00:01
We are going to estimate the integral from 0 to 2 of 1 over square root of 4 plus x cubed, first bytraps.
00:17
So the rule with 4 sub intervals in part a, and then in part b, by simpson's rule with 2 subintervals.
00:26
We will round off the results to 4 decimal places.
00:32
So first let's define the function f of x equals, 1 over square root of 4 plus x q which is the function we are integrating here now we consider that function defined on the closed interval from 0 to 2 so in bar a you will use trapsoid a rule and the general expression is tn equal and we use n's of intervals b minus a over 2 n times f at x0 first note plus 2f at x1 second note plus 2 f at x2 plus up to 2 f at x0 2 f at x2 plus f at xn this is the general expression for trapsoidal rule with n sub intervals in this case, we are going to use n equal 4, and then with this we say that step size h, which is defined as b minus a over n, is equal to 2 minus 0.
02:06
That is the upper limit of integration minus the lower limit over n, which is 4.
02:12
So we get 2 over 4, which is 1 1 .5.
02:16
So the step size h is 1 .5.
02:19
With that, we can now write the expression for the note.
02:23
Are given by x i equal the lower limit of integration zero plus i times h and that becomes i over two because h is one half and that for index i from zero up to four so is zero one two three and four and with that we can now say that the trapsoiler rule with four of intervals is equal to let's see that this factor b minus a over 2n is one half or is the half of this b minus a over n that is h so it's each half times f at x0 plus 2 f at x1 plus 2 f at x1 plus 2 f at x2 plus 2 f at x3 and that's the last image with factor 2 because we had to stop that at xn minus 1.
03:34
And for n equals 4, xn minus 1 is x3.
03:38
So the last term is plus f at the last node, which is in this case, x4.
03:47
So this is the expression for trapezoidal rule with n and with force of intents.
03:52
Now we put the value, so t4 is h, which is 1 1ïżœ, divided by 2 times f the first note is 0 plus 2 f at we know the notes are given by this expression here height i half so x1 is 1 half plus 2 f at for i equal 2 we get 2 over 2 is 1 plus 2 f at x 3 for i three we get three halves plus now the image of x4 that is two which is the upper limit of integration we have this expression for the terms of the rule with four subintables and so t4 is one -fourth times now we apply the function which is one over square root of four plus x q so here is one over square root of 4 plus 0 cube is a square root of 4 which is 2 so f at 0 is 1 1 1â2 plus 1 half plus 1 half cube and remember the function is 1 over this square root but the factor 2 here multiplied by the function gives us 2 over the square root the next term will be 2 over the square root the next term will be 2 over square root of 5 because it's 4 plus 1 cube under 2 is square root plus 2 over square root of 4 plus 3 1 halves square they're cube so cube plus and now the last term is 1 over square root of 4 plus 2 cube 2 cube is 8 so it's 2 root of 12 now we use a calculator to find that t4 the trapesoider rule with four sub intervals approximately equal to 0 .851.
06:35
Now in part b, the simpsons rule with two sub intervals and general expression for simpsons rule with n sub intervals is b minus a over 6 n times the image of the first node plus the image of the last node, x n plus two times the image of x1 plus up to the image of x n minus 1 and that plus 4 times the sum of the images of the midpoints the first one is x0 plus x1 over 2 up to the last one is xn minus 1 plus xn over 2 and that's the expression journal expression for simpson's rule then if you use if we use two sub -intervals, then h is b minus a over n will be 2 minus 0 over 2, that is 2 over 2 equal 1.
07:48
So here, h is 1...