The formula
in solving the integral of the infinity of 3 is ∫3∞?(1)÷((x−2)(3/2))dx
Substitute the numbers given
then solve
limn→inf∫3n(1/((n−2)(3/2))dn
limn→inf[−2/(n−2−−−−−√)−(−2/3−2−−−−√)
=0+2=2
Solve for the integral of 2 when 2 is approximate to 0.
Transpose 2 from the other side to make it -2
∫∞3(x−2)−3/2dx=(x−2)−1/2−1/2+C
(x−2)−1/2=1x−2−−−−√
0−(3−2)−1/2−1/2=2
