Approximating Integrals

1. Feb 15, 2009

hisotaso

1. The problem statement, all variables and given/known data
Integral: cos(x^2) from 0 to 1
(a) Approximate the integral using Mn (Midpoint Rule) and Tn (Trapezoid Rule) with n=8
(b) Estimate the errors involved in the approximations in part (a).
(c) How large do we choose n so that the approximations are accurate to within 0.00001?

2. Relevant equations
(ai) Tn = $$\Sigma$$$$^{8}_{i=1}$$[cos(x$$^{2}$$)$$\Delta$$x/2] = (.0625)[f(0)+2f(1/8)+2f(2/8)+2f(3/8)+2f(4/8)+2f(5/8)+2f(6/8)+2f(7/8)+f(1)]
(aii) Mn = $$\Sigma$$$$^{8}_{i=1}$$[cos(x$$^{2}$$)$$\Delta$$x] $$\bar{xi}$$=xi-1/2$$\Delta$$x $$\Rightarrow$$ Mn = (.125)[f(1/16)+f(3/16)+f(5/16)+f(7/16)+f(9/16)+f(11/16)+f(13/16)+f(15/16)]

(bi)$$\left|ET\right|$$=$$\frac{K(b-a)^{2}}{12n^{2}}$$
(bii)$$\left|EM\right|$$=$$\frac{K(b-a)^{2}}{24n^{2}}$$
My understanding is for ET and EM, K is the largest value of $$\left|f''(x)\right|$$, which i can find by graph or by finding and analyzing the zeroes of f'''(x). The reason I am posting here is another site has the solution to this problem, but with K=6, which doesn't match with my results. I am stopped on part b.
3. The attempt at a solution
(ai)=0.902333
(aii)=0.905619