### 4 Answers

- rotchmLv 71 month agoFavorite Answer
Hints: Recalling your high school algebra, rewrite your integrand

as y^(-1/2) - y^(1/2).

Agree?

Now, since you know how to integrate yⁿ , the most basic integral you learned,

just integrate the above. What do you get?

Surely you can take it from there?

If not, ask for further help.

Another hint: its very impolite and unnecessary here to be anon.

Hopefully no one will spoil you the answer thereby depriving you from your personal enhancement; that would be very inconsiderate of them.

- JOHNLv 71 month ago
Integrand is y^(-1/2) – y^(1/2)

Indefinite integral is 2y^(1/2) – (2/3)y^(3/2)

Definite integral is (4 – (2/3)(8) – (2 – 2/3) = -4/3 – 4/3 = -8/3.

- Iggy RockoLv 71 month ago
4

∫ (1 - y)/√y dy =

1

4

∫ y^(-1/2) - y^(1/2) dy =

1

................................4

2y^(1/2) - 2y^(3/2)/3 | =

................................1

2(4^(1/2)) - 2(4^(3/2))/3 - [2(1^(1/2)) - 2(1^(3/2))/3] =

4 - 16/3 - [2 - 2/3] =

12/3 - 16/3 - 6/3 + 2/3 =

-8/3