Find the average value of 𝑓(𝑥, 𝑦) = 𝑥
2𝑦 over the region 𝑅 which is a rectangle with vertices
(−1, 0), (−1, 5), (1, 5), (1, 0).
The area of the rectangle is,
A(R)=2⋅5=10A(R)=2\cdot5=10A(R)=2⋅5=10
The average value of function over the rectangle RRR is evaluated as,
fave=1A(R)∬Rf(x,y)dA=110∫−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=fave=A(R)1∬Rf(x,y)dA=101∫−11∫05x2ydydx=
=120∫−11x2y2∣05dx=2520x3/3∣−11=2⋅253⋅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}=201∫−11x2y2∣05dx=2025x3/3∣−11=3⋅202⋅25=65
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