Why Are Improper Integrals Undefined?

[tandoorichicken]

Why are these two integrals undefined?
1) \int_{-1}^{1} \frac{\,dx}{x^{\frac{4}{3}}}

2) \int_{3}^{6} \frac{\,dx}{5-x}

I got real answers for both, the first one 0, and the second one ln(2), but I think I'm in serious violation of the Fundamental Theorem of Calculus.

 

[outandbeyond2004]

division by zero - notice that x = 0 blows up the first integrand and x = 5 the second one likewise. Even with these points excluded, you get pretty big answers (not the answers that you got). Infinity in fact.

 

[ShawnD]

texing these is tricky, have a look at the source for these!

\int_{-1}^{1} \frac{\,dx}{x^{\frac{4}{3}}} = \frac{x^{\frac{-1}{3}}}{\frac{-1}{3}}|^1_{-1}

\frac{-3}{x^{\frac{1}{3}}}|^1_{-1}

The problem is that the function crosses over an asymtote. What happens when x is 0? Is the function infinity? How do you add infinity?

\int_{3}^{6} \frac{\,dx}{5-x} = -\ln|5 - x| |^6_5

That log function there, what happens when x = 5? What exponent on e will give you a value of 0?