Evaluate #int_0^4 ##int_(sqrtx)^2 sqrt(1+y^3) dydx#=?

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Answer 1 · 2018-06-24T16:03:29

#= 52/9#

#int_(x=0)^4 int_(y=sqrtx)^2 sqrt(1+y^3) dydx#

Switch order of integration:

#= int_(y=0)^2 int_(x=0)^(y^2) sqrt(1+y^3) dxdy#

#= int_(y=0)^2 [ x sqrt(1+y^3) ]_(x=0)^(y^2) dy#

#= int_(y=0)^2 y^2 sqrt(1+y^3) dy#

#= int_(y=0)^2 d/dx [ 2/9 (1+y^3)^(3/2) ] dy#

#=2/9 [ (1+y^3)^(3/2) ]_(y=0)^2 = 52/9#