Question #276446

Find the average value of 𝑓(π‘₯, 𝑦) = π‘₯


2𝑦 over the region 𝑅 which is a rectangle with vertices


(βˆ’1, 0), (βˆ’1, 5), (1, 5), (1, 0).

Expert's answer

The area of the rectangle is,


A(R)=2β‹…5=10A(R)=2\cdot5=10


The average value of function over the rectangle RR is evaluated as,


fave=1A(R)∬Rf(x,y)dA=110βˆ«βˆ’11∫05x2ydydx=f_{ave}=\frac{1}{A(R)}\iint_{R}f(x,y)dA =\frac{1}{10}\int_{-1}^{1}\int_{0}^{5}x^2ydydx=


=120βˆ«βˆ’11x2y2∣05dx=2520x3/3βˆ£βˆ’11=2β‹…253β‹…20=56=\frac{1}{20}\int_{-1}^{1}x^2y^2|^5_0dx=\frac{25}{20}x^3/3|^1_{-1}=\frac{2\cdot25}{3\cdot20}=\frac{5}{6}




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