Question #140342

Find the average value of f(x,y)=2 x^4 y^5 over the rectangle R with vertices (−6,0),(−6,1),(6,0),(6,1).
Average value =

Expert's answer

Average value of function:


f‾=1S∬Rf(x,y)dS,here:S=∬RdS\overline{f}=\frac{1}{S}\iint_{R}f(x,y)dS, here: S=\iint_{R}dS


S=∫−66∫01dydx=∫−66(y∣01)dx=∫−66(1−0)dx=S=\int_{-6}^{6}\int_{0}^{1}dydx=\int_{-6}^{6}(y|_{0}^{1})dx=\int_{-6}^{6}(1-0)dx=


=x∣−66=6−(−6)=12=x|_{-6}^{6}=6-(-6)=12


∬Rf(x,y)dS=∫−66∫012x4y5dydx=\iint_{R}f(x,y)dS=\int_{-6}^{6}\int_{0}^{1}2x^4y^5dydx=


=∫−66(2x4y66∣y=0y=1)dx=∫−6613x4dx==\int_{-6}^{6}(2x^4\frac{y^6}{6}|_{y=0}^{y=1})dx=\int_{-6}^{6}\frac{1}{3}x^4dx=


=115x5∣−66=115(65−(−6)5)=2∗6515=\frac{1}{15}x^5|_{-6}^{6}=\frac{1}{15}(6^5-(-6)^5)=\frac{2*6^5}{15}


f‾=2∗6515∗12=2∗635=86.4\overline{f}=\frac{2*6^5}{15*12}=\frac{2*6^3}{5}=86.4



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