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Midpoint Rule for Integrals

  1. Jul 21, 2008 #1
    1. The problem statement, all variables and given/known data

    Integral Upper Limit 14, Lower Limit 2, [squareroot(x^2+1)]dx ; n=6

    2. Relevant equations

    3. The attempt at a solution
    Midpoints : 3,5,7,9,11,13
    width of subintervals: (14-2)/6=2

    2([squareroot(3^2+1)]+[squareroot(5^2+1)] ......

    is that on the right track?
  2. jcsd
  3. Jul 23, 2008 #2
    Yes, this is the right track. You make it look so hard. You are adding up areas of rectangles right?

    -Each rectangle has a height and a width.
    -The height is determined by using some point on the interval that the rectangle is on, in this case, the midpoint. So you take the midpoint and plug it into the function to get the height.

    Once you get that, all you need is the width which was 2. So just 2 * f(x_i) where each x_i is a midpoint.
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