00:02
Given the following graph of the function y equal f of x and n equals 6, we want to answer the following questions about the area under the curve from x equals 0 to x equals 6.
00:17
The first one is used the trapeizer rule to estimate the area and then using simpson's rule to estimate the area.
00:25
We have here the graph of the function y equal f of x and we can see this function is positive.
00:34
On the interval 06 and for that reason we can talk about area under the curve given by the integral from 0 to 6 of this function so that's the first observation we get made here the area under the curve under the curve y equal f of x is given by the intro from 0 to 6 well i got to save before that the curve y equal of x over the interval 06, which is the interval mentioned here, from x equal 0 to x equal 6, that area is given by the interval from 0 to 6 of the function f, and the equality between the area under the curve over that interval 06, and this integral comes from the fact that the function is positive over the interval 06.
02:02
So if we approximate this interval, we are approximating the area, and that's what we're going to do.
02:07
We approximate the interval using a trapzother rule on simpson's rule, with n equals 6, that is, with 6 of intervals.
02:18
So n equals 6 implies that h, which is a step size, the common distance between any 2, consecutive notes or what is the same, the length of the common length of all the sub -intervals we are considering is given by the length of the interval of integration, 6 minus 0, over the number of sub -intervals and 6.
02:46
And that is 6 over 6 equal 1.
02:50
So the step size, age, is 1.
02:54
And because of that, the nodes we get to be using for these approximations are x -0 equals 0, which is the left and point of the interval of integration.
03:08
Then x -1 is x -0 plus h, that is 1.
03:13
X2 is x1 plus h, that is 2, and so on.
03:18
X -3 is 3, x4 is 4, x -5 is 5, and x6 is 6, which is the writing point of the interval of integration.
03:32
So we can say that we have a table of values that we can deduce from this graph here.
03:42
So the notes where we want to know the values of the functions are 0, 1, 2, 3, 4, 5, and 6.
03:58
At 0, we can see here the function is equal to 1.
04:05
At 1, the function is equal to 2, this value here.
04:12
At 2, 2 is here.
04:15
The image is 4 over here.
04:18
Here we have 3.
04:21
So we have 4.
04:22
At 3, which is here.
04:25
The image is 2.
04:27
At 4, over here, the image is in the middle between 1 and 2.
04:33
That is 1 .5.
04:37
At 5 here the image is 1 and at 6 let me modify this little bit put that over here at that 6 we have 6 is 3 so this is the table of value of the function at the nodes we are using in both approximations and with that information we can find out the approximation to the area so in part 1, we use a trapezoidal rule with 6 of intervals, t6.
05:14
And that's defined as each half times f at the first node x0 plus 2 times the sum of the images of the internal nodes.
05:26
That is, f at x1 plus f at x2 plus f at x3 plus f at x4 plus f.
05:39
At x5 and all those images are multiplied by 2 there are sum multiplied by 2 plus and i put it down here f at the last node x 6 so that's it and now we put the values to calculate the 6 and that is 1 half because each remember is equal to 1 then 1 half times and f0 is 0 here we have the nodes so x0 is 0 plus 2 times f at 1 plus f at 2 plus f at 3 plus f at 4 plus f at 5 plus f at 6 and now we use the table value we found 4 so at 6 is so at 0 is 1 plus 2 at 1 plus 2 at 1 we have the value 2 plus at 3 at 2 the image is 4 let's see right here plus at 3 the image is 2 at 4 the image is 1 .5 at 5 the image is 1 .5 it's very far a bit here correct and at 6 the value the image is 3 but that image is now multiplied by 1 it's only these images of the internal notes that are multiplied by 2 in the sum so 26 becomes 1 half times 1 plus 2 times now we have 4 times 2 6 plus 4 plus 2 is 8 plus 1 is 9 but plus 1 is 9 plus 1 .5 is 10 .5 plus 3 and that is one half times one plus let me stay keep with this per rackets here so it's one plus two times 10 10 .10 .5 is 21 plus 3 so this is 25 5 half that is 12 .5.
08:49
So this 6 is 12 .5.
08:54
That's the number you got to enter this one here in the text box reserve for the answer of part 1...