Take the 2-minute tour ×
Mathematica Stack Exchange is a question and answer site for users of Mathematica. It's 100% free, no registration required.

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?

EDIT:

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.

share|improve this question
add comment

2 Answers

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]
share|improve this answer
add comment

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.

share|improve this answer
    
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. –  Mark Aug 26 '13 at 0:52
    
Edited the OP to give an example –  Mark Aug 26 '13 at 1:09
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.