Đáp án:
\[\int\limits_0^1 {{x^5}.{{\left( {1 - {x^3}} \right)}^6}dx} = \frac{1}{{168}}\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
t = 1 - {x^3} \Rightarrow \left\{ \begin{array}{l}
{x^3} = 1 - t\\
dt = - 3{x^2}dx\\
x = 0 \Rightarrow t = 1\\
x = 1 \Rightarrow t = 0
\end{array} \right.\\
\int\limits_0^1 {{x^5}.{{\left( {1 - {x^3}} \right)}^6}dx} \\
= \frac{{ - 1}}{3}.\int\limits_0^1 {{x^3}.{{\left( {1 - {x^3}} \right)}^6}.\left( { - 3{x^2}dx} \right)} \\
= \frac{{ - 1}}{3}.\int\limits_1^0 {\left( {1 - t} \right).{t^6}.dt} \\
= \frac{1}{3}.\int\limits_0^1 {\left( {1 - t} \right).{t^6}dt} \\
= \frac{1}{3}.\int\limits_0^1 {\left( {{t^6} - {t^7}} \right)dt} \\
= \frac{1}{3}.\mathop {\left. {\left( {\frac{{{t^7}}}{7} - \frac{{{t^8}}}{8}} \right)} \right|}\nolimits_0^1 \\
= \frac{1}{3}.\left( {\frac{1}{7} - \frac{1}{8}} \right)\\
= \frac{1}{{168}}
\end{array}\)
