Posted by **Rachel** on Sunday, January 30, 2011 at 3:26pm.

Find the area enclosed by the following functions from x=0 to x=2:

y=sqrt*(x+2)

y=1/(1+x)

I got

2/3(8)-ln3-2/3(sqrt*8)

can someone please verify this?

- Calculus -
**helper**, Sunday, January 30, 2011 at 5:11pm
2/3(8)-ln3-2/3(sqrt*8)

My answer,

16/3 - ln 3 - 4/3(sqrt(2))

You have 16/3,

2/3 (8) = 16/3

Where you went wrong,

You have, 2/3 (sqrt(8))

It should be,

2/3 (0 + 2)^3/2

2/3 (2)^3/2

2/3 (sqrt(2))^3

2/3 (2(sqrt(2)))

4/3 (sqrt(2))

- Calculus -
**let me help**, Sunday, January 30, 2011 at 5:33pm
Perfect, thanks!

I agree that your answer is more simplified..

but my answer is definitely not wrong because

2/3(sqrt(8)) and 4/3(sqrt(2)) are the same.

Because.. 8=4x2 and sqrt of 4 is 2.. so you take 2 out of the sqrt to be 2. 2x2 is 4. so you end up with 4/3(sqrt(2)). Do you agree?

- Calculus -
**helper**, Sunday, January 30, 2011 at 5:51pm
I guess it depends on your teacher, because

(2)^3/2 means (sqrt(2))^3

I've never seen anyone interpret this as

(sqrt(2^3)).

The first way will always get you the correct, simplified answer. If I were you, I would get in the habit of doing this the first way.

Radicals should always be in the lowest terms. Any teacher I had would take off half credit for your answer.

Good luck.

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