Deriving the trapezium rule from the formula used to find the area of a trapezium?
I have to derive the trapezium rule. In other words going from h/2(a+b) (the formula to get the area of a trapezium) to h/2[f(0)+2(f(1)+f(2)+...+f(n-1))+f(n)] (the trapezium rule)
I really hope that makes sense…
Yes it makes sense. You are talking about finding areas under a curve numerically rather than by integration.
The first strip will have sides f(0) and f(1) so its area is
(h/2)[f(0) + f(1)]
The second strip has sides f(1) and f(2) so its area is
(h/2)[f(1) + f(2)]
Continue in this way up to the penultimate strip area of
(h/2)[f(n-2) + f(n-1)]
and the final strip with area of
(h/2)[f(n-1) + f(n)]
Adding these up you see that (h/2) is a common factor. All the function terms appear twice except f(0) and f(n) which only appear once. That is why the long bracket starts
f(0) + 2f(1) + 2f(2) . . .
and ends
. . . 2f(n – 2) + 2f(n – 1) + f(n)
P.S. You have missed out some of the 2’s in your long bracket.
Yes it makes sense. You are talking about finding areas under a curve numerically rather than by integration.
The first strip will have sides f(0) and f(1) so its area is
(h/2)[f(0) + f(1)]
The second strip has sides f(1) and f(2) so its area is
(h/2)[f(1) + f(2)]
Continue in this way up to the penultimate strip area of
(h/2)[f(n-2) + f(n-1)]
and the final strip with area of
(h/2)[f(n-1) + f(n)]
Adding these up you see that (h/2) is a common factor. All the function terms appear twice except f(0) and f(n) which only appear once. That is why the long bracket starts
f(0) + 2f(1) + 2f(2) . . .
and ends
. . . 2f(n – 2) + 2f(n – 1) + f(n)
P.S. You have missed out some of the 2’s in your long bracket.
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