I need help with a calculus problem, please help with steps?

Question

The question is evaluate or show that the integral diverges: ∫ (from upper bound 3 to lower bound 0) (1/x) dx. The answer is ∞ but I don't understand how they got that answer. Please help!

Answer 1

∫ (0 to 3, (1/x) dx )

For one thing, this is an improper integral, because (1/x) is not defined at 0. For that reason, you have to change this into a limit. More specifically, the limit as t approaches 0 from the right.

lim (t to 3, (1/x) dx )
t -> 0+

Solving the integral, we get

lim (ln|x| ) {evaluated from t to 3}
t -> 0+

Plugging in our integration bounds,

lim ( ln|3| - ln|t| )
t -> 0+

Since we are dealing with positive numbers (with the bounds being from 0 to 3), we don't even need those absolute value signs, because for positive numbers, |x| = x.


lim ( ln(3) - ln(t) )
t -> 0+

As t approaches 0 from the right, ln(t) approaches negative infinity. Also, ln(3) is a finite term, so we have the form

[ln(3) - (-∞)], or [ln(3) + ∞], which is ∞.

Answer 2

1/x is not define at the point 0. Therefore is not possible to integrate 1/x about 0.

Answer 3

The integral of 1/x is ln(x), right?

So your answer is ln(3) minus ln(0).

But if you look at the graph of ln(x), you'll find that it shoots downward quite rapidly as x approaches zero from the positive side. ln(3)-(-infinity)=infinity.

Does that work for you?