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

##### 1 Answer
Aug 29, 2015

The improper integral ${\int}_{0}^{16} \frac{1}{16 - x} \mathrm{dx}$ diverges (no integral exists).

#### Explanation:

Because $16$ is not in the domain of $\frac{1}{16 - x}$, a textbook may say that the integral is "not defined" -- meaning that the text has not yet defined improper integrals.

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:

${\int}_{0}^{16} \frac{1}{16 - x} \mathrm{dx} = {\lim}_{b \rightarrow {16}^{-}} {\int}_{0}^{b} \frac{1}{16 - x} \mathrm{dx}$

= lim_(brarr16^-) -ln(16-x) ]_0^b

$= {\lim}_{b \rightarrow {16}^{-}} \left(- \ln \left(16 - b\right) + \ln \left(16\right)\right)$

As $b \rightarrow {16}^{-}$,

the argument $\left(16 - b\right) \rightarrow {0}^{+}$,

so we have ln(16-b)rarr-oo), and

the integral diverges.