I came across a problem when evaluating the following integral:

$\int_0^t \sqrt{9 x^4+1} \, dx$

Now, when I evaluate that, I get

ConditionalExpression[t Hypergeometric2F1[-(1/2), 1/4, 5/4, -9 t^4], t >= 0]

The issue is if t is negative. If I do this $\int_0^{-5} \sqrt{9 x^4+1} \, dx$ I get

-5 Hypergeometric2F1[-(1/2), 1/4, 5/4, -5625]

That seems okay, since it's a number. However, if I do this $\int_0^t \sqrt{9 x^4+1} \, dx\text{/.}t\to -5$ I get "Undefined".

The issue is that I need to keep t symbolic, but still be able to use negative numbers later.

I don't understand why t must be positive, especially since Mathematica seems to be able to compute it when it's negative. Does anyone know why this is happening, or what I can do to fix it?


I'm actually looking for a general solution for any function. This code is being used to calculate arc length of any function (in this case, x^3). This is just an example of it not working.

Here is my actual code:

baseFunction = Function[x,x^3]
secondFunction = Sin
r[t_] = FullSimplify[p0[t] + norm[t]*a*secondFunction[b*Integrate[Sqrt[1 + Derivative[1][baseFunction][x]^2], {x, 0, t}]], Element[t, Reals]]

p0 and norm are 2D vector functions, and a and b are constants. I want to be able to use any function for baseFunction, though, which is why I need a general solution.


2 Answers 2


How about

Integrate[Sqrt[9 x^4 + 1], {x, 0, t}, Assumptions -> {t < 0}]

t Hypergeometric2F1[-(1/2), 1/4, 5/4, -9 t^4]

which can be evaluated at t=-5 to give

% //. t -> -5
-5 Hypergeometric2F1[-(1/2), 1/4, 5/4, -5625]

Use Assuming, in order to obtain the solution first

Assuming[Element[t, Reals] && t > 0,Integrate[Sqrt[9 x^4 + 1], {x, 0, t}]] /. t -> -5

Mathematica graphics

I can't answer now why t has to be assumed positive, but can subs a negative value for it later without looking more into this. But the above seems to do what you wanted. Actually the above used t>0 and not t>=0 and it worked.

  • $\begingroup$ Thanks! The only issue, though, and I probably should have mentioned this in the OP, is that the function to integrate over can be anything. This particular example is finding arc length of the function x^3. So I'm actually looking for a general solution to this problem. $\endgroup$
    – Mark
    Aug 26, 2013 at 0:52
  • $\begingroup$ Edited the OP to give an example $\endgroup$
    – Mark
    Aug 26, 2013 at 1:09

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