Question

Find the average value of the function f(x,y) over the plane region r. f(x, y) = xey; r is the triangle bounded with vertices (0, 0), (6, 0), and (6, 6)

Answer 1

The average value of
`f(x,y)` over
`R` is the ratio

`(\displaystyle \iint_R f(x,y)\,dA)/(\displaystyle \iint_R dA)`

The denominator is the area of
`R`, 1/2•6•6 = 18.

We can parameterize
`R` by the set

`R = \left\{(x,y)\in\Bbb R^2 ~:~ 0 \le x \le 6 \text{ and } 0 \le y \le x \right\}`

so that the integral in the numerator is

`\displaystyle \iint_R f(x,y) \, dA = \int_0^6 \int_0^x xe^y \, dy \, dx \\\\ ~~~~~~~~ = \int_0^6 x(e^x - e^0) \, dx \\\\ ~~~~~~~~ = \int_0^6 xe^x \, dx - \int_0^6 x\,dx \\\\ ~~~~~~~~ = \int_0^6 xe^x\,dx - \frac12(6^2 - 0^2) \\\\ ~~~~~~~~ = \int_0^6 xe^x\,dx - 18`

For the remaining integral, integrate by parts.

`\displaystyle \int u\,dv = uv-\int v\,du`

Let

`u = x \implies du = dx`

`dv = e^x \, dx \implies v = e^x`

Then

`\displaystyle \int_0^6 xe^x \, dx = xe^x\bigg|_(x=0)^(x=6) - \int_0^6 e^x \, dx`

so that

`\displaystyle \iint_R f(x,y) \, dA = \left(6e^6 - (e^6 - e^0)\right) - 18 = 5e^6 - 17`

Hence the average value is

`f_(\rm ave) = \boxed{(5e^6 - 17)/(18)}`