Find the average value of the given function f(x, y)
Over the plane region R. ; R is the region bounded by the graph of y = 2x and y = 0 from x = 1 to x = 3.

Question

Find the average value of the given function f(x, y)
Over the plane region R.  ; R is the region bounded by the graph of y = 2x and y = 0 from x = 1 to x = 3.

Quizplus · Accepted Answer

We need to find the average value of the function over the region R, which is given by:

$\frac{1}{Area(R)} \iint_R f(x,y) dA$

The region R is bounded by the graph of y=2x and y=0, from x=1 to x=3. This means that the region R is a trapezoid with base lengths of 2 and 6, and a height of 3.

Area(R) = 1/2(base1 + base2)*height = 1/2(2+6)*3 = 9

Next, we need to evaluate the double integral over R:

$\iint_R f(x,y) dA$

$= \int_1^3 \int_0^{2x} x^2y^3 e^{y^2}dy dx$

$= \int_1^3 \left[\frac{1}{2}x^2e^{y^2}\right]_0^{2x} dx$

$= \int_1^3 \frac{1}{2}x^2(e^{4x^2}-1) dx$

This integral can be evaluated using u-substitution with u=4x^2, resulting in:

$= \frac{1}{16}(e^{36}-e^4-31)$

Therefore, the average value of f(x,y) over R is:

$\frac{1}{Area(R)} \iint_R f(x,y) dA = \frac{\frac{1}{16}(e^{36}-e^4-31)}{9} \approx 11.02$

So the best choice is B.

Find the Average Value of the Given Function F(x, Y)