Approximating Integrals

[hisotaso]

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 = $$\Sigma$$$$^{8}_{i=1}$$[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.

The Attempt at a Solution

(ai)=0.902333
(aii)=0.905619

 

[lippyka]

For (bi) and (bii) I'm stuck here. (ci) Get the ET and EM smaller than 0.00001.A:For part b:The error of trapezoid rule is $$ET = \frac{K (b-a)^2}{12n^2},$$where $K$ is the maximum value of the second derivative of $f$. In this case, $f(x) = \cos(x^2)$. Then $f''(x) = -2 x \sin (x^2)$. So, we have to find the maximum value of $f''(x)$. Since $f''(x)$ is an oscillatory function and the interval of integration is $[0,1]$, we can conclude that $K = 2$. Also, $b-a = 1$. Now, plugging these values in the equation, we get $$ET = \frac{2 (1)^2}{12n^2} = \frac{1}{6n^2}.$$Hence, the error of the Trapezoid Rule is $$ET = \frac{1}{6n^2}.$$Similarly, for the Midpoint Rule, we have$$EM = \frac{K (b-a)^2}{24n^2} = \frac{1}{12n^2}.$$Hence, the error of the Midpoint Rule is $$EM = \frac{1}{12n^2}.$$For part c:Let us assume that we want the error to be less than $0.00001$. Then, we have to make sure that both $ET$ and $EM$ are less than $0.00001$. The errors of Trapezoid Rule and Midpoint Rule are$$ET = \frac{1}{6n^2}\quad\text{and}\quad EM = \frac{1}{12n^2}.$$From these two equations, we can say that $ET = 2EM$. Hence, for $ET < 0.00001$, we need $2EM < 0.00001$ or $EM < 0.000005$. So, from the equation for $EM$, we get$$\frac{1}{