# What is the integral from 0 to 16 of #1/(16-x) dx#?

The improper integral

Or, if you have extended the definition of integral to include improper integral, then the textbook should say that the integral diverges.

If your course has covered logarithms and improper integrals, then here is the work:

the integral diverges.

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The integral from 0 to 16 of ( \frac{1}{16-x} ) dx is ( \ln|16-x| ) evaluated from 0 to 16, which equals ( \ln|16-16| - \ln|16-0| ), simplifying to ( \ln|0| - \ln|16| ), further simplifying to ( -\ln|16| ), which is approximately ( -2.7726 ).

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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.

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