What is the integral of #x/(x-2)# from 0 to 3?

Answer 1

Depending on where you are in your learning of calculus, the best answer may be "Is not defined" or the best answer may be "The integral diverges".

The definite integral over interval #[a,b]# of #f# is initially defined For a function #f# that includes #[a,b]# in its domain. That is: we start with a function #f# that is defined for all #x in [a,b]#
(Some treatments initially require the stronger condition: "Let #f# be continuous on #[a,b]#".)

Therefore, using the intial definition (before you learn about "Improper Inetgrals") the best answer is this integral is not defined.

Improper integrals extend the initial definition by allowing the endpoints #a,b# to be outside the domain of #f# (but on the 'edge' so we can look for limits) or for the interval to lack left and/or right endpoints (infinite intervals).

General Method

Replace the "problem number" (or #+- oo#) with a variable and evaluate the limit and the variable approaches the "problem number" (or #+- oo#).

If the limit exists, then we say that the integral converges. If the limit fail to exist (possibly by "being" infinite), then then integral diverges.

Example 1 To try to evaluate #int_0^3 x/(x-2) dx#, we split the problem into the two improper integrals:
#int_0^3 x/(x-2) dx = int_0^2 x/(x-2) + int_2^3 x/(x-2)#,
#int_2^3 x/(x-2) dx# is not defined by the first definition of definite integral, but the extension of the definition to improper integral requires us to evaluate:
#lim_(ararr2^+) int_a^3 x/(x-2) dx#
(Eric gives and excellent description of finding the indefinite integral, so I won't go through that again, though I'll write #ln(x-2)^2# in place of #2ln(x-2)#)
#lim_(ararr2^+) int_a^3 x/(x-2) dx =lim_(ararr2^+) (x+ln(x-2)^2)|_a^3#
#color(white)"sssssssssssssssss"# # = lim_(ararr2^+)((3+ln1)-(a+ln(a-2)^2))#
#color(white)"sssssssssssssssss"# # = lim_(ararr2^+)(3-a-ln(a-2)^2)#
But this limit does not exist. (#lim_(a rarr 2^+) ln(a-2) = -oo#)

So the integral diverges.

Because one of the two integrals needed diverges, there is no need to check the other.

#int_0^3 x/(x-2) dx# diverges. (Some would say "does not exist".)

That answers the question this is posted under.

Example 2

It will probably be helpful to many students to see an example of an improper integral that converges.

#int_0^1 1/sqrtx dx# is not defined in the initial definition of definite integral.

When we extend the definition to Improper integrals (I always want to say " so-called Improper Integrals")

We try to evaluate:

#lim_(ararr0^+) int_a^1 1/sqrtx dx#
#lim_(ararr0^+) int_a^1 1/sqrtx dx = lim_(ararr0^+) (2sqrtx)|_a^1#
#color(white)"ssssssssssssss"# # = lim_(ararr0^+) (2-2sqrta)#
#color(white)"ssssssssssssss"# # = 2#
The integral converges to #2# and we write:
#int_0^1 1/sqrtx dx = 2#
(Opinions vary on whether we should say "the integral equals #2#" or stick with "converges to #2#".)

If you have additional questions of would like to see additional examples, post new questions or send me a note and I'll suggest a question.

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Answer 2

To solve the integral of ( \frac{x}{x-2} ) from 0 to 3, we can split the integral into two parts using partial fraction decomposition. The integral will be equal to the integral of ( \frac{x-2+2}{x-2} ) from 0 to 3. This simplifies to the integral of ( \left(1 + \frac{2}{x-2}\right) ) from 0 to 3. Integrating each part separately gives ( x + 2\ln|x-2| ). Evaluate this expression at 3 and 0, then subtract the result at 0 from the result at 3. After substitution, the result will be ( (3 + 2\ln|1|) - (0 + 2\ln|2|) ), which simplifies to ( 3 + 2\ln(1) - 2\ln(2) ). Finally, ( \ln(1) = 0 ), so the result is ( 3 - 2\ln(2) ), approximately 0.3069.

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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.

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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