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Definite Integral by Definition

  1. Dec 9, 2015 #1
    1. The problem statement, all variables and given/known data
    Let A be the area of the region that lies under the graph of f(x) = 2x 2 + 5 between x = 0 and x = 4. Find an expression for A using n rectangles. Then evaluate this expression.

    2. Relevant equations
    Answer is 188/3
    h= (4/n)

    3. The attempt at a solution
    2943aeaecd3d29cb25c8faaa4fe7e4d4.png

    The problem I am having is I do not know how to simplify this limit further after factoring out the h.
     
    Last edited by a moderator: Dec 9, 2015
  2. jcsd
  3. Dec 9, 2015 #2

    BvU

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    ##f(a+h)## is not ##5+h## but ##2(a+h)^2+5## !?

    Aren't you supposed to find ##A(n)## ?
     
  4. Dec 9, 2015 #3
    You are right about 2(a+h)^2+5. So you would suggest to convert this limit all to "n" and evaluate the limit as n ---> infinity?
     
  5. Dec 9, 2015 #4

    BvU

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    If you have faithfully rendered the problem statement, the answer is an expression. Something like ##\displaystyle A = \sum_{i=1}^n \ ... \ ##

    The way I read it you want to work out e.g. the middle riemann sum

    300px-MidRiemann2.svg.png
     
  6. Dec 11, 2015 #5

    Ray Vickson

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    I cannot read your formula for f(x). Is it ##f(x) = 2x^2 + 5##? ##f(x) = (2x)^2 + 5##? Something else? If it really does involve the second power (and if you don't want to use either LaTeX or the formula template in the top green panel, then you need to use "^" instead, as in 2 x^2 + 5 or (2x)^2 + 5.
     
  7. Dec 11, 2015 #6

    BvU

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    Well, the book answer helps you out, Ray :wink:
     
  8. Dec 11, 2015 #7

    Ray Vickson

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    Well, of course I can guess what the OP means, but the whole point is that I should not need to. He/she needs to adhere to the standard notation; that's why it was invented.
     
  9. Dec 12, 2015 #8

    BvU

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    Agree.
    But it looks as if Newt is out of the picture ?
     
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