Due to a symmetry of the cosine we can just double the integral from 0 to 1

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SUMMARY

The integral of the absolute value of the cosine function, specifically ∫|cos(πx/2)|dx over the interval [0,2], can be simplified using symmetry properties. The integral can be expressed as I = 2∫|cos(πx/2)|dx from 0 to 1 due to the function's symmetry about the midpoint of the interval. This property extends to any interval of length 2, allowing for the integral to be calculated over any equivalent interval, such as [3,4] or [7,9], without changing the result.

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tmt1
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Hello,

∫|cos(px/2)|dx between [0,2]

I encountered this rule. How does this apply to other intervals of say [3,4],[7,9] etc.

Are the numbers both halved?

so [3,4] becomes [1.5,2] etc?

Also, does this rule apply to all symmetrical functions?

Thank you,

Tim
 
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tmt said:
Hello,

∫|cos(px/2)|dx between [0,2]

I encountered this rule. How does this apply to other intervals of say [3,4],[7,9] etc.

Are the numbers both halved?

so [3,4] becomes [1.5,2] etc?

Also, does this rule apply to all symmetrical functions?

Thank you,

Tim

Assuming you mean:$$ I=\int_0^2 |\cos(\pi x/2)| \;dx = 2\int_0^1 |\cos(\pi x/2)| \;dx $$.

You can do this because the integrand is symmetric about the mid-point of the interval $$(0,2)$$. But also the integrand is periodic with period $$2$$, which means that its integral over any interval of a integer number of periods is equal to its integral over any other interval of the same length. That is, for any real $$a$$ :

$$\phantom{xxxxxxx} \int_0^2 |\cos(\pi x/2)| \;dx=\int_a^{a+2} |\cos(\pi x/2)| \;dx$$

.
 

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