Solve Complex Integration: Find 2.36 Area of y=-x/2e+1/e+e & y=e^x/2

In summary, the area bound by the curve and the lines AB and BC is given by:A=\int_0^2 -\frac{2}{e}x+\frac{e^2+4}{e}-e^{\frac{x}{2}}\,dx.
  • #1
minimoocha
7
0
The area of two lines that I need to find is 2.36, however i need this in exact form. The lines are y=-x/2e+1/e+e the other line is y=e^x/2
Since y=-x/2e+1/e+e is on top it is the first function.
A=(the lower boundary is 0 and the top is 2) -x/2e+1/e+e-e^x/2

If you could please help!
 
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  • #2
Hello, and welcome to MHB! :)

I am assuming the linear function is:

\(\displaystyle f(x)=-\frac{x}{2e}+\frac{1}{e}+e\)

And the exponential function is:

\(\displaystyle g(x)=\frac{e^x}{2}\)

Are those correct?
 
  • #3
MarkFL said:
Hello, and welcome to MHB! :)

I am assuming the linear function is:

\(\displaystyle f(x)=-\frac{x}{2e}+\frac{1}{e}+e\)

And the exponential function is:

\(\displaystyle g(x)=\frac{e^x}{2}\)

Are those correct?
[DESMOS]{"version":7,"graph":{"viewport":{"xmin":-10,"ymin":-12.5,"xmax":10,"ymax":12.5}},"randomSeed":"61268199dd3570cc9c59f2ef62b0245a","expressions":{"list":[{"type":"expression","id":"1","color":"#c74440","latex":"y=e^{\\frac{x}{2}}"},{"type":"expression","id":"3","color":"#388c46","latex":"y=-\\frac{x}{2e}+\\frac{1}{e}+e"}]}}[/DESMOS]

I am trying to find the area created by those two lines, however I can't integrate this, thank you for responding and for the welcome :D
 
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  • #4
Okay, so my assumptions were correct. So we wish to find this area (given that \(0\le x\le2\)):

mhb_0013.png


So, we need to find the \(x\)-coordinate of the point of intersection. Hence, we need to solve:

\(\displaystyle -\frac{x}{2e}+\frac{1}{e}+e=\frac{e^x}{2}\)

We cannot get a solution in terms of elementary functions, and in fact we find:

\(\displaystyle x=-W\left(e^{3+2e^2}\right)+2+2e^2\)

Where \(W\) is the product log function. Using a numeric root finding technique, we find:

\(\displaystyle x\approx1.7124201115416355473\)

And so the area would be approximately given by:

\(\displaystyle A\approx\int_0^{1.7124201115416355473} -\frac{x}{2e}+\frac{1}{e}+e-\frac{e^x}{2}\,dx+\int_{1.7124201115416355473}^2 \frac{e^x}{2}-\left(-\frac{x}{2e}+\frac{1}{e}+e\right)\,dx\)

Can you proceed?
 
  • #5
MarkFL said:
Okay, so my assumptions were correct. So we wish to find this area (given that \(0\le x\le2\)):

View attachment 9785

So, we need to find the \(x\)-coordinate of the point of intersection. Hence, we need to solve:

\(\displaystyle -\frac{x}{2e}+\frac{1}{e}+e=\frac{e^x}{2}\)

We cannot get a solution in terms of elementary functions, and in fact we find:

\(\displaystyle x=-W\left(e^{3+2e^2}\right)+2+2e^2\)

Where \(W\) is the product log function. Using a numeric root finding technique, we find:

\(\displaystyle x\approx1.7124201115416355473\)

And so the area would be approximately given by:

\(\displaystyle A\approx\int_0^{1.7124201115416355473} -\frac{x}{2e}+\frac{1}{e}+e-\frac{e^x}{2}\,dx+\int_{1.7124201115416355473}^2 \frac{e^x}{2}-\left(-\frac{x}{2e}+\frac{1}{e}+e\right)\,dx\)

Can you proceed?
Actually, because I do not do very hard maths, it is just 0<x<2
That is wayyy to complicated for me to understand. Desmos deformed it because my graphing calculator says different.
What I am doing is for my maths assignment:

So I had to find the equation of the normal for the point (2,e)
Which was = to the eq of the line y=-\frac{x}{2e}+\frac{1}{e}+e
So I need to find the area of e^x/2 and - from the normal
Which hopefully i am right would = to normal-curve = the area of the weird triangle, because i need it on exact form.

The original question is: The diagram shows part of the curve with the equation y=e^x/2. Find the exact area bound by the curve and the lines AB and BC.
THANK YOU THOUGH! I APPRECIATE IT A LOT!
 

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  • #6
Okay, this is a different problem. I didn't notice the difference between my assumptions and the functions you have in the live calculator above. Sorry for the confusion.

We are given the curve:

\(\displaystyle y=e^{\frac{x}{2}}\)

Let's find the derivative:

\(\displaystyle \frac{dy}{dx}=\frac{1}{2}e^{\frac{x}{2}}[\)

And so the slope of the normal line is:

\(\displaystyle m=\left.-\frac{dx}{dy}\right|_{x=2}=-\frac{2}{e}\)

And so the normal line is given by:

\(\displaystyle y=-\frac{2}{e}(x-2)+e\)

And thus, the shaded area is given by:

\(\displaystyle A=\int_0^2 -\frac{2}{e}(x-2)+e-e^{\frac{x}{2}}\,dx\)

Let's clean that up a bit:

\(\displaystyle A=\int_0^2 -\frac{2}{e}x+\frac{e^2+4}{e}-e^{\frac{x}{2}}\,dx\)

So, our integrand has 3 types of expressions...one with \(x\) to the first power, a constant, and an exponential. Do you have any thoughts on how to find the anti-derivative?
 
  • #7
yes.
A=[-x^2e+ ((e^3+4x)/e^2))-1/2e^-0.5x]

I think?
 
  • #8
I find:

\(\displaystyle A=\int_0^2 -\frac{2}{e}x+\frac{e^2+4}{e}-e^{\frac{x}{2}}\,dx=\left[-\frac{1}{e}x^2+\frac{e^2+4}{e}x-2e^{\frac{x}{2}}\right]_0^2\)

Do you see that if you differentiate the anti-derivative within the brackets, you get the integrand? I used the power rule and the rule for exponential functions to get it.

Can you proceed?
 
  • #9
that makes much more sense- I can proceed :)
\[ A=((-1/e)2^2+(2^2+4/e)2-2e^2/2)-((-1/e)0^2+(0^2+4/4)(0)-2e^0/2)) \]
which simplifies to
\[ A=((-1/e)(4))+((8/e)(2)-2e^1)-(2e^0) \]
this is as far as i got :)
\( A=((-1/e)(4))+((8/e)(2)-2e)-(2) \)
 
  • #10
You are making some mistakes in your application of the FTOC.

\(\displaystyle A=\left(-\frac{1}{e}2^2+\frac{e^2+4}{e}2-2e^{\frac{2}{2}}\right)-\left(-\frac{1}{e}0^2+\frac{e^2+4}{e}0-2e^{\frac{0}{2}}\right)\)

\(\displaystyle A=-\frac{4}{e}+\frac{2e^2+8}{e}-2e+2=\frac{2(e+2)}{e}\)
 
  • #11
MarkFL said:
You are making some mistakes in your application of the FTOC.

\(\displaystyle A=\left(-\frac{1}{e}2^2+\frac{e^2+4}{e}2-2e^{\frac{2}{2}}\right)-\left(-\frac{1}{e}0^2+\frac{e^2+4}{e}0-2e^{\frac{0}{2}}\right)\)

\(\displaystyle A=-\frac{4}{e}+\frac{2e^2+8}{e}-2e+2=\frac{2(e+2)}{e}\)
Sorry I've been staring at maths all day. Thank you so much for the help! You are a lifesaver!
 

FAQ: Solve Complex Integration: Find 2.36 Area of y=-x/2e+1/e+e & y=e^x/2

What is complex integration?

Complex integration is a mathematical technique used to find the area under a curve in the complex plane. It involves breaking down a complex function into simpler parts and using integration rules to solve for the area.

How do you find the area under a complex curve?

To find the area under a complex curve, you can use the complex integration technique. This involves breaking down the complex function into simpler parts and using integration rules to solve for the area. In some cases, it may also involve using geometric concepts such as the unit circle to visualize the complex curve and its area.

What is the formula for complex integration?

The formula for complex integration is similar to that of regular integration, but it involves using complex numbers and variables. It can be written as ∫f(z)dz, where f(z) is the complex function and dz represents the infinitesimal change in the complex variable z.

How do you solve for the area of a complex curve using integration?

To solve for the area of a complex curve using integration, you can follow these steps:

  • Break down the complex function into simpler parts
  • Use integration rules to solve for the area of each part
  • Add the areas of each part together to get the total area under the curve

Can complex integration be used to find the area of any complex curve?

Yes, complex integration can be used to find the area of any complex curve as long as the function is well-defined and continuous. However, the process can be more complicated for certain types of curves and may require advanced techniques such as contour integration.

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