.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).
R=[a,b]×[c,d]=[−1,1]×[0,5]favg=1(b−a)(d−c)∫cd∫abf(x,y)dxdy=1(1−(−1))(5−0)∫05∫−11x2ydxdy=110∫0513[x3]−11ydy=13012[y2]05=160×25=512R=[a,b]×[c,d]=[-1,1]×[0,5]\\ f_{avg}=\frac{1}{(b-a)(d-c)}\int_c^d\int_a^b f(x,y)dxdy\\ =\frac{1}{(1-(-1))(5-0)}\int_0^5\int_{-1}^1x^2ydxdy\\ =\frac{1}{10}\int_0^5 \frac{1}{3}[x^3] _{-1}^1ydy\\ =\frac{1}{30}\frac{1}{2}[y^2]_0^5\\ =\frac{1}{60}×25\\ =\frac{5}{12}R=[a,b]×[c,d]=[−1,1]×[0,5]favg=(b−a)(d−c)1∫cd∫abf(x,y)dxdy=(1−(−1))(5−0)1∫05∫−11x2ydxdy=101∫0531[x3]−11ydy=30121[y2]05=601×25=125
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