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monroe17
Group Title
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?
 one year ago
 one year ago
monroe17 Group Title
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?
 one year ago
 one year ago

This Question is Closed

monroe17 Group TitleBest ResponseYou've already chosen the best response.0
nevermind :) got it!
 one year ago

monroe17 Group TitleBest ResponseYou've already chosen the best response.0
I inputted the function wrong..
 one year ago

campbell_st Group TitleBest ResponseYou've already chosen the best response.1
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.
 one year ago

campbell_st Group TitleBest ResponseYou've already chosen the best response.1
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.
 one year ago

campbell_st Group TitleBest ResponseYou've already chosen the best response.1
oops should be \[\frac{0.25}{2}\] at the start
 one year ago

campbell_st Group TitleBest ResponseYou've already chosen the best response.1
i got 0.361469
 one year ago

monroe17 Group TitleBest ResponseYou've already chosen the best response.0
yeah I figured it out. Thanks.
 one year ago
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