00:02
In part a of this problem we are going to show that the average of the left and right end point approximations give rise to formula 2 for the trapezoidal approximation.
00:15
And in part p, we have a function f defined a non -negative on the interval ab, which is subdivided into ends of intervals of the same length by using the points x0 equals a, less than x1 less than up to x n minus 1 less than x sub n which is equal to b.
00:43
Here we want to find the area of the trapezoid under the line segment joining the points x sub k f of x sub k and x sub k plus 1 f of x sub k plus 1 f of x sub k plus 1 and above the interval x of k x of k plus 1 then we will prove that the right side of formula 2 is the sum of these trapezoidal areas so we start in part a by remembering that formula 2 given on chapter 7 .7 of the test book is the trapezoidal approximation t sub n equals b minus a over 2 n times y 0 plus 2 times y sub 1 plus 2 plus 2 plus up 2 times what y sub n minus 1 where y sub k equals f of x of k for k from zero up to n so this is the formula of the trapezoidal approximation and we are going to show in part eight that the average of the left and right end point approximations give rise to this formula too so we start by remembering those and left and right end point approximation so the left end point approximation is given by integral from a to b of f is just equal or approximately equal to b b minus a over n times y0 plus y1 plus up to y sub n minus 1 that is because we are taking the left 10 points, the images of the left end points of the sub -interval, so the last, the first one is y -0 and the last one is y -7 minus 1.
04:00
Then we have the right endpoint approximation, so the integral from a to b is approximately equal to b minus a over n times.
04:28
In this case, we take the right endpoints of the sub -interval, so the first one is y1 then y2 up to y sub n so we can take the average of these two formulas is b minus a over n is one half we can say a half times b minus a over n times y0 plus y1 up to y sub n plus y sub n plus y sub n plus b minus a over n times y1 plus y2 up 2 y sub n that is the sum of these two expressions this one here and this one here over 2 or which is the same and multiply by one half so this is one half times now this term b minus a over n is a common factor we take it out of the square bracket so we get b minus a over n times one half is over 2 n times y 0 plus y1 plus up to n minus 1 plus y1 plus y 2 plus up to y sub n that's equal to b minus a over 2 n times now we see that's equal to b minus a over 2 n times now we see that this term is one time in this whole sum.
06:50
This term appears twice because it's here and here.
06:58
Y2, which is the term that comes here next, appears twice because it's there and there.
07:07
And so on.
07:08
So we are going to have finally that y of n minus 1, which comes before this term here, before this term here, we have y sub n minus 1, which is here...