Mumbai University > Mechanical Engineering > Sem 4 > Applied Mathematics IV

Marks: 5M

Year: Dec 2014

$$Mean = E(X) = \int^{\infty }_{-\infty }{x\ f\left(x\right)dx}$$ $$ = \int^1_0{x\ 6\left(x-x^2\right)dx}$$ $$ = 6 \int^1_0{\left(x^2-\ x^3\right)dx}$$ $$ = 6 {[\frac{x^3}{3}-\frac{x^4}{4}]\ }^1_0$$ $$= 6 { (1/3 - ¼ ) - (0-0)} = 6/12$$ $$ = 1/2$$ Consider $$f\left(x^2\right)=\ \int^{\infty }_{-\infty }{x^2\ .\ f\left(x\right)dx}$$ $$= 6\int^1_0{\left(x^3-\ x^4\right)dx}$$ $$= 6 {[\frac{x^4}{4}-\frac{x^5}{5}\ ]\ }^1_0$$ $$= 6 [(1/4 - 1/5) - 0]$$ $$= 6/20$$ $$= 3/10$$ $$\mathrm{\therefore } Variance = E(X^2)-\ {[E\left(X\right)]}^2 = 3/10 - {(\frac{1}{2})}^2$$ $$= 3/10 - 1/4$$ $$= 5/100 = 1/20$$ $$\boldsymbol{\mathrm{\therefore }} Mean = 1/2 = 0.5$$ $$Variance = 1/20 = 0.05$$