Applying Preben's corrections, and adding an additional boundary condition f(150) = 1, another error occurs:

> ics := f(0) = 0, D(f)(150) = 0, D(f)(0) = 0, f(150) = 1;
   res := dsolve({ics, ode1}, numeric);

Error, (in dsolve/numeric/bvp) system is singular at left endpoint, use midpoint method instead

Maple is right: the system is indeed singular at the left endpoint because of the boundary condition f(0) = 0 and the fact that the highest-order derivative is multiplied by f(eta) in the differential equation.  But the error message is slightly strange, because there is no "midpoint" method to use here.  What should be used, I think, are method=bvp[midrich] or method = bvp[middefer].  However, both of those also produce an error:

Error, (in dsolve/numeric/bvp) initial Newton iteration is not converging

Perhaps a suitable approximate solution could be provided with the approxsoln option, or a continuation method could be used.  Or perhaps there simply isn't a solution to this boundary value problem.

It could be that some of the citers actually know what they are doing, but just wanted a citable source for the Trapezoid Rule, and maybe an article in a medical journal looks better to them than a calculus textbook.  Pre-med students, at least around here, do take a calculus course where they would meet the Trapezoid Rule.  Of course more sophisticated methods would be somewhat better if their function is smooth, but the improvement is probably not significant in these applications, and might be nullified by noisy data.

@mhrdmaple : with L as above:

> L := [6, 24, 60, 120, 210];
   g:= gfun[guessgf](L,t, [ogf]);
    coeftayl(g[1],t,nops(L));

                336

@mhrdmaple : with L as above:

> L := [6, 24, 60, 120, 210];
   g:= gfun[guessgf](L,t, [ogf]);
    coeftayl(g[1],t,nops(L));

                336

@bhornsby89 : OK, so if you're choosing an entering or leaving variable, and V is the list of variables corresponding to those numbers, your entering or leaving variable will be V[leastpos(L)].

@bhornsby89 : OK, so if you're choosing an entering or leaving variable, and V is the list of variables corresponding to those numbers, your entering or leaving variable will be V[leastpos(L)].

@chaynes75268 : A couple of problems with what you have:

1) f : x -> x^2 instead of f := x -> x^2

2) PI in one place instead of Pi

3) If you want to add up a definite number of terms rather than use a closed-form summation formula, use add instead of sum.

4) Your a[k] and b[k] would be easier to use as functions rather than table entries.  So

  a:= k -> ...   and b:= k -> ....

As for putting them in an Array, try something like

Array([seq(a(0)/2 + add(a(k)*cos(k*x) + b(k)*sin(k*x), k=1....n), n = 1 .. 10)]);

@chaynes75268 : A couple of problems with what you have:

1) f : x -> x^2 instead of f := x -> x^2

2) PI in one place instead of Pi

3) If you want to add up a definite number of terms rather than use a closed-form summation formula, use add instead of sum.

4) Your a[k] and b[k] would be easier to use as functions rather than table entries.  So

  a:= k -> ...   and b:= k -> ....

As for putting them in an Array, try something like

Array([seq(a(0)/2 + add(a(k)*cos(k*x) + b(k)*sin(k*x), k=1....n), n = 1 .. 10)]);

@levy : Do you mean something like this?

> plots[implicitplot3d]( cos(theta[1])*tan( theta[2])+ arccos(theta[3]) = 1, theta[1] = 0 .. 2*Pi,
      theta[2] = -Pi/2+0.01 .. Pi/2 - 0.01, theta[3] = -1 .. 1, grid = [30, 30, 30], axes=box,
      style=patchnogrid, lightmodel=light2);

@levy : Do you mean something like this?

> plots[implicitplot3d]( cos(theta[1])*tan( theta[2])+ arccos(theta[3]) = 1, theta[1] = 0 .. 2*Pi,
      theta[2] = -Pi/2+0.01 .. Pi/2 - 0.01, theta[3] = -1 .. 1, grid = [30, 30, 30], axes=box,
      style=patchnogrid, lightmodel=light2);

The questions about the Mean Value Theorem and 3 intervals don't make sense: there is no "other solution".  The function f(x) = x^3+2*x+1 is increasing for -infinity < x < infinity, so there is only one real solution.  I don't think you're telling us the actual question. 

The questions about the Mean Value Theorem and 3 intervals don't make sense: there is no "other solution".  The function f(x) = x^3+2*x+1 is increasing for -infinity < x < infinity, so there is only one real solution.  I don't think you're telling us the actual question. 

Sorry: that would work for most commands, but diff is rather special. 

> `diff/omega_2`:= (v,t) -> alpha_2(v)*diff(v,t);

Sorry: that would work for most commands, but diff is rather special. 

> `diff/omega_2`:= (v,t) -> alpha_2(v)*diff(v,t);

See also <http://oeis.org/A008918>