How do you integrate by substitution #int (9-y)sqrty dy#?

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Answer 1 · 2018-03-19T17:04:47

The answer is #=6y^(3/2)-2/5y^(5/2)+C#

We need

#intx^ndx=x^(n+1)/(n+1)+C(x!=-1)#

Let #u=sqrty#, #=>#, #du=(dy)/(2sqrty)#

Therefore,

#int(9-y)sqrtydy=int(9-u^2)sqrty*2sqrtydu#

#=2int(9u^2-u^4)du#

#=2*9u^3/3-2/5u^5#

#=6y^(3/2)-2/5y^(5/2)+C#

Answer 2 · 2018-03-19T17:05:07

#2/5y^(3/2)(15-y)+c#

#I=int(9-y)sqrtydy#

to do this by substitution:

#u=sqrty=>u=y^(1/2#

#=>du=1/2y^(-1/2)dy#

#dy=2y^(1/2)du#

#I=int(9-u^2)y^(1/2)2y^(1/2)du#

#=int(9-u^2)ydu#

#=2int(9-u^2)u^2du#

#=2int(9u^2-u^4)du#

#=2(3u^3-u^5/5)+c#

#=2/5u^3(15-u^2)=c#

#=2/5y^(3/2)(15-y)+c#

note for this integral it would be more efficient to multiply the brackets out and integrate directly using the power rule

ie

#int(9-y)y^(1/2)dy=int(9y^(1/2)-y^(3/2))dy#

etc