Calculus
posted by Rachel .
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)ln32/3(sqrt*8)
can someone please verify this?

2/3(8)ln32/3(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)) 
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? 
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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