Question

How do you integrate by substitution `int 2piy(8-y^(3/2))dy`?

Answer 1

`int 2piy(8-y^(3/2))dy = 2pi{ 4y^2 - 2/7y^(7/2)} + C`

Explanation:

You do not need to use a substitution as the integral can be calculated "as is":

`int 2piy(8-y^(3/2))dy = 2piint 8y-y^(5/2)dy`
`:. int 2piy(8-y^(3/2))dy = 2pi{ 8y^2/2 -y^(7/2)/(7/2)} + C`
`:. int 2piy(8-y^(3/2))dy = 2pi{ 4y^2 - 2/7y^(7/2)} + C`