Results 1 to 7 of 7
Like Tree3Thanks
  • 1 Post By MarkFL
  • 2 Post By Prove It

Math Help - Sketching and Finding an Area between curves

  1. #1
    Newbie
    Joined
    Mar 2013
    From
    Queensland
    Posts
    10

    Sketching and Finding an Area between curves

    f(x) = 1/2 e^x and g(x) = sech(x). I have already worked out that there are no x intercepts for either graph and that f(x) intercepts the y axis at 1/2 and g(x) intercepts it at 1.

    when i tried to find the intercept between the two curves I got the x value = (log 3) / 2, making the y value = 0 which i know is incorrect. i believe my x value is right i just dont know what to do to get the right y value.

    the area needed to be calculated is between the graphs between 0 and the intersection point, since i think its log 3 / 2, i worked it out to be (3+3 sqrt(3)+pi)/6. just wondering if that is correct
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor MarkFL's Avatar
    Joined
    Dec 2011
    From
    St. Augustine, FL.
    Posts
    1,988
    Thanks
    734

    Re: Sketching and Finding an Area between curves

    You have the correct x-value of intersection, to find the y-value, you could use:

    y=\frac{1}{2}e^{\frac{\ln(3)}{2}}=?

    Use exponential and logarithmic identities...what do you find?

    As for the area in question, it is not correct. What did you find for the anti-derivative to use in the FTOC?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Mar 2013
    From
    Queensland
    Posts
    10

    Re: Sketching and Finding an Area between curves

    to find the area i just integrated sech(x)- 1/2 e^x between 0 and log 3 / 2, and that was the result I was given.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor MarkFL's Avatar
    Joined
    Dec 2011
    From
    St. Augustine, FL.
    Posts
    1,988
    Thanks
    734

    Re: Sketching and Finding an Area between curves

    When you say "that was the result I was given", do you mean you are using software to compute the definite integral?

    How did you enter the upper limit of integration? Does your program take log to have an implied base of e? How is the argument for the log function being interpreted? I would use (ln(3))/2.

    I would think you are expected to compute the definite integral by hand and use the software as a check only.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Mar 2013
    From
    Queensland
    Posts
    10

    Re: Sketching and Finding an Area between curves

    I used wolframalpha, to see if i was on the right track with regards to my thinking and both ways of in putting the log function give 1/6 (3-3*sqrt(3)+pi)
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor MarkFL's Avatar
    Joined
    Dec 2011
    From
    St. Augustine, FL.
    Posts
    1,988
    Thanks
    734

    Re: Sketching and Finding an Area between curves

    Yes, that's what it returns for me as well...this is different than the result you cited before. It only returned a decimal approximation the first time for me using ln, but when I used log it gave the exact result.
    Thanks from OldMate
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,513
    Thanks
    1404

    Re: Sketching and Finding an Area between curves

    Quote Originally Posted by OldMate View Post
    f(x) = 1/2 e^x and g(x) = sech(x). I have already worked out that there are no x intercepts for either graph and that f(x) intercepts the y axis at 1/2 and g(x) intercepts it at 1.

    when i tried to find the intercept between the two curves I got the x value = (log 3) / 2, making the y value = 0 which i know is incorrect. i believe my x value is right i just dont know what to do to get the right y value.

    the area needed to be calculated is between the graphs between 0 and the intersection point, since i think its log 3 / 2, i worked it out to be (3+3 sqrt(3)+pi)/6. just wondering if that is correct
    The two functions intercept where \displaystyle \begin{align*} f(x) = g(x) \end{align*}, i.e.

    \displaystyle \begin{align*} \frac{1}{2}e^x &= \textrm{sech}\,{(x)} \\ \frac{e^x}{2} &= \frac{2}{e^x + e^{-x}} \\ e^x \left( e^x + e^{-x} \right) &= 4 \\ e^{2x} + 1 &= 4 \\ e^{2x} &= 3 \\ 2x &= \ln{(3)} \\ x &= \frac{1}{2}\ln{(3)} \end{align*}

    And to find the corresponding y-value...

    \displaystyle \begin{align*} f \left( \frac{1}{2}\ln{(3)} \right) &= \frac{1}{2}e^{\frac{1}{2}\ln{(3)}} \\ &= \frac{1}{2}e^{\ln{\left[ \left( 3 \right) ^{\frac{1}{2}} \right]} } \\ &= \frac{1}{2} \left( 3^{\frac{1}{2}} \right) \\ &= \frac{\sqrt{3}}{2} \end{align*}
    Thanks from MarkFL and OldMate
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Finding area between two bounded curves
    Posted in the Calculus Forum
    Replies: 1
    Last Post: December 5th 2012, 07:06 PM
  2. Finding area between curves?
    Posted in the Calculus Forum
    Replies: 3
    Last Post: May 22nd 2011, 03:51 PM
  3. finding the area between two curves
    Posted in the Calculus Forum
    Replies: 3
    Last Post: April 18th 2010, 03:29 PM
  4. sketching and finding area of the region?
    Posted in the Calculus Forum
    Replies: 0
    Last Post: September 24th 2009, 02:14 PM
  5. Finding the Area Between 2 Curves
    Posted in the Calculus Forum
    Replies: 4
    Last Post: February 2nd 2008, 10:33 AM

Search Tags


/mathhelpforum @mathhelpforum