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Anonymous Mon 04 Jan 13:45:21 2021 No.12543868 View Reply Original Report

Quoted By: >>12543938 >>12543947

can anyone help with this problem i know trapezoidal rule but this function is confusing.

Anonymous

Anonymous Mon 04 Jan 2021 14:15:35 No.12543938 Report

Quoted By: >>12543976

>>12543868

This function has rotational symmetry around the point $(\frac{1}{2}, \frac{1}{2})$ .

As long as your composite trapezoidal cells are mirror-symmetrically spaced around the line $x = \frac{1}{2}$ , then the identity can be used to simplify the composite trapezoidal approximation.

Anonymous

Anonymous Mon 04 Jan 2021 14:20:52 No.12543947 Report

Quoted By: >>12543969

>>12543868
integral from 0 to 1 of g(x) is equal to integral from 0 to 1 of g(1-x) . This is true for any g.

We add these together and get integral from 0 to 1 of (g(x) + g(1-x)) . From your relation that gives integral from 0 to 1 of 1, which gives 1. Dividing by 2 we recover integral from 0 to 1 of g(x) as 1/2.

Anonymous

Anonymous Mon 04 Jan 2021 14:28:14 No.12543969 Report

Quoted By:

>>12543947
but i need to verify it using trapezoidal rule

Anonymous

Anonymous Mon 04 Jan 2021 14:31:18 No.12543976 Report

Quoted By:

>>12543938
explain to me like im a retard please.

Anonymous

Anonymous Mon 04 Jan 2021 15:02:22 No.12544026 Report

Quoted By:

this is not a homework help board

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