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November 29, 2014

November 29, 2014

Posted by **COFFEE** on Wednesday, June 13, 2007 at 2:33am.

Use Simpson's Rule with n = 10 to estimate the arc length of the curve.

y = tan x, 0 <or= x <or= pi/4

.. this is what i did:

y' = sec(x)^2

(y')^2 = [sec(x)^2]^2

[f'(x)]^2 = sec(x)^4

Integral of

sqrt( 1 + sec(x)^4 ) dx

from x=0 to x=pi/4

deltaX = (pi/4 - 0) / 10

deltaX = (pi/4) / 10

deltaX = pi/40

= (pi/40)/3 [f(0) + 4f(pi/40) + 2f(2pi/40) + 4f(3pi/40) + 2f(4pi/40) + 4f(5pi/40) + ... + 4f(9pi/40) + (pi/4)]

= (pi/120) [sqrt(1+sec(0)^4) + 4sqrt(1+sec(pi/40)^4) + ..etc..etc..

and after calculating.. i got these:

= (pi/120) [1.414 + 5.674 + 2.864 + 5.822 + 2.981 + 6.161 + 2.837 + 6.802 + 3.652 + 7.991 + 0.785]

= (pi/120) [46.983]

= 1.230012064

= 1.23

........ and this answer was wrong. can someone point out how i calculated this wrong or if i missed something? i thought i did everything by the book & notes. thanks :)

never mind, I figured it out...1.277995

- calc: simpson's rule & arc length -
**dude**, Friday, March 19, 2010 at 12:16amdude me too i cant lol

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