Double Integral Problem

**r2d2** · January 14th 2011, 05:43 AM

If ∫−1to3 f(x) dx = −3 and ∫−4to−2 g(x) dx = −2, what is the value of ∫∫D f(x)g(y) dA where D is the square: −1 ≤ x ≤ 3, −4 ≤ y ≤ −2?

I am confused on where to start for this problem.

BTW, -1to3, and -4to-2 are the integral bounds for their respected variables.

Thanks!

**Prove It** · January 14th 2011, 05:57 AM

Your double integral is written as

$\displaystyle \int_{-1}^3{\int_{-4}^{-2}{f(x)\,g(y)\,dy}\,dx}$

$\displaystyle = \int_{-1}^3{f(x)\,\int_{-4}^{-2}{g(y)\,dy}\,dx}$

$\displaystyle = \int_{-1}^3{-2f(x)\,dx}$

$\displaystyle = -2\int_{-1}^3{f(x)\,dx}$

$\displaystyle = -2(-3)$

$\displaystyle = 6$ .

**r2d2** · January 14th 2011, 06:03 AM

Ok great!

They key I know understand is breaking the problem up, recognizing the integral bound and going from there.

Thanks a lot!

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