# HI there I need help in my calculs question a) find the approximate T8 and M8 for the integral from upper bound(1) to zero ( lower) cos (x^2) dxb) estimate the errors in the approximation for part...

HI there I need help in my calculs question

a) find the approximate T8 and M8 for the integral from upper bound(1) to zero ( lower) cos (x^2) dx

b) estimate the errors in the approximation for part a

C) how large do we have to choose n so that the appriximations Tn and Mn to the integral in part (A) are accurate to within 0.0001?

thanks in advance!

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### 1 Answer

Following the rules for the Midpoint Rule and Trapezoidal Rule, which involve breaking the area underneath the curve into even-width rectangles or trapezoids. The rectangles in the midpoint rule rely on a height determined by the value of the function at the x-value between the edges of the rectangle.

Apologies for the brevity, but I have had this page go buggy on me three times already. If you want me to add HOW to get M_8 and T_8, then reply back. Refer to the book below to get the formula.

`∆x = (b-a)/n = 0.125`

`barx_1 = 0.0675, barx_2 = 0.1875 ... barx_8 = 0.9375`

`M_8 =0.905619957`

`T_8 =0.902332843`

b) To calculate the estimated error, we use the second derivative of the function and the approximations:

`|E_M|=(K(b-a)^3)/(24n^2)` and `|E_T|=(K(b-a)^3)/(12n^2)`

where K is the largest magnitude of the second deriviative between a and b.

`d^2/(dx^2)cos(x^2) =-2 (2 x^2 cos(x^2)+sin(x^2))`

a glance at the resulting graph shows a local minimum at x=1, so

`Klt=|f''(1)|=3.8441511932`

Since (b-a)=1, the error estimates are easier to calculate:

`E_M lt= (3.84...)/(24*8^2) =0.0025027026`

`E_T lt=(3.84...)/(12*8^2) =0.005005405199`

c) Now that we have K and desired error, we can leave n as our unknown, solve for it, and round up to the nearest integer.

`E_M lt= 0.0001`

so, `E_M lt= (3.84...)/(24*n^2) lt=0.0001`

thus `n = sqrt((3.84...)/(24*0.0001))=40.02161496`

so n=41 will insure that we have an error less than 0.0001 for the midpoint rule.

`E_T lt= 0.0001`

so, `E_T lt=(3.84...)/(12*n^2) lt=0.0001`

thus `n=sqrt((3.84...)/(12*0.0001))=56.59911066`

so n=57 will insure that we have an error less than 0.0001 for the trapezoidal rule.

We can see that the midpoint rule is better at approximating the integral in these cases. Also, judging from how Error is calculated here:

`n_(E_T) = sqrt(2)*n_(E_M) `

and

`E_T = 2*E_M`

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