Trapizoid and Simpson's Rule

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lamaclass

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I'm not sure how to go about solving this problem:

Given the integral e^cos x dx, limits: 2TT and 0, determine the number of subintervals, n, needed to approximate the integral to within 0.0001 using the Trapezoid Rule and Simpson's Rule.

I believe I heard my teacher mention something about needing to find the 4th derivative of the equation which I figured out and I do know the formulas for both rules. However I am not sure how to apply them for this problem. I found the 4th derivative to be:

3 e^cos x cos^2 x + e^cos x cos x + sin^4 x e^cos x - 4 sin^2 x e^cos x - 6 sin^2 x cos x

Any help is greatly appreciated!
 
\(\displaystyle \int_{0}^{2}e^{cos(x)} \ dx \ = \ 3.45435489652, \ (TI-89)\)

Trapezoid:

\(\displaystyle Let \ f(x) \ = \ e^{cos(x)}, \ f^{2}(x) \ = \ [sin^{2}(x)-cos(x)]e^{cos(x)}, \ (TI-89)\)

\(\displaystyle |max| \ of \ f^{2}(x) \ = \ f^{2}(0) \ = \ |-1(e^{1})| \ = \ e\)

\(\displaystyle E \ \le \ \frac{(b-a)^{3}}{12n^{2}}[max|f^{2}(x)|]\)

\(\displaystyle Hence, \ E \ \le \ \frac{8e}{12n^{2}} \ = \ \frac{2e}{3n^{2}}\)

\(\displaystyle Ergo, \ \frac{2e}{3n^{2}} \ \le \ \frac{1}{10,000} \ \implies \ \frac{3n^{2}}{2e} \ \ge \ 10,000\)

\(\displaystyle n \ \ge \ \sqrt\frac{20,000e}{3} \ \ge \ 134.817 \ = \ 135, \ so \ when \ n \ = \ 135, \ we \ will \ be \ within \ .0001 \ units\)

\(\displaystyle I'll \ let \ you \ do \ Simpson, \ use \ f^{4}(x) \ = \ [cos^{4}(x)+2cos^{3}(x)-cos(x)[4sin^{2}(x)+1]-5sin^{2}(x)+2]e^{cos(x)}\)

\(\displaystyle Have \ fun. \ Note: \ With \ the \ advent \ of \ calculators, \ Simpson \ and \ the \ others \ are \ obsolete.\)
 
\(\displaystyle Afterthought: \ If \ you \ do \ Simpson's, \ I \ think \ n \ = \ 12 \ units, \ shows \ that \ Simpson's \ is \ by \ far \ the\)

\(\displaystyle better \ in \ a \ close \ calculation, \ the \ problem \ being \ the \ f^{4}(x) \ derivative, \ which \ can \ get \ quite \ "hairy".\)

\(\displaystyle However, \ as \ I \ stated \ above, \ other \ than \ grunt \ work \ given \ by \ your \ Professor/a, \ the \ modern\)

\(\displaystyle \ calculator, \ for \ all \ praticable \ purposes, \ makes \ these \ kind \ of \ problems \ obsolete.\)
 
Thanks a ton for your help Glenn! :D I have to agree with you on Simpson's Rule being more accurate and that the TI-89 makes these formulas unecessary. Just to clarify when applying Simpson's Rule, I would just use the Error formula for this problem too right?
 
\(\displaystyle Simpson's \ Error\)

\(\displaystyle E \ \le \ \frac{(b-a)^{5}}{180n^{4}}[max|f^{4}(x)|], \ a \ \le \ x \ \le \ b\)

\(\displaystyle Hence, \ E \ \le \ \frac{32(4e)}{180n^{4}} \ = \ \frac{32e}{45n^{4}}\)

\(\displaystyle \frac{32e}{45n^{4}} \ < \ \frac{1}{10,000} \ \implies \ \frac{45n^{4}}{32e} \ > \ 10,000\)

\(\displaystyle n \ > \ \bigg[\frac{320,000e}{45}\bigg]^{1/4} \ > \ 11.79 \ = \ 12\)

\(\displaystyle I \ trust \ now, \ that \ you \ get \ it.\)

\(\displaystyle Post Script: \ Don't \ ask \ me \ to \ prove \ these \ two \ theorems, \ I'll \ leave \ that \ to \ the \ purists.\)
 
BigGlenntheHeavy said:
for all praticable purposes, makes these kind of problems obsolete
Click to expand...

What about the "practicable" purpose of torturing calculus students? :)
 
mmm444bot, practicable, was a bad choice of words, as Simpson's rule is still applied when the function is poorly defined, for example, to find the area of a lake by taking equally space distances along its width, knowing the length of the lake and then applying \(\displaystyle A \ = \ \int_{a}^{b}w(x)dx\).
 
BigGlenntheHeavy said:
Simpson's rule is still applied when the function is poorly defined
Click to expand...

Hi Glenn:

I didn't read your statement to mean that Simpson's Method itself is obsolete, but rather that its application to the type of problem posted is not really done in the real world, anymore.

My claim is that the application of Simpson's Rule to these types of problems still finds use as a device of student torture. I mean, we must find our entertainment somewhere, right?

(heh, heh, heh, heh, heh)

An unrelated comment: I once found the area of a lake's surface by weighing a piece of aluminum foil.
 
OK mmm444bot, I'll bite, how did you find the area of a lake's surface by weighing a piece of aluminum foil? If this is a joke, I don't get it.
 
It's no joke, and I actually needed to weight two pieces of foil.

Started with a satellite image of the lake (with known scale).

Used image to cut matching shape out of foil.

Carefully cut one square inch of foil.

Weighed both to the nearest 10,000th gram in the microbiology lab next to my calculus classroom.

Used weights to determine area of foil lake surface, in square inches.

Used known-scale to convert from square inches to square miles.

Classmates spent 45 minutes carefully drawing four-dozen strips and measuring their dimensions, followed by summing. That method seemed too tedious.

My instructor later told me that my result was nearly 4% more accurate than the others' mean result, but I got extra credit for neither increased precision nor my ingenuity.

Will you give me something extra? :wink:
 
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