Đáp án:
\[I = \int\limits_0^1 {x.\sqrt {1 - {x^2}} dx} = \dfrac{1}{3}\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
I = \int\limits_0^1 {x.\sqrt {1 - {x^2}} dx} \\
t = 1 - {x^2} \Rightarrow \left\{ \begin{array}{l}
dt = \left( {1 - {x^2}} \right)'dx = - 2xdx\\
x = 0 \Rightarrow t = 1\\
x = 1 \Rightarrow t = 0
\end{array} \right.\\
\Rightarrow I = \int\limits_1^0 {\sqrt t .\dfrac{{ - dt}}{2}} = - \int\limits_0^1 {\sqrt t .\dfrac{{ - dt}}{2}} = \dfrac{1}{2}\int\limits_0^1 {\sqrt t dt} = \dfrac{1}{2}.\int\limits_0^1 {{t^{\dfrac{1}{2}}}dt} \\
= \mathop {\left. {\dfrac{1}{2}.\dfrac{{{t^{\dfrac{1}{2} + 1}}}}{{\dfrac{1}{2} + 1}}} \right|}\nolimits_0^1 = \mathop {\left. {\dfrac{1}{3}{t^{\dfrac{3}{2}}}} \right|}\nolimits_0^1 = \dfrac{1}{3}.\left( {{1^{\dfrac{3}{2}}} - {0^{\dfrac{3}{2}}}} \right) = \dfrac{1}{3}
\end{array}\)
Vậy \(I = \int\limits_0^1 {x.\sqrt {1 - {x^2}} dx} = \dfrac{1}{3}\)
