Refer to Exercise 48 . Find the average population density inside the rectangular area R.

Name: Problem 49, Chapter 8: Applied Calculus for the MLSS A Brief Approach
Uploaded: 2021-10-30T23:40:02-08:00
Duration: 7 min 52 s
Channel: Uma Kumari
Description: Refer to Exercise 48 . Find the average population density inside the rectangular area R.

Refer to Exercise 48 . Find the average population density...

Key Concepts

Integration Over Rectangular Regions

When the region of interest is a rectangle, the limits of integration are constant, making it simpler to set up and evaluate the double integral. This key concept leverages the properties of rectangular domains, where variables can often be separated, facilitating the computation of integrals necessary for finding average values and accumulative quantities.

Average Value of a Function

This concept involves finding the mean or average of a function over a given region. In multivariable calculus, the average value of a function f(x, y) over a region R is computed by integrating f over R and then dividing by the area of R. This provides a single number that represents the typical value of the function across the entire region.

Double Integration

Double integration is a method used to calculate the accumulation of quantities over a two-dimensional area. In the context of finding an average population density, double integrals allow us to sum up the population density values across a region, and then averaging them by dividing by the total area. This is one of the fundamental techniques in multivariable calculus for processing functions over regions.

Transcript

00:01 In this question we have a function f x comma y is equal to 10 ,000 e to the power y divided by 1 plus 0 .5 x.

00:16 This function represents the population density of a coastal town.

00:21 Here, x and y are in miles as shown in the figure.

00:27 X lies between minus 10 and 10 and the y lies between minus 4 and 0 we are required to evaluate the average population density inside the rectangular area are and r is defined by x comma y x lies between minus 5 and 5 and 5 and y and y lies between minus 5 and y lies between minus 2 and 0.

01:05 So let's see how to solve this question.

01:07 So first of all, let's find the value of double integral are function f x comma y da and this will be equals to integration x is equals to minus 5 to x is equal to 5 multiplied by integration y is equals to minus 2 to 0 and we have the function 10 ,000 e to the power y divided by 1 plus 0 .5 x, d .y, dx.

01:51 Since here we have the symmetry, therefore we can write this as integration as 2 into integration 025 multiplied by integration minus 2 to 0 ,000 e to the power y divided by 1 plus 0 .5 x, d .x.

02:19 And now let's integrate this function with respect to y.

02:23 So consider all other terms as a constant.

02:27 So we can write 2 into integration 0 to 5, 10 ,000 divided by 1 plus 0 .5 x.

02:41 And the integration of e to the power y, dy, will be equals to e to the power y having the limits minus 2 to 0 dx.

02:52 Now substitute these limits, so we will have 2 into integration 025, 10 ,000 divided by 1 plus 0 .5 x, e to the power 0 minus e to the power minus 2 dx.

03:15 And now let's integrate this function with respect to x.

03:20 So we can take out the constants.

03:22 So we will have 2 into 10 ,000 multiplied by e to the power 0 minus e to the power minus 2.

03:32 Integration 0 to 5 1 upon 1 plus 0 .5 x d x...

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