a)
since for F(a) =ln(a) =0
a =e0 =1
and F(b) =ln(b) =1
b=e1 =e
therefore a=1 and b=e
b)
P(X>2 )=1-P(X<2) =1-ln(2) =1-0.6931 =0.3069
c)
f(x) =(d/dx)*F(x) =(d/dx)*ln(x) =1/x
d)
E(x)= xf(x) dx = x*(1/x) dx= dx = x|e1 =e-1 =1.7183
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