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Approximations rules of a function

  1. Oct 12, 2011 #1
    1. The problem statement, all variables and given/known data

    Given the following integral and value of n, approximate the integral using the methods indicated (round your answers to six decimal places):

    [itex]
    \int_0^1 e^{-3x^2} dx
    [/itex]

    [itex]
    n = 4
    [/itex]


    (a) Trapezoidal Rule

    (b) Midpoint Rule
    This is the only one I'm having trouble with.

    (c) Simpson's Rule


    2. Relevant equations
    [itex]
    \Delta x = \frac{b-a}{n}
    [/itex]

    Trapezoid Rule
    [itex]
    \Delta X \cdot \frac{1}{2} \cdot [ f(0) + 2f(\frac{1}{4}) + 2f(\frac{1}{2}) + 2f(\frac{3}{4}) + f(1) ]
    [/itex]

    Midpoint Rule <- incorrect
    [itex]
    \Delta X \cdot [ f(\frac{0 + \frac{1}{4}}{2}) + f(\frac{ \frac{1}{4} + \frac{1}{2} }{2}) + f(\frac{\frac{1}{3} + \frac{3}{4}}{2}) +f(\frac{\frac{3}{4} + 1}{2})]
    [/itex]

    Simpson's Rule
    [itex]
    \Delta X \cdot \frac{1}{3} \cdot [ f(0) + 4f(\frac{1}{4}) + 2f(\frac{1}{2}) + 4f(\frac{3}{4}) + f(1) ]
    [/itex]

    3. The attempt at a solution
    I set variable for my f-values.
    [itex]
    a = f(0) = 1
    [/itex]

    [itex]
    b = f(\frac{1}{4}) \approx 0.82903
    [/itex]

    [itex]
    c = f(\frac{1}{2}) \approx 0.47237
    [/itex]

    [itex]
    d = f(\frac{3}{4}) \approx 0.18498
    [/itex]

    [itex]
    e = f(1) \approx 0.04979
    [/itex]

    So I plugged everything in and I get

    Midpoint Rule
    [itex]
    \Delta X \cdot [ f(\frac{0 + \frac{1}{4}}{2}) + f(\frac{ \frac{1}{4} + \frac{1}{2} }{2}) + f(\frac{\frac{1}{3} + \frac{3}{4}}{2}) +f(\frac{\frac{3}{4} + 1}{2})]
    [/itex]

    (a) Trapezoidal Rule
    0.5028176513

    (b) Midpoint Rule
    0.5028176513

    (c) Simpson's Rule
    0.5042135205

    However, my midpoint rule answer is wrong. I do think it's strange that I'm getting exactly the same answer for B as I am for A, but I can't seem to see what I'm doing wrong.
     
    Last edited: Oct 12, 2011
  2. jcsd
  3. Oct 12, 2011 #2

    Mark44

    Staff: Mentor

    I think you should be getting different values for the trapezoid and midpoint approximations. Your formula for the midpoint rule is wrong, I'm pretty sure. Instead of averaging the function values at the left and right ends of the subinterval (which is what the trapezoid rule does), it should evaluate the function at the midpoint of the subinterval. For your work, you want to evaluate [itex][f(1/8) + f(3/8) + f(5/8) + f(7/8)]\cdot \Delta x[/itex].
     
  4. Oct 12, 2011 #3

    Mark44

    Staff: Mentor

    Also, your formula for the Trapezoid Rule is wrong. This is what it should be:

    [tex]\Delta X \cdot \frac{1}{2} \cdot [ f(0) + 2f(\frac{1}{4}) + 2f(\frac{1}{2}) + 2f(\frac{3}{4}) + f(1) ][/tex]
     
  5. Oct 12, 2011 #4
    Thanks, Mark44. I must have transcribed my trapezoid rule incorrectly.

    Thanks for the clarification on my midpoint rule. I'm gonna give that a try!
     
  6. Oct 12, 2011 #5
    I think you were absolutely right, Mark but I plugged this in but I'm getting something just above 0.3, which is way too low.
     
  7. Oct 12, 2011 #6

    Mark44

    Staff: Mentor

    I think you made an error. I'm getting something right around .5.
     
  8. Oct 12, 2011 #7
    It took some mathematical gymnastics, but I eventually got the right number. I'm running into a strange problem with these rules. I don't seem to need to multiply by dx for some reason I can't fathom.
     
  9. Oct 12, 2011 #8

    Mark44

    Staff: Mentor

    Then I think you're making another mistake. You have to multiply by [itex]\Delta x[/itex], the width of each subinterval.
     
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