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Find the average value of f over the region D. f(x, y) = 3xy, D is the triangle with vertices (0, 0), (1, 0), and (1, 9)

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oladayoademosu

The average value of f over the region D is 243/4

To answer the question, we need to know what the average value of a function is

What is the average value of a function?

The average value of a function f(x) over an interval [a,b] is given by

[tex]\frac{1}{b - a} \int\limits^b_a {f(x)} \, dx[/tex]

Now, given that we require the average value of f(x,y) = 3xy over the region D where D is the triangle with vertices (0, 0), (1, 0), and (1, 9).

x is intergrated from x = 0 to 1 and the interval is [0,1] and y is integrated from y = 0 to y = 9

So, [tex]\frac{1}{b - a} \int\limits^b_a {f(x,y)} \, dA = \frac{1}{1 - 0} \int\limits^1_0 \int\limits^9_0 {3xy} \, dxdy \\= \frac{3}{1} \int\limits^1_0 {x} \,dx\int\limits^9_0 {y} \,dy\\ = \frac{3}{1} [\frac{x^{2} }{2} ]^{1}_{0}[\frac{y^{2} }{2} ]^{9}_{0} \\= 3[\frac{1^{2} }{2} - \frac{0^{2}}{2} ] [\frac{9^{2} }{2} - \frac{0^{2}}{2} ] \\= 3[\frac{1}{2} - 0 ][\frac{81}{2} - 0 ]\\= \frac{81}{2} X3 X \frac{1}{2} \\= \frac{243}{4}[/tex]

So, the average value of f over the region D is 243/4

Learn more about average value of a function here:

https://brainly.com/question/15870615

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