Aragorn 15  I have corrected something in your worksheet. The procedure  orbit  now works successfully for  mode=1  and  mode=2  . 
 

Download Project_(1)_new1.mw

restart;
plots:-arrow([1,2], scaling=constrained, caption="first:1 \n second:2", captionfont=[times,16]);

What does  f  and  g  mean in your code? These are undefined. You probably meant that  f = Fresnelf  and  g= Fresnelg . 

Here's the corrected code:

restart;
a := 0;  
b := 1;
f:=Fresnelf:
g:=Fresnelg: 
p := plot({f(x), g(x)}, x = a .. b);
A := [a, g(a)];
B := [a, f(a)];                    
C := [b, g(b)];
D_ := [b, f(b)];                  
left_line := plot([A, B]);
right_line := plot([D_, C]);
plots[display](p, left_line, right_line, view = [0 .. b + 0.5, 0 .. 0.5]);

The left_line is just one point, because  A=B .

Alternatively, you can use the  alias  command as a shorthand:

alias(f=Fresnelf):
alias(g=Fresnelg): 

See help on  alias  command for details .

This will be a curve in space obtained from the intersection of the inclined plane  F  and the cylindrical surface  G:

restart;
F:=z=x+y;
G:= x^5+x^4*y+x^3*y+2*x^2*y^2+x*y^5+y^5-1=0;
A:=plots:-intersectplot(F, G, x=-5..5, y=-5..5, z=-5..5, thickness=3, numpoints=64000):
B:=plot3d(x+y, x=-5..5, y=-5..5, style=surface, color="LightBlue", view=-5..5):
plots:-display(A, B, axes=normal, orientation=[-40,60], lightmodel=light2);  

restart;
ls:=["a", "b", "c", "d", "e"]:
selectremove(t->ListTools:-Search(t,ls)::odd, ls);

The easiest way to solve the problem is to show your graphs in the  x  and  y ranges (using the  view  option), where the differences are clearly visible:

restart:
with(plots):
f1:=x->0.95*x;
f2:=x->0.99*x;
f3:=x->0.991*x;
p1:=plot(f1(x),x=0..10,color=red):
p2:=plot(f2(x),x=0..10,color=blue):
p3:=plot(f3(x),x=0..10,color=green):
P1:=display(p1,p2,p3, caption="Plot 1", captionfont=[times,16]); 
P2:=display(p1,p2,p3, view=[9.5..10,9.2..9.7],caption="Plot 2", captionfont=[times,16]);
display(<P1 | P2>);

Download plotting1.mw

Edit.

Maple finds the answer through elliptic functions. Unfortunately, the result is quite cumbersome:

restart;
y:=Int((k/2*x^2+x+c)/sqrt(lambda^2*(1+k*x)^4 - (k/2*x^2+x+c)^2), x);
value(y):
simplify(%);

Download Int.mw

Edit. In my opinion, the value of this overall result is rather dubious. With some parameters and variable  x  values, you can get  0 or negative numbers under the root. It is much more reliable to first set the values of the parameters, and only then calculate the integral.

Use  add  instead of  sum  when summing a specific finite number of terms. The  sum  command is used if at least one of the summation limits is a symbol or the infinity:

restart;
seq(add(x^r*(add((-1)^s*binomial(r, s)*pochhammer((a+s+1)*d, n), s = 0 .. r))/factorial(r), r = 0 .. n), n = 0 .. 4);

It is easy to prove that your second graph is empty. Probably you rewrote something wrong. I modified the second equation a bit to make the graph look like your second graph in the picture:

Edit. Let us prove that the equation  abs(z-I) - abs(z+1)  =  2  has no solutions for complex numbers  z=x+I*y . This is most easily done using the fact that   abs(z1 - z2)  is the distance between the points  z1  and  z2 . Consider 3 points  z1=I ,  z2=-1 and  z . It follows from the triangle inequality that  abs(z-z1) - abs(z-z2) < abs(z1-z2) = sqrt(2)<2

Download z.mw

If you need to do this more than once, then (as a development of vv's idea) a simple procedure  Sort  can be used to reduce typing:

restart;
S:=[seq(_C||k=c[k], k=0..10)]:
sol:=dsolve(diff(y(x),x) = x+y(x),y(x));

Sort:=s->sort(subs(S,s)):

Sort(sol); 

Use  piecewise  instead of  if .. fi  :

Download if_new.mw

I do not know the reasons for this behavior, but to immediately get the desired position of the constants use the  sort  command:

restart;
ode:=diff(diff(y(x),x),x)+8*diff(y(x),x)+25*y(x) = 1:
sol:=sort(dsolve(ode));

This code in Maple 2018.2

Here is another proof without complex numbers using the well-known rotation matrix by an angle. We take the unit vector  <1,0>  and rotate it first by the angle of  2*x , we get the vector  X . Rotating the same vector  <1,0>  sequentially, first by the angle  x , and then again by x , we get a vector  Y . Obviously  X=Y . At the same time, we proved another formula  cos(2*x)=cos(x)^2-sin(x)^2 .

Download proof.mw

You are probably interested in calculating the area of the region bounded by a red closed curve below (this is not an ellipse of course).
It's easy to do numerically:

Download int.mw

To solve your example, it is enough to know the simplest definitions related to the joint distribution of two random variables  X  and  Y . Let's introduce the notations:  f__XY  is the density function of the joint distribution of  X  and  Y ,  f__X  is the density function of  X ,  f__Y  is the density function of  Y ,  f__X_Y  is the density function of the conditional distribution of  X  given  Y ,   f__Y_X  is the density function of the conditional distribution of  Y  given  X .

Download XY.mw