Trying to integrate a non one-to-one parametric function

[BinBinBinBin]

Hi, I'm working on an independent research project - and am trying to integrate this (with respect to x between some arbitrary m and infinite).

http://www.wolframalpha.com/input/?i=+x+=(t+2)/(1+e^(t-r)),+y=(e^(-t^2/2))/sqrt(2*pi)

If you graph this as a parametric eqn (set r to 2 or 3), the problem is that it is not a one-to-one mapping. I want to find the area under the curve (and the part where there are two values of x, I want to include that twice.

Is there any way I can do this?

In the end I want to have the integral be a function of r (m is constant) that I can use elsewhere.

I want to see how the area under the curve changes when I vary r

Any suggestions?

 

[Mark44]

When we find the area "under a curve" it's implied that the lower boundary of the region is the horizontal axis. For your parametric curve, with r = 3, the graph is nearly a loop, so "under the curve" doesn't make much sense.

 

[jackmell]

Hi, I'm working on an independent research project - and am trying to integrate this (with respect to x between some arbitrary m and infinite).

http://www.wolframalpha.com/input/?i=+x+=(t+2)/(1+e^(t-r)),+y=(e^(-t^2/2))/sqrt(2*pi)

If you graph this as a parametric eqn (set r to 2 or 3), the problem is that it is not a one-to-one mapping. I want to find the area under the curve (and the part where there are two values of x, I want to include that twice.

Is there any way I can do this?

In the end I want to have the integral be a function of r (m is constant) that I can use elsewhere.

I want to see how the area under the curve changes when I vary r

Any suggestions?

First just get a bearing on it. This is what it looks to me:

x(t)=\frac{t+2}{1+e^{t-2}}
y(t)=\frac{e^{-t^2/2}}{\sqrt{2\pi}}

Ok, right off the bat, I'm going to plot it parametrically and since you want to integrate it, in the interest of just trying something even if it's wrong, I want to integrate it "twice" as you suggest where it's not one-to-one. That means I need to find where it's not. That's where the black dot is. Since \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}, then the point at which it has an infinite slope is when dx/dt=0. Alright then, then I'll integrate the red to the t-value where \frac{dx}{dt}=0, and the blue from that spot to infinity:

\mathop\int\limits_{\text{red}} y(x)dx=\mathop\int_{-\infty}^{\frac{dx}{dt}=0} y(t)\frac{dx}{dt} dt
\mathop\int\limits_{\text{blue}} y(x)dx=\mathop\int_{\frac{dx}{dt}=0}^{\infty} y(t)\frac{dx}{dt} dt

I don't know, is that right guys? I need to check.

 

[JJacquelin]

Hi !

There is no difficulty to integrate on a closed area.
In fact, the difficulty is that the antiderivative cannot be expressed in terms of a finite number of standard functions. So you cannot give the result on the form of a formula.
The only way is a numerical integration, giving a numerical result for each numerical value of the parameter r.