Given p.d.f. is f(x) = \(\begin{cases} { 2e }^{ -2x },>0 \\ 0,\quad otherwise \end{cases}\)
Expected life of this piece of equipment is 
\(E(X)=\int _{ -\infty }^{ \infty }{ x.f(x)dx=\int _{ 0 }^{ \infty }{ x.{ 2e }^{ -2x } } } \)
\(=2\int _{ 0 }^{ \infty }{ { xe }^{ -2x }dx=2\left[ \frac { 1! }{ { 2 }^{ 2 } } \right] } \)
[\(\because \int _{ 0 }^{ \infty }{ { x }^{ n }{ e }^{ -ax }dx=\frac { n! }{ { a }^{ n+1 } } } \) gamma integral here n = 1, a = 2]
\(E(X)=\frac { 2 }{ 4 } =\frac { 1 }{ 2 } \)

 LHS = V[aX] = E[aX - E(aX)]2
= E(aX-a.E(X)]2[∵E(aX) = a.E(X)]
= E[aX-a.E(X)]2[∵E(aX) = a.E(X)]
= E[a{X-E(X)}]²
= a².E[X-E(X)]² = a²E(X-\(\bar { x }\))²
= a².[E(X)²-[E(X)]²]
= a².V(X) [∵V(X) = E(X²)-[E(X)]²]
= RHS Hence proved.

Given X is a random variable
∴ Y = 2X + 1 is also a random variable
Given Var (X) = 5
∴ Var (Y) = Var (2X + 1) = 22 Var (X)
[∵ Var (aX + b) = a² Var (X)]
= 4(5)
∴ Var (Y) = 20

If X is a continuous random variable and f(x) is the value of its probability density function at x, then the expected value of X is
\(E(X)=\int _{ -\infty }^{ \infty }{ x.(x)dx } \)

Var (X) = E(X²) - [E(X)]² where
\(E({ X }^{ 2 })=\begin{cases} \sum { { x }^{ 2 }p(x)\text {for discrete random variable} } \\ \int _{ -\infty }^{ \infty }{ { x }^{ 2 }P(x)\text{dx for continuous random variable} } \end{cases}\)