Average Value Function

[kenhung123]

Totally do not know what this is. Anyone care to explain? The text doesn't describe how its derived, what its used for and how it works but just the height of a rectangle is equal to the area in the interval [a,b]
 

[98.40_for_sure]

Are you talking about the average of the area under the curve? by drawing a rectangle approximately the same area as the area under the curve between two points.

It isn't a useful thing, but i have seen it in previous exams, so i would want to be comfortable with doing it.

[TrueTears]

consider rectangles with an arbitrary width of 1 unit, sum this over the integral and divide by the total width and thus you get an average.

This is exactly what the integral does (it sums over the interval [a,b]) then by dividing (b-a) you are getting an "average"

[kenhung123]

Are you talking about the average of the area under the curve? by drawing a rectangle approximately the same area as the area under the curve between two points.

It isn't a useful thing, but i have seen it in previous exams, so i would want to be comfortable with doing it.

Nah not approximation, I can't explain it coz I don't know what it is lol but yea its called "average value of a function"

[98.40_for_sure]

Yeah it's in the essentials textbook and the formula thing is

[kenhung123]

Yea so how is it derived and whats the point of it?

[TrueTears]

is the sum of the rectangles of an arbitrary unit of 1 from a to b.

dividing by b- a finds the average from a to b.


QED

[98.40_for_sure]

How it is derived is above the scope of the year 12 course, if you're interested, then it is what TrueTears said.

It is just like finding an approximation for the area under the curve between two points; b & a.

Basically all you need to know is the formula itself, and how to use it

[TrueTears]

well... if you know how to work out the average of 3 numbers say a,b,c then you should be able to understand this, the concepts are the same.

[kenhung123]

is the sum of the rectangles of an arbitrary unit of 1 from a to b.

dividing by b- a finds the average from a to b.


QED

I don't get it so your dividing the area with an infinite number of rectangles from [a,b] by (b-a) they would just give us the sum of the heights?

[TrueTears]

how do u normally find out the average of say... 1,2,3,...,n ?

you sum from 1 to n, 1+2+3+4+...+n

then divide by how many numbers you added, ie, n numbers

use the same principle here.

we sum from a to b, namely,

how many numbers did we sum? we summed b-a numbers (since the width of the rectangles are unit length there are b-a rectangles)

now divide.

[fady_22]

is the sum of the rectangles of an arbitrary unit of 1 from a to b.

dividing by b- a finds the average from a to b.


QED

I don't get it so your dividing the area with an infinite number of rectangles from [a,b] by (b-a) they would just give us the sum of the heights?

In simple terms:

The integral from a to b is equal to the area under the curve, and also the area of a rectangle with width b-a and a height of the average value of the function from a to b.

So to find this height, you would have to DIVIDE by the width of the rectangle (if A=length*width, then length =A/width).

This gives you the average value of the function.

[kenhung123]

Thanks guys I appreciate your help!