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

Answer 1

The improper integral #int_0^16 1/(16-x) dx# diverges (no integral exists).

Because #16# is not in the domain of #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 1/(16-x) dx = lim_(brarr16^-) int_0^b 1/(16-x) dx#

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

#= lim_(brarr16^-) (-ln(16-b) + ln (16))#

As #brarr16^-#,

the argument #(16-b)rarr0^+#,

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

the integral diverges.

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Answer 2

Scarlett Allison

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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Answer from HIX Tutor

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.