- #1

hisotaso

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## Homework Statement

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?

## Homework Equations

(ai) Tn = [tex]\Sigma[/tex][tex]^{8}_{i=1}[/tex][cos(x[tex]^{2}[/tex])[tex]\Delta[/tex]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 = [tex]\Sigma[/tex][tex]^{8}_{i=1}[/tex][cos(x[tex]^{2}[/tex])[tex]\Delta[/tex]x] [tex]\bar{xi}[/tex]=xi-1/2[tex]\Delta[/tex]x [tex]\Rightarrow[/tex] 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)[tex]\left|ET\right|[/tex]=[tex]\frac{K(b-a)^{2}}{12n^{2}}[/tex]

(bii)[tex]\left|EM\right|[/tex]=[tex]\frac{K(b-a)^{2}}{24n^{2}}[/tex]

My understanding is for ET and EM, K is the largest value of [tex]\left|f''(x)\right|[/tex], 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.

## The Attempt at a Solution

(ai)=0.902333

(aii)=0.905619