anonymous
  • anonymous
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?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
nevermind :) got it!
anonymous
  • anonymous
I inputted the function wrong..
campbell_st
  • campbell_st
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.

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campbell_st
  • campbell_st
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.
campbell_st
  • campbell_st
oops should be \[\frac{0.25}{2}\] at the start
campbell_st
  • campbell_st
i got 0.361469
anonymous
  • anonymous
yeah I figured it out. Thanks.

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