# Calculus

posted by on .

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 - ,

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 - ,

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 - ,

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.

### Answer This Question

 First Name: School Subject: Answer:

### Related Questions

More Related Questions

Post a New Question