# How do you evaluate the integral #int 1/x# from [0,1]?

It's undefined.

By signing up, you agree to our Terms of Service and Privacy Policy

Diverging...

So hence the integral is diverging...

By signing up, you agree to our Terms of Service and Privacy Policy

The integral ( \int_{0}^{1} \frac{1}{x} , dx ) is improper because it has a singularity at ( x = 0 ). To evaluate it, we need to use the limit definition of the integral:

[ \int_{0}^{1} \frac{1}{x} , dx = \lim_{a \to 0^+} \int_{a}^{1} \frac{1}{x} , dx ]

This becomes:

[ \lim_{a \to 0^+} \int_{a}^{1} \frac{1}{x} , dx = \lim_{a \to 0^+} \left[ \ln|x| \right]_{a}^{1} ]

[ = \lim_{a \to 0^+} \left( \ln|1| - \ln|a| \right) ]

[ = \lim_{a \to 0^+} \left( 0 - \ln|a| \right) ]

[ = \lim_{a \to 0^+} -\ln|a| ]

[ = -\infty ]

Therefore, the integral ( \int_{0}^{1} \frac{1}{x} , dx ) diverges to negative infinity.

By signing up, you agree to our Terms of Service and Privacy Policy

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

- 98% accuracy study help
- Covers math, physics, chemistry, biology, and more
- Step-by-step, in-depth guides
- Readily available 24/7