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Exact value of the are of the region bounded by

  1. Jan 11, 2012 #1
    Question:
    Find the exact value of the are of the region bounded by:
    x^3, the x-axis and x=1 and x=4

    Answer is 3.75

    I tried finding the anti derivative so 1/4(x)^4, and therefore I've got 1/4(4)^4 - 1/4, which isn't the correct answer
     
    Last edited: Jan 11, 2012
  2. jcsd
  3. Jan 11, 2012 #2

    Dick

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    You are integrating x^3 for x from 0 to 1, right? So the antiderivative is, yes, x^4/4 and the result is 1^4/4-0^4/4=1/4. That's what you are doing yes? You've got a bunch of what I assume are 'dashes' in the post that look like 'minuses'. That looks ok to me, and I don't see how 3.75 could be correct.
     
  4. Jan 11, 2012 #3
    I apologize, I meant x=1 and x=4 and I deleted all the dash (there was just one)
     
  5. Jan 11, 2012 #4

    Dick

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    That's different! So you did evaluate the antiderivative between 4 and 1. Then 4^4/4-1/4 is correct. And it's not 3.75. It's 63.75. A typo?
     
  6. Jan 11, 2012 #5
    the answer in the book clearly states "3 and 3/4 units^2". If the answer is wrong than that's pretty awful, since it was the first problem. Other answers seem fine. Thanks for your help though.

    If you don't mind another question I have is
    Find the exact value of the are of the region bounded by:
    e^x, the x-axis, the y-axis and x=1

    I have no what to do if its bounded by both x and y axes
     
  7. Jan 11, 2012 #6

    Dick

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    The 'y-axis' is x=0, right? That makes it pretty similar to the previous question.
     
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