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17Calculus - Improper Integrals - Internal Discontinuity

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On this page we discuss the third and last type of improper integral, integrals that are not continuous inside the limits of integration.

Case 3 - Discontinuity Inside the Interval

Cases 1 and 2 above were very similar and dealing with them used the similar technique of taking the limit of one of ends of the interval. Now, Case 3 may seem very different but handling it is very straightforward and, if you understand Cases 1 and 2, you will be able to deal with Case 3 easily. Case 3 involves discontinuities INSIDE the interval, i.e. not at one of the endpoints or limits of integration. To handle this case, you just break the integral again (like you did in Case 2) but this time you break it AT THE POINT OF DISCONTINUITY (not at just any point). When you do, you end up with something like this.

Under These Conditions

\(a\) and \(b\) are finite

\(f(x)\) is continuous on the interval \([a,b]\) but discontinuous at \(x=c\), where \(a < c < b\)

This Equation Holds

\(\displaystyle{ \int_{a}^{b}{f(x) ~ dx} = }\) \(\displaystyle{ \int_{a}^{c}{f(x) ~ dx} + \int_{c}^{b}{f(x) ~ dx} }\)

Since \(f(x)\) is discontinuous at \(x=c\), you now have two integrals that fit Case 2. So you can use those techniques to evaluate the integrals. Pretty cool, eh? An example might be \(\displaystyle{ \int_{-1}^{1}{1/x~dx} = \int_{-1}^{0}{1/x~dx} + \int_{0}^{1}{1/x~dx} }\).

Additional Situations You Will See
1. Another minor twist might be that you have an infinite limit at one or both ends AND a discontinuity in the interval. In this case, break the integral at the discontinuity and drop back to Cases 1 and 2 until you have only one 'problem' in each integral. You probably also know what to do with multiple discontinuities, right? Just break the integral at each discontinuity.
2. You may also have more than one discontinuity in the interval. As long as you have a finite number of discontinuties, just break the integral into more than two integrals at each discontinuity. Make sure that you end up with only one 'problem' point for each integral.

Okay, try these practice problems before going on.

Practice

Unless otherwise instructed, evaluate these integrals using proper notation. Give your answers in exact, simplified and factored form.

\(\displaystyle{ \int_{-2}^{3}{ \frac{1}{x^3} dx } }\)

Problem Statement

Evaluate \(\displaystyle{ \int_{-2}^{3}{ \frac{1}{x^3} dx } }\) giving your answer in exact, simplified and factored form. Make sure to use correct notation in every step of your work.

Solution

The Organic Chemistry Tutor - 1307 video solution

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\(\displaystyle{ \int_{-1}^{1}{ x^{-2/3} ~dx } }\)

Problem Statement

Evaluate \(\displaystyle{ \int_{-1}^{1}{ x^{-2/3} ~dx } }\) giving your answer in exact, simplified and factored form. Make sure to use correct notation in every step of your work.

Solution

MIT OCW - 1480 video solution

video by MIT OCW

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\(\displaystyle{ \int_{0}^{33}{ \frac{1}{\sqrt[5]{x-1}} dx } }\)

Problem Statement

Evaluate \(\displaystyle{ \int_{0}^{33}{ \frac{1}{\sqrt[5]{x-1}} dx } }\) giving your answer in exact, simplified and factored form. Make sure to use correct notation in every step of your work.

Solution

The Organic Chemistry Tutor - 3565 video solution

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\(\displaystyle{ \int_{-1}^{1}{ \frac{e^x}{e^x-1} ~dx } }\)

Problem Statement

Evaluate \(\displaystyle{ \int_{-1}^{1}{ \frac{e^x}{e^x-1} ~dx } }\) giving your answer in exact, simplified and factored form. Make sure to use correct notation in every step of your work.

Solution

In the video, he incorrectly evaluates \(\ln|e^x-1|\) at \(x=-1\) as \(\ln(2)\). It should be \(\ln|e^{-1}-1| = \ln|1/e-1|\). This could be simplified to \(\ln|1-e|-1\). However, this mistake does not change his final answer, which is correct, that the integral evaluates to \(-\infty\) and therefore the integral diverges.

PatrickJMT - 1462 video solution

video by PatrickJMT

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\(\displaystyle{ \int_1^{11}{ \frac{dx}{(x-3)^{2/3}} } }\)

Problem Statement

Evaluate \(\displaystyle{ \int_1^{11}{ \frac{dx}{(x-3)^{2/3}} } }\) giving your answer in exact, simplified and factored form. Make sure to use correct notation in every step of your work.

Final Answer

\(3(2^{1/3})+6\)

Problem Statement

Evaluate \(\displaystyle{ \int_1^{11}{ \frac{dx}{(x-3)^{2/3}} } }\) giving your answer in exact, simplified and factored form. Make sure to use correct notation in every step of your work.

Solution

The domain of the integrand is all reals except for \(x=3\). The discontinuity is within the interval of integration, so we need to break the integral into two integrals at that point.
\(\displaystyle{\int_1^{11}{ \frac{dx}{(x-3)^{2/3}} } = }\) \(\displaystyle{ \int_1^{3}{ \frac{dx}{(x-3)^{2/3}} } + \int_3^{11}{ \frac{dx}{(x-3)^{2/3}} } }\)
Now we rewrite the integrals with limits, so that we can integrate using the FTC.
\(\displaystyle{\lim_{b\to3^-}{ \left[ \int_1^{b}{ \frac{dx}{(x-3)^{2/3}} } \right] } + \lim_{a\to3^+}{ \left[ \int_a^{11}{ \frac{dx}{(x-3)^{2/3}} } \right] } }\)
To save space, we will integrate the indefinite integral.

\(\displaystyle{ \int{ \frac{dx}{(x-3)^{2/3}} } }\)

\( u=x-3 \to du=dx \)

\( \int{ (u)^{-2/3} ~du } \)

\(\displaystyle{ \frac{u^{1/3}}{1/3} }\)

\( 3(x-3)^{1/3} \)

Now, plug this result back into the limit equations.

\(\displaystyle{ 3\lim_{b\to3^-}{ (x-3)^{1/3}|_1^b } + 3\lim_{a\to3^+}{ (x-3)^{1/3}|_a^{11} } }\)

\(\displaystyle{ 3\lim_{b\to3^-}{ [ (b-3)^{1/3} - (1-3)^{1/3} ] } + }\) \(\displaystyle{ 3\lim_{a\to3^+}{ [ (11-3)^{1/3} - (a-3)^{1/3} ] } }\)

\(\displaystyle{ -3(-2)^{1/3}+3(8)^{1/3} = 3(2^{1/3})+6 }\)

Final Answer

\(3(2^{1/3})+6\)

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Practice Instructions

Unless otherwise instructed, evaluate these integrals using proper notation. Give your answers in exact, simplified and factored form.

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