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Traditional integration of X^3

  1. Aug 23, 2014 #1

    I'm trying to prove that the integral of x^3 (x cubed) between the limits of a (lower limit) and b (upper limit) is:

    (b^4)/4 - (a^4)/4

    I'm using the traditional method of dividing the area into n rectangles (where n tends to infinity). Hence the width of 1 rectangle is (b-a)/n

    The x coordinate (left side of each thin rectangle) of any rectangle is: a + (k-1) * ((b-a)/n)

    I can prove other integrations using this 'traditional' approach but cannot get the correct answer here.

    Does anyone have a link to the proof or can provide it?

    Thanks. This is driving me mad.
  2. jcsd
  3. Aug 23, 2014 #2
    Maybe you could post your calculation that gave you the wrong answer.
  4. Aug 23, 2014 #3

    Simon Bridge

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    So you divide the area into N rectangles width ##\Delta x : N\Delta x = b-a##
    The area of the nth rectangle is ##A_n=(x_n)^3\Delta x:x_n=a+n\Delta x##

    The area between a and b is the sum:$$A=\frac{b-a}{N}\sum_{n=0}^N \left(a+n\frac{b-a}{N}\right)^3 $$ ... expand the cubic and sum each term separately.
    Then take the limit as ##N\to\infty##

    This what you tried?
    Where did you get stuck?

    Or just google for "riemann sum for x^3" ;)
  5. Aug 23, 2014 #4

    Ray Vickson

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    It is probably a lot less messy to first get ##F(c) =\int_0^c x^3 \, dx## using the traditional approach, then getting your answer as ##F(b) - F(a)## (assuming ##0 < a < b##). To get ##F(c)## you just need to perform summations of the form ##\sum_{n=1}^N n^3##.
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