00:01
The problem is determine whether each integral is convergent or divergent.
00:05
Evaluated those that are convergent.
00:10
For this improper integral, by definition, we can write this integral as integral from negative 2 to 0, 1 over x to force, dx, plus 0 to 3, 1 over x2 4th, dx, plus 0 to 3, 1 over x to 4th, 1 over x2 4th, d x this one x is equal to zero.
00:42
The denominator of this function is 0.
00:46
This is an improper integral.
00:50
We need to compute the two improper integrals.
00:59
Now we compute the first one...