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
<h3>What is the average value of a function?</h3>
The average value of a function f(x) over an interval [a,b] is given by

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, ![\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}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7Bb%20-%20a%7D%20%5Cint%5Climits%5Eb_a%20%7Bf%28x%2Cy%29%7D%20%5C%2C%20dA%20%3D%20%5Cfrac%7B1%7D%7B1%20-%200%7D%20%5Cint%5Climits%5E1_0%20%5Cint%5Climits%5E9_0%20%7B3xy%7D%20%5C%2C%20dxdy%20%5C%5C%3D%20%5Cfrac%7B3%7D%7B1%7D%20%5Cint%5Climits%5E1_0%20%7Bx%7D%20%5C%2Cdx%5Cint%5Climits%5E9_0%20%7By%7D%20%5C%2Cdy%5C%5C%20%3D%20%20%5Cfrac%7B3%7D%7B1%7D%20%5B%5Cfrac%7Bx%5E%7B2%7D%20%7D%7B2%7D%20%5D%5E%7B1%7D_%7B0%7D%5B%5Cfrac%7By%5E%7B2%7D%20%7D%7B2%7D%20%5D%5E%7B9%7D_%7B0%7D%20%20%5C%5C%3D%203%5B%5Cfrac%7B1%5E%7B2%7D%20%7D%7B2%7D%20-%20%5Cfrac%7B0%5E%7B2%7D%7D%7B2%7D%20%5D%20%5B%5Cfrac%7B9%5E%7B2%7D%20%7D%7B2%7D%20-%20%5Cfrac%7B0%5E%7B2%7D%7D%7B2%7D%20%5D%20%5C%5C%3D%203%5B%5Cfrac%7B1%7D%7B2%7D%20-%200%20%5D%5B%5Cfrac%7B81%7D%7B2%7D%20-%200%20%5D%5C%5C%3D%20%20%5Cfrac%7B81%7D%7B2%7D%20X3%20X%20%5Cfrac%7B1%7D%7B2%7D%20%5C%5C%3D%20%20%5Cfrac%7B243%7D%7B4%7D)
So, the average value of f over the region D is 243/4
Learn more about average value of a function here:
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