[1,2] + [3,4] *~ 3 *~ (x-1)^2;

Enter your integral in unevaluated form, i.e. as Int with a capital "I", and with symbolic e_0 as the upper limit:

t:= Int((1+121/304*e^2)^(1181/2299)*(1-e^2)^(-3/2)*e^(29/19), e= 0 .. e_0);

Then, simply,

plot(t, e_0= 0 .. 0.9);

Note to other readers: In this case, I needed to make sure that Maple did not try to evaluate the integral symbolically.  There is a very serious bug such that evaluating the integral (i.e. value(t)) causes the Maple kernel to utterly and completely crash after using about 11 seconds of processor time.

plots:-display(

   [seq](

      plots:-odeplot(

         dsolve(eval(sys, n= p), numeric)

        ,[x,g(x)]

        ,0..12

      )

     ,p in [0.1e-1, .1, .7, 1, 10]

   )

);

You could use the spreadsheet package.  For example,

> with(Spread):

> CreateSpreadsheet("mySS");

> M:= <<a,b>|<c,d>>:

> SetMatrix("mySS",M):

For example,

plot(1+x, x= -2..2, linestyle= dash, thickness= 3, color= green);

The order of the options does not matter.  For more details see ?plot,options ---probably my most frequently visited help page.

Apparently, fsolve has trouble with double integrals presented in the Int(..., [x= ..., y= ...]) form as opposed to the Int(Int(..., x= ...), y= ...) form.  Note the difference in the following two examples. 

restart:
J:= Int(a*x+y, [x= 0..1, y= 0..1]):
fsolve(J=0, a);
Error, (in fsolve) {x, y} are in the equation, and are not solved for

restart;
J:= Int(Int(a*x+y, x= 0..1), y= 0..1):
fsolve(J=0, a);
                          -1.000000000

Yet, for both forms type(J, freeof(x)) returns true. 

You can use the coeff command on either the series form or on the polynomial form that you had before converting it to a series.  If S is your series, then coeff(S,xi,0) and coeff(S,xi,2) will extract the expressions in your example.