Question

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

Answer 1

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

Explanation:

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

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

Explanation:

`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