Antiderivatives math problem

[rowdy3]

Find the following.
∫ (2y^(1/2) - 3y^(2)) / 6y ; dy
It's number 34 if you want to see it.Thanks.
http://pic20.picturetrail.com/VOL1370/5671323/23643016/396306428.jpg
I did
∫ [ ( 2√y - 3y² ) / ( 6y ) ] dy

= ∫ { [ (2√y ) / (6y) ] - [ (3y²) / (6y) ] } dy

= (1/3) ∫ ( 1/ √y ) dy - (1/2) ∫ y dy

= (1/3) [ 2√y ] - (1/2) [ y²/2 ] + C

= (2/3) √y - (1/4) y² + C ...... Ans.
Is that right?

 

[LCKurtz]

Find the following.
∫ (2y^(1/2) - 3y^(2)) / 6y ; dy
It's number 34 if you want to see it.Thanks.
http://pic20.picturetrail.com/VOL1370/5671323/23643016/396306428.jpg
I did
∫ [ ( 2√y - 3y² ) / ( 6y ) ] dy

= ∫ { [ (2√y ) / (6y) ] - [ (3y²) / (6y) ] } dy

= (1/3) ∫ ( 1/ √y ) dy - (1/2) ∫ y dy

= (1/3) [ 2√y ] - (1/2) [ y²/2 ] + C

= (2/3) √y - (1/4) y² + C ...... Ans.
Is that right?

Looks good to me. You can always check by differentiating it.