Finding average value from a double integral

Formula for average value with a double integral

We can estimate the average value of a region of level curves by using the formula

$f_{ave}=\frac{1}{A(R)}\int\int_Rf(x,y)\Delta{A}$

where $A(R)$ is the area of the rectangle defined by $R=[x_1,x_2]\times[y_1,y_2]$, and where the double integral gives the volume under the surface $f(x,y)$ over the region $R$.

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If an equation is not given, we can use the provided diagram to estimate the value of the function at the midpoints.

Finding average value over a specific region using a double integral with a multivariable function

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Using level curves to estimate the average value over a region

Example

Use the sketch of level curves to estimate the region’s average value where $m=n=2$ and $R=[0,2]\times[0,3]$.

The question is asking us to use a sketch of the level curves that represent our function to estimate the average value of the region depicted and defined by $R=[0,2]\times[0,3]$. We’ll need to use the formula for average value:

$f_{ave}=\frac{1}{A(R)}\int\int_Rf(x,y)\Delta{A}$

We can start by solving for $A(R)$, the total area of the region $R$.

$A(R)=(2)(3)$

$A(R)=6$

Next, we can solve for $\Delta{A}$, which is the area of one of the subrectangles. Based on the figure we were given, the width of each subrectangle is $1$ and the height is $3/2$. Which means that the area of each subrectangle is

$\Delta{A}=(1)\left(\frac32\right)$

$\Delta{A}=\frac32$

If an equation is not given, we can use the provided diagram to estimate the value of the function at the midpoints.

Next we need the midpoints of each subrectangle.

Using $R=[0,2]\times[0,3]$, $m=n=2$ and our diagram, we can see that the midpoints of the four smaller rectangles are

$\left(\frac12,\frac34\right)$, $\left(\frac32,\frac34\right)$, $\left(\frac12,\frac94\right)$ and $\left(\frac32,\frac94\right)$

Since we don’t have an equation for our function, we can use the diagram of its level curves to estimate the functions values at each midpoint.

At $\left(\frac12,\frac34\right)$, the function is approximately $19$.

At $\left(\frac32,\frac34\right)$, the function is approximately $35$.

At $\left(\frac12,\frac94\right)$, the function is approximately $20$.

At $\left(\frac32,\frac94\right)$, the function is approximately $26$.

Now we have everything we need to find an estimate for the function’s average value over the region. We’ll use the formula

$f_{ave}=\frac{1}{A(R)}\int\int_Rf(x,y)\Delta{A}$

but since we don’t have the function, we’ll use the midpoint rule formula instead of the double integral, such that the formula becomes

$f_{ave}=\frac{1}{A(R)}\left[\Delta{A}\left(f(x_1,y_1)+f(x_2,y_2)+f(x_3,y_3)+f(x_4,y_4)\right)\right]$ 

$f_{ave}=\frac{1}{6}\left[\frac32\left(19+35+20+26\right)\right]$

$f_{ave}=\frac{1}{6}\left[\frac32\left(100\right)\right]$

$f_{ave}=\frac{300}{12}$

$f_{ave}=25$

Using midpoint rule, this is our estimate for the function’s average value over the region.

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