To find theaveragevalue of the function \( f(x, y) = 2x^2y \) over the given rectangle \( R \) withvertices(-4,0), (-4,2), (4,2), and (4,0), we calculate thedouble integralof the function over the rectangle and divide it by the area of therectangle.The average value of a function over a region is calculated by taking the integral of the function over the region and dividing it by the area of the region. IN this case, we have the function \( f(x, y) = 2x^2y \) and the rectangle \( R \) withvertices(-4,0), (-4,2), (4,2), and (4,0).To find the average value of \( f \) over \( R \), we calculate the double integral of \( f \) over \( R \):\(\text{avg} = \frac{1}{\text{Area}(R)} \iint_R f(x, y) \, dx \, dy\).The area of the rectangle \( R \) is calculated as the product of the length and width:\(\text{Area}(R) = (\text{length})(\text{width}) = (4 - (-4))(2 - 0) = 8 \times 2 = 16\).Now, we evaluate the double integral of \( f \) over \( R \):\(\iint_R f(x, y) \, dx \, dy = \int_{-4}^{4} \int_{0}^{2} 2x^2y \, dy \, dx\).We first integrate with respect to \( y \):\(\int_{0}^{2} 2x^2y \, dy = [x^2y^2]_{0}^{2} = 4x^2\).Now weintegratewith respect to \( x \):\(\int_{-4}^{4} 4x^2 \, dx = [\frac{4}{3}x^3]_{-4}^{4} = \frac{4}{3}(4^3 - (-4)^3) = \frac{4}{3}(64 - (-64)) = \frac{4}{3} \times 128 = \frac{512}{3}\).Finally, we divide the double integral by the area of \( R \) to obtain the average value:\(\text{avg} = \frac{1}{16} \times \frac{512}{3} = \frac{32}{3} \approx 10.667\).Therefore, the average value of the function \( f(x, y) = 2x^2y \) over the rectangle \( R \) isapproximately10.667.Learn more aboutintegratehere:-brainly.com/question/33073166#SPJ11...

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