Asked by Rasheda
I'm a little confused with this integration problem: If the definite integral from 0 to 2 of (e^(x^2)) is first approximated by using two inscribed rectangles of equal width and then approximated by using the trapezoidal rule with n=2, the difference between the two approximations is what?
Answers
Answered by
Steve
x f(x)
0 1
1 e
2 e^4
So, if there are 2 rectangles of width 1, then the area, using left-sides is
1*1 + 1*e = e+1 = 3.718
using right-sides, it's
1*e + 1*e^4 = 57.316
Using the trapezoidal rule, we have
1(1+e)/2 + 1(e+e^4)/2 = 30.517
Kind of a coarse approximation.
0 1
1 e
2 e^4
So, if there are 2 rectangles of width 1, then the area, using left-sides is
1*1 + 1*e = e+1 = 3.718
using right-sides, it's
1*e + 1*e^4 = 57.316
Using the trapezoidal rule, we have
1(1+e)/2 + 1(e+e^4)/2 = 30.517
Kind of a coarse approximation.
Answered by
Rasheda
Thank you!
Answered by
took50benadryls
Steve's answer is unfortunately incorrect. 30.517 was wrong when picked.
Answered by
Dom
The answer is 26.80
Answered by
David
Steve is correct. He gave us both the LH & RH approximations as well as the trapezoid approximation
You just needed to subtract Trapezoid & the LH which gives 30.517-3.718 = 26.80
You just needed to subtract Trapezoid & the LH which gives 30.517-3.718 = 26.80
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