Sketching and Finding an Area between curves

• Apr 21st 2013, 03:56 AM
OldMate
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
• Apr 21st 2013, 05:06 AM
MarkFL
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?
• Apr 21st 2013, 05:52 AM
OldMate
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.
• Apr 21st 2013, 01:44 PM
MarkFL
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.
• Apr 21st 2013, 10:13 PM
OldMate
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)
• Apr 21st 2013, 10:23 PM
MarkFL
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.
• Apr 22nd 2013, 01:48 AM
Prove It
Re: Sketching and Finding an Area between curves
Quote:

Originally Posted by OldMate
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*}