[m]D(X) = \int_{-\infty}^{+\infty}x^2p(x)dx - (\int_{-\infty}^{+\infty}xp(x)dx)^2[/m]
В нашем случае функция p(x) = f задается системой:
{ f = 0; x < -7/9
{ f = 162x + 126; -7/9 ≤ x < -2/3
{ f = 0; x ≥ -2/3
Поэтому формула дисперсии примет такой вид:
[m]D(X) = \int_{-7/9}^{-2/3}x^2(162x + 126)dx - (\int_{-7/9}^{-2/3}x(162x + 126)dx)^2 =[/m]
[m] =\int_{-7/9}^{-2/3}(162x^3 + 126x^2)dx - (\int_{-7/9}^{-2/3} (162x^2+126x)dx)^2 =[/m]
[m]=(\frac{162x^4}{4}+\frac{126x^3}{3})|_{-7/9}^{-2/3} - ((\frac{162x^3}{3}+\frac{126x^2}{2})|_{-7/9}^{-2/3})^2 =[/m]
[m]= (\frac{81x^4}{2}+ 42x^3)|_{-7/9}^{-2/3} - ((54x^3+63x^2)|_{-7/9}^{-2/3})^2 =[/m]
[m]=(\frac{81}{2}\frac{(-2)^4}{3^4}+ 42\frac{(-2)^3}{3^3}) - (\frac{81}{2}\frac{(-7)^4}{9^4}+ 42\frac{(-7)^3}{9^3}) - (54\frac{(-2)^3}{3^3}+63\frac{(-2)^2}{3^2} - 54\frac{(-7)^3}{9^3}-63\frac{(-7)^2}{9^2})^2 =[/m]
[m]= (\frac{81}{2}\frac{16}{81}+ 42\frac{-8}{27}) - (\frac{81}{2}\frac{2401}{6561}+ 42\frac{-343}{729}) - (54\frac{-8}{27}+63\frac{4}{9} - 54\frac{-343}{729}-63\frac{49}{81})^2 =[/m]
[m]= (8- \frac{112}{9}) - (\frac{21609}{1458}- \frac{28812}{1458}) - (\frac{-432}{27}+\frac{756}{27} - \frac{-2 \cdot 343}{27}-\frac{21 \cdot 49}{27})^2 =[/m]
[m]= \frac{72-112}{9} + \frac{7203}{1458} - (\frac{-432}{27}+\frac{756}{27} + \frac{686}{27}-\frac{1029}{27})^2 =-\frac{40}{9} + \frac{2401}{486} - (\frac{-19}{27})^2 =[/m]
[m]= -\frac{40}{9} + \frac{2401}{486}- \frac{361}{729} = -\frac{2160}{486} + \frac{2401}{486} - \frac{361}{729} = \frac{241}{486} - \frac{361}{729} =[/m]
[m]=\frac{241 \cdot 3}{486 \cdot 3} - \frac{361 \cdot 2}{729 \cdot 2} = \frac{723}{1458} - \frac{722}{1458} = \frac{1}{1458} ≈ 0.00068[/m]
В ответе можно смело написать: [b]0.00[/b]