   Chapter 7.9, Problem 5CP ### Calculus: An Applied Approach (Min...

10th Edition
Ron Larson
ISBN: 9781305860919

#### Solutions

Chapter
Section ### Calculus: An Applied Approach (Min...

10th Edition
Ron Larson
ISBN: 9781305860919
Textbook Problem
1 views

# Find the average value of f ( x ,   y )   =   4   −   1 2 x   − 1 2 y over the region 0   ≤ x ≤   2  and  0   ≤   v   ≤   2.

To determine

To calculate: The average value of f(x,y)=412x12y over the region 0x2 and 0y2.

Explanation

Given information:

The provided function is f(x,y)=412x12y

The region 0x2 and 0y2

Formula used:

The average value of integrable function z=f(x,y) over the region R with area A is

Average value=1ARf(x,y)dxdy

Calculation:

Consider equation of function,

f(x,y)=412x12y

The region R: triangle with vertices (0,0),(0,2),(2,2),(2,0).

The area of region R is 4 square units.

Now apply, the formula of the average value of integrable function f(x,y)=412x12y over the region triangle with vertices (0,0),(0,2),(2,2),(2,0) with area 4 units is

Average value=140202(412x12y)</

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