# How Accurate Are Midpoint and Trapezoid Rule Approximations for cos(x^2)?

• hisotaso
In summary: Solving for $n$, we get$$n > \left(\frac{1}{0.000005\times 12}\right)^{1/2} \approx 58.14.$$Therefore, to approximate the integral to within $0.00001$, we need to choose $n > 58$.In summary, we need to use the Midpoint Rule and Trapezoid Rule with at least 58 subintervals to approximate the integral of cos(x^2) from 0 to 1 to within 0.00001. The maximum error for both rules can be found by using the largest value of the second derivative of cos(x^2), which
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

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}{

## What is an integral approximation?

An integral approximation is a method used to estimate the value of a definite integral. It involves dividing the integral into smaller parts and using various techniques to approximate the area under the curve.

## Why is it important to approximate integrals?

Integrals are used in various areas of science and engineering to calculate quantities such as area, volume, and average values. However, in many cases, it is not possible to find an exact solution for an integral. Approximating integrals allows us to still obtain a close estimate of the value without having to solve the integral exactly.

## What are the different methods for approximating integrals?

There are several methods for approximating integrals, including the Midpoint Rule, Trapezoidal Rule, and Simpson's Rule. These methods use different techniques to estimate the area under the curve, such as using rectangles, trapezoids, or parabolas.

## How accurate are integral approximations?

The accuracy of an integral approximation depends on the method used and the number of intervals used to divide the integral. Generally, the more intervals used, the more accurate the approximation will be. However, some methods, such as Simpson's Rule, are generally more accurate than others.

## What are the applications of integral approximation?

Integral approximation is used in a wide range of fields, including physics, engineering, economics, and statistics. It is particularly useful in cases where an exact solution is not possible or practical to obtain, such as in real-world problems with complex functions.

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