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monroe17

  • 2 years ago

trapezoidal rule of integral from 0 to 1 of e^(-6x^2)dx n=4 I got 2.016753867... and it says it's wrong. I split it up as.. delta x/2=1/8 x_0=0 x_1=0.25 x_2=0.50 x_3=0.75 x_4=1 help?

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  1. monroe17
    • 2 years ago
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    nevermind :) got it!

  2. monroe17
    • 2 years ago
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    I inputted the function wrong..

  3. campbell_st
    • 2 years ago
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    so if you draw it I've calcuated the function values and they are on top of the offset. |dw:1359660333907:dw| and from here you can just apply the formula for area of each trapeziod then sum the areas... I've always hated the formula.

  4. campbell_st
    • 2 years ago
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    but if you want to use the formula you would have \[\int\limits_{0}^{1}e^{-6x^2} dx \approx \frac{0.5}{2}[1 + 2 \times 0.687289 + 2 \times 0.22313+ 2 \times 0.034218 + 0.002479]\] and its just a case of evaluating.

  5. campbell_st
    • 2 years ago
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    oops should be \[\frac{0.25}{2}\] at the start

  6. campbell_st
    • 2 years ago
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    i got 0.361469

  7. monroe17
    • 2 years ago
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    yeah I figured it out. Thanks.

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