Respuesta :

LammettHash
[tex]\dfrac{\displaystyle\iint_Rf(x,y)\,\mathrm dx\,\mathrm dy}{\displaystyle\iint_R\mathrm dx\,\mathrm dy}[/tex]
[tex]\dfrac{\displaystyle\int_{y=0}^{y=3}\int_{x=-5}^{x=5}3x^2y\,\mathrm dx\,\mathrm dy}{\displaystyle\int_{y=0}^{y=3}\int_{x=-5}^{x=5}\mathrm dx\,\mathrm dy}[/tex]
[tex]\displaystyle\frac1{10}\left(\int_{y=0}^{y=3}y\,\mathrm dy\right)\left(\int_{x=-5}^{x=5}x^2\,\mathrm dx\right)[/tex]
[tex]\dfrac{75}2[/tex]
facundo3141592

We want to find the average value of f(x, y) on the given region.

The average value of f(x, y) in the given rectangle is 25.

So we want to find the average value of f(x, y) = 3*x^2*y on the rectangle with vertices (−5, 0), (−5, 3), (5, 3), (5, 0).

This will be given by the quotient between the integral of the function in that region and the area of the region, so we have:

[tex]\frac{\int\limits^5_{-5} \int\limits^3_0 {2*x^2*y} \, dxdy }{\int\limits^5_{-5} \int\limits^3_0 {} \, dxdy} \\\\\\\frac{\int\limits^5_{-5} \int\limits^3_0 {2*x^2*y} \, dxdy }{3*(5 - (-5))}\\\\\frac{\int\limits^5_{-5} \int\limits^3_0 {2*x^2*y} \, dxdy }{30}[/tex]

Where we just solved the integral in the denominator, which just gives the area of the rectangle.

Now we need to solve the integral in the numerator, we can do it one variable at the time.

[tex]\frac{\int\limits^5_{-5} {2*x^2}dx\int\limits^3_0 {y} \, dy }{30}\\\\\frac{[(2/3)*(5^3 - (-5)^3)]*[(1/2)*3^2]}{30} = 25[/tex]

This means that the average value of f(x, y) in the given rectangle is 25.

If you want to learn more, you can read:

https://brainly.com/question/21846827

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