Probability and Statistics question, need help?

the density function of X is
f(x)
= x^(α-1) * e^(-x/β) / [ β^α * Γ(α) ]
= x^(2-1) * e^(-x/0.5) / [ 0.5^2 * Γ(2) ]
= x * e^(-2x) / [ 0.25 * (2-1)! ]
= 4x * e^(-2x)

∫ f(x) dx
= ∫ 4x * e^(-2x) dx
= 4 * ∫ x * e^(-2x) dx
= 4 * ∫ x * [ 1/(-2) ] * d e^(-2x)
= (-2) * ∫ x d e^(-2x)
= (-2) * [ x*e^(-2x) - ∫ e^(-2x) dx ]
= (-2) * [ x*e^(-2x) + (1/2)e^(-2x) ] + C
= - 2x*e^(-2x) - e^(-2x)
= - ( 2x + 1 )*e^(-2x) + C

Let F(x) = - ( 2x + 1 )*e^(-2x) , then
∫ f(x) dx , from x = a to x = b
= F(b) - F(a)

(a)
P( 0 ≦ X ≦ 1 )
= ∫ f(x) dx , from x = 0 to x = 1
= F(1) - F(0)
= - 3e^(-2) - (-1)
= 1 - 3e^(-2)
≒ 0.594 ..... Ans

(a)
P( 2 ≦ X < ∞ )
= ∫ f(x) dx , from x = 2 to x = ∞
= F(∞) - F(2)
= 0 + 5e^(-4) ..... please see Note below
= 5e^(-4)
≒ 0.092 ..... Ans

Note :
F(x) = - ( 2x + 1 )*e^(-2x) = - ( 2x + 1 ) / e^(2x)
as x → ∞ , by the L'Hopital's rule , we have
F → - 2 / [ 2 * e^(2x) ] → 0
Thus, F(∞) → 0