Results 1 to 9 of 9

Math Help - Finding the average value of a(x) on an interval.

  1. #1
    Member
    Joined
    Oct 2009
    Posts
    128

    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).



    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?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by Lord Darkin View Post
    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).



    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
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Oct 2009
    Posts
    128
    Quote Originally Posted by CaptainBlack View Post
    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.

    New integral:
    <br />
\frac{\int_{0}^2 A(x)\;dx}{2}




    I checked my notes (which I just got today ) 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"?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by Lord Darkin View Post
    Okay, thanks.

    New integral:
    <br />
\frac{\int_{0}^2 A(x)\;dx}{2}




    I checked my notes (which I just got today ) 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
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Oct 2009
    Posts
    128
    Quote Originally Posted by CaptainBlack View Post
    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.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by Lord Darkin View Post
    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

    <br />
A(x)=2xe^{-2x^2}<br />

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

    <br />
\frac{\int_{0}^2 A(x)\;dx}{2}<br />

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

    CB
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Member
    Joined
    Oct 2009
    Posts
    128
    Quote Originally Posted by CaptainBlack View Post
    You have been told that

    <br />
A(x)=2xe^{-2x^2}<br />

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

    <br />
\frac{\int_{0}^2 A(x)\;dx}{2}<br />

    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.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Member
    Joined
    Oct 2009
    Posts
    128
    Quote Originally Posted by Lord Darkin View Post
    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.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by Lord Darkin View Post
    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
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Average speed over an interval.
    Posted in the Calculus Forum
    Replies: 8
    Last Post: January 24th 2010, 12:49 PM
  2. average velocity over the time interval
    Posted in the Calculus Forum
    Replies: 3
    Last Post: September 1st 2009, 08:15 PM
  3. Finding Average Value
    Posted in the Calculus Forum
    Replies: 7
    Last Post: December 2nd 2007, 10:00 PM
  4. average value on an interval
    Posted in the Calculus Forum
    Replies: 2
    Last Post: January 8th 2007, 02:19 AM
  5. finding the average
    Posted in the Algebra Forum
    Replies: 1
    Last Post: July 31st 2005, 10:00 PM

Search Tags


/mathhelpforum @mathhelpforum