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

Marks: 5M

Year: May 2014

For any probability density function $$\int^{\infty }_{-\infty }{f\left(x\right)dx=1}$$ $$\mathrm{\therefore } \int^1_0{\left(A+Bx\right)=1}$$ $$\mathrm{\therefore } {\left[Ax+\ \frac{Bx^2}{2}\right]}^1_0=1$$ $$\mathrm{\therefore } [A(1)+\ \frac{B{\left(1\right)}^2}{2}]-0=1$$ $$\mathrm{\therefore } [A + \frac{B}{2}]=1----(1)$$

Also, Mean $$E(X) =\int^{\infty }_{-\infty }{f\left(x\right)dx}$$ $$\mathrm{\therefore } \frac{1}{3}\ =\ \int^1_0{x\left[A+Bx\right]dx}$$ $$\mathrm{\therefore } \frac{1}{3}\ =\ \int^1_0{\left[Ax+B\ x^2\right]dx}$$ $$\mathrm{\therefore } \frac{1}{3}\ =\ {[\ \frac{Ax^2}{2}+\frac{Bx^3}{3}]\ }^1_0$$ $$\mathrm{\therefore } \frac{1}{3}\ =\ [\frac{A{\left(1\right)}^2}{2}+\frac{B{\left(1\right)}^3}{3}]-[0+0]$$ $$\mathrm{\therefore } \frac{1}{3}\ =\left[\frac{A}{2}+\ \frac{B}{3}\right]---------\left(2\right)$$ Solving (1) and (2) simultaneously we get ,

A = 2 and B = -2