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 =
1
Expert's answer
2020-11-09T19:59:45-0500

Average value of function:


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


S=6601dydx=66(y01)dx=66(10)dx=S=\int_{-6}^{6}\int_{0}^{1}dydx=\int_{-6}^{6}(y|_{0}^{1})dx=\int_{-6}^{6}(1-0)dx=


=x66=6(6)=12=x|_{-6}^{6}=6-(-6)=12


Rf(x,y)dS=66012x4y5dydx=\iint_{R}f(x,y)dS=\int_{-6}^{6}\int_{0}^{1}2x^4y^5dydx=


=66(2x4y66y=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=


=115x566=115(65(6)5)=26515=\frac{1}{15}x^5|_{-6}^{6}=\frac{1}{15}(6^5-(-6)^5)=\frac{2*6^5}{15}


f=2651512=2635=86.4\overline{f}=\frac{2*6^5}{15*12}=\frac{2*6^3}{5}=86.4



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