00:01
Hello students from the above given question first we have to write the trepezoidal rule okay so therefore the trepezoidal rule will be represoidal rule for n is equal to 4 for n is equal to 4 the rule will be t4 is equal to for sorry t4 is equal to i'll highlight it so t4 is equal to del x by 2 into f of x not plus 2 into f of x 1 plus 2 into f of x 2 plus 2 into f of x 3 plus f into x power x by x 4 okay so this is the rule for the when n is equal to 4 okay so it is given that the interval is it is given that the interval is 5 comma 11 so therefore by finding del we can use del x is equal to by using this interval so 11 minus 5 by 4 so which is 6 by 4 so what is the y have substitute 11 minus 5 by 4 so this is the x2 minus x 1 okay x2 minus x1 or you can take it as y minus x by n is the in given 4 okay so 6 by 4 will be equal to 3 by 2 so which is equal to 1 .5 so thus we can write it as thus we can write it as x not is equal to 5 so we will substitute one by one okay so therefore x1 is equal to 5 plus lambda x which is equal to 5 plus 1 .5 which is equal to 6 .5.
01:46
Next substituting x2 will be given as 6 .5 which is the x1 plus del x so which is equal 0 .5 plus 1 .5 which is equal to 8.
02:00
So, and for x3, we will be getting it as 8 plus del x.
02:04
So, which is equal to 8 plus 1 .5, which is equal to 9 .5, which is equal to 10 .5 plus del x, so which is equal to 9 .5 plus 1 .5, which is equal to 11.
02:19
Okay.
02:20
So, you are having that, you are being given that d, so you have been given that f of x is equal to 6x.
02:32
Okay, so therefore, f of x not is equal to f of 5, right? so which is equal to 6 by, we are substituting x in the place of 5.
02:43
So 6 by 5 which is equal to 1 .2.
02:46
Then by substituting f of x1, we'll be getting it as f of 6 .5, which is equal to 6 by 6 .5, which is 0 .923.
02:56
8.
02:58
Okay, then substituting the third term, f of x2 is equal to f of 8, which is equal to 6 by 8, which is 0 .75.
03:07
Then substituting x, x3, which is f of 9 .5, which is equal to 6 by 9 .5, you'll be getting it as 0 .631579.
03:20
Okay, then substituting the fourth one, x power 4, x4, which is f of 11.
03:27
6 by 11.
03:28
So we get it as 0 .545454.
03:35
So these are the values we have substituted.
03:38
Okay.
03:39
So now we are going to write the trepezoidal approximation.
03:47
So for tripisoidal approximation where n is equal to 4.
03:53
So for this you'll be writing integral 5 to 11, 6x by x into dx, the given f of x, which is equal, equal to del x by 2 into f of x not plus 2 into f of x 1 plus 2 f into x2 plus 2 f into x2 plus 2 f into x 3 plus f into x of 4.
04:20
So now we are going to substitute the values in the above equation so which is 1 .5 by 2 into 1 .2 plus 2 into 0 .92 plus 2 into 0 .75 plus 0 .63175 into 2 plus 0 .549 into 2 plus 0 .545.
04:47
Okay.
04:49
So i'll be getting it as 1 .5 by 2 will be 0 .75.
04:53
So into by multiplying these with 2 we'll be getting it as 1 .84616 plus 1 .5 plus 1 .25 plus 1 .26315.
05:06
Plus 0 .545545 plus 1 .2.
05:13
Okay, so by adding all these values, we'll be getting it as 0 .75 into 6 .3, sorry, 6 .354 -725.
05:24
So, we're multiplying it will be 4 .7608...