Answer
$4$
Work Step by Step
Our aim is to integrate the integral as follows:
$ Area =\iint_D dA $
$\int^{2}_{0} \int^{4}_{2x}\space dy \space dx =\int^{2}_{0}(4-2x) dx $
or, $=[4x-x^2]^2_0$
or, $=(4)(2)-(2)^2$
or, $=4$
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