# Finding the average value of a(x) on an interval.

• Oct 18th 2009, 07:23 PM
Lord Darkin
Finding the average value of a(x) on an interval.
I'm given a problem with a diagram. Here is my bad recreation of it (there's really a point of inflection at the upper corners of the rectangle ...). The curve is y=e^-2x^2 (this means that it's x squared and then x squared is multiplied by -2) and the vertices of this rectangle inscribed under the curve are (x, 0) and (-x, 0).

http://img23.imageshack.us/img23/450...ulusfrq.th.jpg

This question had 2 other parts that I was able to solve correctly.

-Find A(1) with A(x) being the area of the rectangle inscribed under this curve y=e^-2x^2

A(1) is 2/e^2

Then the next question was the max value of a(x) which was e^-0.5 (I got this by taking the derivative).

Then I'm told to find the average value of a(x) in the interval 0 =< x =< 2.

Does this mean that I have to include the stuff (the part of the rectangle above in the diagram that's to the left of the y axis) that's from 0 to -x in the average area, or do I ignore that?
• Oct 18th 2009, 09:19 PM
CaptainBlack
Quote:

Originally Posted by Lord Darkin
I'm given a problem with a diagram. Here is my bad recreation of it (there's really a point of inflection at the upper corners of the rectangle ...). The curve is y=e^-2x^2 (this means that it's x squared and then x squared is multiplied by -2) and the vertices of this rectangle inscribed under the curve are (x, 0) and (-x, 0).

http://img23.imageshack.us/img23/450...ulusfrq.th.jpg

This question had 2 other parts that I was able to solve correctly.

-Find A(1) with A(x) being the area of the rectangle inscribed under this curve y=e^-2x^2

A(1) is 2/e^2

Then the next question was the max value of a(x) which was e^-0.5 (I got this by taking the derivative).

Then I'm told to find the average value of a(x) in the interval 0 =< x =< 2.

Does this mean that I have to include the stuff (the part of the rectangle above in the diagram that's to the left of the y axis) that's from 0 to -x in the average area, or do I ignore that?

The average value of $A(x)$ over the interval $x \in [a,b]$ is:

$\frac{\int_{x=a}^b A(x)\;dx}{b-a}$

CB
• Oct 20th 2009, 04:15 PM
Lord Darkin
Quote:

Originally Posted by CaptainBlack
The average value of $A(x)$ over the interval $x \in [a,b]$ is:

$\frac{\int_{x=a}^b A(x)\;dx}{b-a}$

CB

Okay, thanks. (Rofl)

New integral:
$
\frac{\int_{0}^2 A(x)\;dx}{2}$

I checked my notes (which I just got today (Rofl) ) and this matches what my notes have. My notes say that I'm supposed to find the area under the curve from 0 to 2 (since the interval is from 0 to 2). So what is the value of this area under the curve that I should put in to subsitute in for "A(x)dx"?
• Oct 20th 2009, 08:28 PM
CaptainBlack
Quote:

Originally Posted by Lord Darkin
Okay, thanks. (Rofl)

New integral:
$
\frac{\int_{0}^2 A(x)\;dx}{2}$

I checked my notes (which I just got today (Rofl) ) and this matches what my notes have. My notes say that I'm supposed to find the area under the curve from 0 to 2 (since the interval is from 0 to 2). So what is the value of this area under the curve that I should put in to subsitute in for "A(x)dx"?

As the base of the rectangle is $2x$, and its height is $\ e^{-2x^2}\text{ }$ we have:

$A(x)=2xe^{-2x^2}$

CB
• Oct 22nd 2009, 03:47 PM
Lord Darkin
Quote:

Originally Posted by CaptainBlack
As the base of the rectangle is $2x$, and its height is $\ e^{-2x^2}\text{ }$ we have:

CB

So now I have ( $2xe^{-2x^2}$)/2 for the average value of A(x). Then I can simplify that to $xe^{-2x^2}$

Is that the final answer? The question asks for the average value of A(x) o ver this interval 0 to 2.
• Oct 22nd 2009, 10:59 PM
CaptainBlack
Quote:

Originally Posted by Lord Darkin
So now I have ( $2xe^{-2x^2}$)/2 for the average value of A(x). Then I can simplify that to $xe^{-2x^2}$

Is that the final answer? The question asks for the average value of A(x) o ver this interval 0 to 2.

You have been told that

$
A(x)=2xe^{-2x^2}
$

you have been told that the avaerage over $(0,2)$ is:

$
\frac{\int_{0}^2 A(x)\;dx}{2}
$

so just do the integral, the answer will be a number.

CB
• Oct 23rd 2009, 04:01 AM
Lord Darkin
Quote:

Originally Posted by CaptainBlack
You have been told that

$
A(x)=2xe^{-2x^2}
$

you have been told that the avaerage over $(0,2)$ is:

$
\frac{\int_{0}^2 A(x)\;dx}{2}
$

so just do the integral, the answer will be a number.

CB

Er, how do I take the integrand of this function? I know I'm supposed to find the derivative preceding this but it's tough with the x in the exp.
• Oct 23rd 2009, 12:07 PM
Lord Darkin
Quote:

Originally Posted by Lord Darkin
Er, how do I take the integrand of this function? I know I'm supposed to find the derivative preceding this but it's tough with the x in the exp.

Never mind, I think I figured out how to do this problem. I figured out how to get the integrand. Then, I wrote the integrand down TWICE, substituting 2 in for x in the first and 0 for x in the second. After some algebra, I got a final answer that had several e's in it, and it made sense according to my calculator.
• Oct 23rd 2009, 02:07 PM
CaptainBlack
Quote:

Originally Posted by Lord Darkin
Er, how do I take the integrand of this function? I know I'm supposed to find the derivative preceding this but it's tough with the x in the exp.

The integrand is the derivative of $-e^{-2x^2}$, so the fundamental theorem of calculus will tell you what the indefinite integral is and so you can evaluate the definite integral.

CB