Answer to Question #140342 in Calculus for Promise Omiponle

Question

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 =

Answer 1

Average value of function:

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

$S=\int_{-6}^{6}\int_{0}^{1}dydx=\int_{-6}^{6}(y|_{0}^{1})dx=\int_{-6}^{6}(1-0)dx=$

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

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

$=\int_{-6}^{6}(2x^4\frac{y^6}{6}|_{y=0}^{y=1})dx=\int_{-6}^{6}\frac{1}{3}x^4dx=$

$=\frac{1}{15}x^5|_{-6}^{6}=\frac{1}{15}(6^5-(-6)^5)=\frac{2*6^5}{15}$

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