Answer to Question #276446 in Calculus for Air

Question

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).

Answer 1

The area of the rectangle is,

$A(R)=2\cdot5=10$

The average value of function over the rectangle $R$ is evaluated as,

$f_{ave}=\frac{1}{A(R)}\iint_{R}f(x,y)dA =\frac{1}{10}\int_{-1}^{1}\int_{0}^{5}x^2ydydx=$

$=\frac{1}{20}\int_{-1}^{1}x^2y^2|^5_0dx=\frac{25}{20}x^3/3|^1_{-1}=\frac{2\cdot25}{3\cdot20}=\frac{5}{6}$