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Evaluate ∬D (x + y)e^(x^2 - y^2)dA where D is the rectangle enclosed by the lines x - y = 0, x - y = 2, x + y = 0, x + y = 3.

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Final answer:

To evaluate the given double integral, we can set up the integral in terms of x and y. We integrate with respect to x first, keeping y constant, then we integrate with respect to y. By evaluating the two integrals, the result of the double integral is 4.

Step-by-step explanation:

To evaluate the given double integral ∫∫D (x + y)e(x2 - y2)dA, where D is the rectangle enclosed by the lines x - y = 0, x - y = 2, x + y = 0, x + y = 3, we can set up the integral in terms of x and y.

We integrate with respect to x first, keeping y constant, then we integrate with respect to y, which will be our changing parameter.

The rectangular region D can be defined as 0 ≤ x ≤ 2 and 0 ≤ y ≤ x.

So the integral becomes:

∫∫D (x + y)e(x2 - y2)dA = ∫02 ∫0x (x + y)e(x2 - y2) dy dx

We first evaluate the inner integral with respect to y:

∫0x (x + y)e(x2 - y2) dy = [(x + y)e(x2 - y2)]

Substituting the limits of integration, we have [(x + x)ex2 - x2] = 2x.

Now we integrate with respect to x:

∫02 2x dx = [(x2)]

Substituting the upper and lower limits, we get [(22)] - [(02)] = 4.

So the evaluated double integral is 4.

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