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I have a double integral of $1dxdy$, where the area is given by $x=[-1,3]$ and $y=[-4,1]$

I can just evaluate the integral by inspection, but I was wondering how to calculate this more properly. I tried rewriting in polar coordinates, but I got stuck since the lengths of the sides don't give useful values of theta.

Does anyone know if it is possible to evaluate this integral by other means then inspection?

Thanks in advance!

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Polar coordinates is not really applicable to that type of region I think. That region is rectangular whereas polar coordinates are more for regions like $1 \leq x^2+y^2 < 4$.

Anyway, since the integrand is 1, the value of the double integral is the area of the region, which I'm sure you've already guessed.

$\int \int_{R} 1 dA = Area(R) = 4 x 5 = 20$

You can also actually evaluate the integral one at a time.

$\int \int_{R} 1 dxdy = \int_{-4}^{1} \int_{-1}^{3} 1 dx dy = \int_{-4}^{1} 4 dy = 200$

Or you can split it up. Forgot the justification. Independence or something.

$\int \int_{R} 1 dxdy = \int_{-4}^{1} 1 dy \int_{-1}^{3} 1 dx$

where R = [-1,3]x[-4,1].

That's all I got.

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