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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.

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