài 6 trang 113 SGK Giải tích 12

##### Hướng dẫn giải

a) Đặt $$u = 1 - x$$.

b) Đặt $$\left\{ \begin{array}{l}u = x\\dv = {\left( {1 - x} \right)^5}dx\end{array} \right.$$

Lời giải chi tiết

a) Đặt $$u = 1 - x \Rightarrow x = 1 - u \Rightarrow dx = - du$$.

Đổi cận: $$\left\{ \begin{array}{l}x = 0 \Rightarrow u = 1\\x = 1 \Rightarrow u = 0\end{array} \right.$$

$$\begin{array}{l}\Rightarrow \int\limits_0^1 {x{{\left( {1 - x} \right)}^5}dx} = - \int\limits_1^0 {\left( {1 - u} \right){u^5}du} \\= \int\limits_0^1 {\left( {{u^5} - {u^6}} \right)du} = \left. {\left( {\frac{{{u^6}}}{6} - \frac{{{u^7}}}{7}} \right)} \right|_0^1 = \frac{1}{6} - \frac{1}{7} = \frac{1}{{42}}\end{array}$$

b) Đặt $$\left\{ \begin{array}{l}u = x\\dv = {\left( {1 - x} \right)^5}dx\end{array} \right. \Rightarrow \left\{ \begin{array}{l}du = dx\\v = - \frac{{{{\left( {1 - x} \right)}^6}}}{6}\end{array} \right.$$

$$\begin{array}{l}\Rightarrow \int\limits_0^1 {x\left( {1 - {x^5}} \right)dx} = - x\left. {\frac{{{{\left( {1 - x} \right)}^6}}}{6}} \right|_0^1 + \frac{1}{6}\int\limits_0^1 {{{\left( {1 - x} \right)}^6}dx} \\= - \frac{1}{6}\left. {\frac{{{{\left( {1 - x} \right)}^7}}}{7}} \right|_0^1 = \frac{1}{{42}} \end{array}$$