Let X be a continuous r.v has a p.f given by: f(x)={(ce^(-2x) , X≥0 0 otherwise)┤ Find the first four moment about the mean

Since  f(x) is a p.d.f, then      ∫_(-∞)^∞〖f(x)〗 dx=1
∴1=c∫_0^∞ e^(-2x)  dx=c/2 [-e^(-2x) ]_0^∞=c/2
c=2
To evaluate the moments, we calculate the m.g.f:
∵M_x (t)=E(e^tx )
∴M_x (t)=∫_(-∞)^∞ e^tx  f(x)dx=2∫_0^∞ e^tx  e^(-2x) dx
=2∫_0^∞ e^(-(2-t)x)  dx=2/(2-t) [-e^(-(2-t)x) ]_0^∞=2/(2-t)   ,    t>2
=1/(1-t⁄2)=(1-t⁄2)^(-1)   
2/(2-t)=1+t/2+(t/2)^2+(t/2)^3+(t/2)^4+⋯
But:
M_x (t)=1+μt+μ_2  t^2/2+μ_3  t^3/3!+μ_4  t^4/4!+⋯
Thus the first four moments are:
μ_1=1/2   ,   μ_2=1/2    ,   μ_3=3/4    ,   μ_4=3/2
The centeral moments are:
∴m_1=0
m_2=μ_2-μ_1^2=1/2-1/4=1/4
m_3=μ_3-3μ_1 μ_2+2μ_1^3=3/4-3(1/2)(1/2)+2(1/8)=1/4
m_4=μ_4-4μμ_3+6μ^2 μ_2-3μ_1^4=3/2-4(1/2)(3/4)+6(1/4)(1/2)-3(1/16)=9/16
c=2