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Wencester
Can somebody on this page tell about trapezoidal rule in calculus?
The Trapezoidal rule is a way of approximating area under a curve. somewhat similar to what you may have done using right sum and left sum that utilized rectangles. in this case we use trapezoids that better refine the partition ( are more accurate) in estimating the area under a curve. later you will see this done with integrals but for now this is your bet bet. \[Deltax \times .5 (y0+2y1+2y2+2y3...+yn)\]
..tnx for that particular information ..... in what i studied in our class trapezoidal rule is a technique in approximating indefinite integrals.. we also tackled about that particular formula... can u give a particular example for that?
The further you zoom in on a curve, the straighter it gets. When you use a greater and greater number of trapezoids and then zoom in on where the trapezoids touch the curve, the tops of the trapezoids get closer and closer to the curve. If you zoom in “infinitely,” the tops of the “infinitely many” trapezoids become the curve and, thus, the sum of their areas gives you the exact area under the curve. This is a good way to think about why integration produces the exact area. integrals do a similar thing by evaluating the integral of a function you are getting the area under the curve essentially you can think of them doing the same thing.
Use the trapezoidal rule with n = 8 to estimate \[\int\limits_{1}^{5}\sqrt{1+x^{2}} dx\] so \[Deltax= (5-1)/(8)\] find values x--1--1.5--2--2.5--3--3.5--4--4.5--5 y--\[\sqrt{2}--2\sqrt{3.25}--\sqrt{5}--\sqrt{7.25}--\sqrt{10}--\sqrt{13.25}--\sqrt{17}--\sqrt{21.25}--\sqrt{26}\] because i could not make a chart i hope this works the "--" is just spacing between terms for X1--X2--X3 ect same for y now simple plug into formula i provided before and you will get 12.76 which evaluates the integral aswell as provide area under curve
ahh yeah we got the same answer i also try it in my notebook while waiting for you but do you know that there is still an error in using trapezoidal rule that is called the truncation error?
Yes the rule is not perfect it is still an appoximation yet it provides a rather close answer
Have u done Simpsons rule yet if i am correct that is more accurate
yup... simpsons rule is more accurate that trapezoidal rule... because Simpson's rule subdivides the area into slices that have little curves , so Simpson's rule is more accurate for functions that have a lot of curvy sections. or what i mean is the interval of that..
thank you very much areyes06 for that particular information about the trapezoidal rule... sorry for the last time that i didn't reply that fast because i really can't verify my email address... thank you once more...
not a problem what so ever and seems you have a grasp on the simpsons rule so u are good to go. good for you i hope i helped and lmk if i can help you out with anything else in the future
yeah thanks.. i really appreciate all your replies... but i want to know simpsons rule further..
Simpson’s rule is a very accurate approximation method. In fact, it gives the exact area for any polynomial function of degree three or less. In general, Simpson’s rule gives a much better estimate than either the midpoint rule or the trapezoid rule.With Simpson’s rule, you approximate the area under a curve with curvy-topped “trapezoids.” The tops of these shapes are sections of parabolas. Parabola are much more accurate due to the fact that they hug the curve of a given function much better then the sharp ends of a trapezoid. simpsons Rule A=1/3 DetlaX (f(x1)+4f(x2)+2f(x3)+.....4f(xn-1)+f(xn)) Procedure is much the same as the trap rule but formula is slightly different
so if i use Simpson’s rule with n = 6 to estimate \[\int\limits_{1}^{4}\sqrt{1+x ^{3}}dx\] then we will get delta x which is \[\frac{ 4-1 }{ 6}\] = .5 ... this will be my 1st step right?
Correct that will give you Delta X Next....?
next ahm... i will make a table... x 1 1.5 2 2.5 3 3.5 4 y=\[\sqrt{1+x ^{3}}\] 1.414 2.092 3 4.077 2.3 6.624 8.062 am i right or there is another process in doing a tble?
Looks good to me and no thats usually the easiest way if you have a graphing calc you may use that by adding the function into Y1 and utilizing the table but thats good whats next?
yeah but i dont have a graphing calculator... ahm \[\int\limits_{1}^{4}\sqrt{1+x ^{3}}\] = .5/3 (1.414 + 4(2.092) +2 (3) +4(4.077)+ 2(2.3) + 4(6.624) +8.062 then the answer will be 12.871
i got 11.87 try the math again to the parens first i could be wrong but i think its 11.87
ay yeah i also got 11.87... sorry can u try to solve it using the trapezoidal rule? so that we can compare our answrs?
are you using a graphing calculator or u are solving it manually?
i have a graphing calc but i am just using it for arithmetic not w/ functions or programs
i also want to have a graphing calculator haha.... ahm so if you use the calculator to solve for this equation what will be the anwr?
so... i think the trpezoidal rule gives much appropriate answer? but then it is ststed that simpsons is the best that we can use
the answer i got was through a trapazoidal program. so naturally it will be similar to the hand work simpsons program gives the 12.871 you originally got. so that is right i am sorry but not matter what when you actually do the integral through the fundamental theorem of calc you will be closer wih the simpsons rule
ahh yeah.... maybe i have typographical error on what i wrote on the solution
its ok no big deal as long as you get the process?
oh i'm very much impress with your answers by the way are you a professor in mathematics?
lol no im not but thank you i am an engineering student you tend to be forced to view calc very in depth in this major
ah so your from what school?
i really appreciate all your answer... hope that i can also help you the next way around... =)
not a problem anytime glad i could help feel free to message me with anyother problem in the future i will always do my best to help out
@Wencester you should close this thread to open a new thread.