Question

Find the average value of the function h(x, y) = x3y2 on the region bounded by f(x) = 2x 1, g(x) = −x2 − 2, x = 0, and x = 1

Answer 1

The average value of the function $h(x,y) = x^{3}\cdot y^{2}$ is approximately 3.1496.

How to find the average of a two-variable function

The average is defined by the following double integral equation:

$\bar h = \frac{\int\limits^1_0 {\int\limits^{f(x)}_{g(x)} {x^{3}\cdot y^{2}} \, dy } \, dx }{(1-0)\cdot [y(1)-y(0)]}$ (1)

If we know that $g(x) = -x^{2}-2$ and $f(x) = 2\cdot x - 1$, then the double average is:

$h = \frac{\frac{1}{3} \int\limits^{1}_{0} {x^{3}\cdot [f(x)-g(x)]^{3}} \, dx }{(1-0)\cdot [f(1)-f(0)-g(1)+g(0)]}$

$h = \frac{\frac{1}{3}\int\limits^{1}_{0} {x^{3}\cdot [2\cdot x-1+x^{2}+2]^{3}} \, dx }{1-1+3-2}$

$h = \frac{1}{3}\int\limits^{1}_{0} {(2\cdot x^{2}-x+x^{3}+2\cdot x)^{3}} \, dx }$

The average value of the function $h(x,y) = x^{3}\cdot y^{2}$ is approximately 3.1496. $\blacksquare$

Remarks

The statement presents mistakes and is poorly formatted:

Find the average value of the function $h(x,y) = x^{3}\cdot y^{2}$ on the region bounded by $f(x) = 2\cdot x - 1$, $g(x) = -x^{2}-2$, $x = 0$ and $x = 1$.

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