Approximate Integral of cos(x^2)

sugarxsweet
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Homework Statement


I am having issues figuring out the error involved in the trapezoidal and midpoint methods for ∫cos(x^2) from 0 to 1 with n=8


Homework Equations


|Et|<= k(b-a)^3/(12n^2)
|Em|<=k(b-a)^3/(24n^2)


The Attempt at a Solution



f(x)=cos(x^2)
f'(x)=-2xsin(x^2)
f''(x)=-4x^2cos(x^2)

f''(1) = -3.844

Is it right given that f''(1) is negative? Just wanted to make sure! Thanks
 
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sugarxsweet said:

Homework Statement


I am having issues figuring out the error involved in the trapezoidal and midpoint methods for ∫cos(x^2) from 0 to 1 with n=8


Homework Equations


|Et|<= k(b-a)^3/(12n^2)
|Em|<=k(b-a)^3/(24n^2)


The Attempt at a Solution



f(x)=cos(x^2)
f'(x)=-2xsin(x^2)
f''(x)=-4x^2cos(x^2)

f''(1) = -3.844

Is it right given that f''(1) is negative? Just wanted to make sure! Thanks

It doesn't matter: k = max{|f''(x)|: 0 ≤ x ≤ 1}.

RGV
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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