# Working with Integrals

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Can anyone show me the working between the integral in the enclosed file & the answer of A = 4/3

First let's sketch the graph for 0≤t≤2п:

Ok, so one loop is the half of this, i.e. 0≤t≤п:

Now we have:

where x=f(t) and y=g(t). Then we have:

or:

A=4/3

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