Integration of discontinuous function

  • #1

Homework Statement



∫ [(x + 2)(x - 2)] / (x - 2) dx from -3 to 3

Homework Equations





The Attempt at a Solution


∫ [(x + 2)(x - 2)] / (x - 2) dx from -3 to 3
= [ (x^2)/2 + 2x ] from -3 to 3


I know that this function is integrable, but I could not apply the 2nd part of fundamental theorem of calculus since the function has a removable discontinuity at x=2. So how do I proceed?
 

Answers and Replies

  • #2
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Think about what the integral tells you: the value of the integral is the area under the curve. Does the removable discontinuity cover any finite amount of area, over which it could affect the value of the integral?

(By the way, formally stated, the fundamental theorem of calculus does in fact apply to functions that have removable discontinuities.)
 
  • #3
575
76
Is it possible you're meant to treat this as an improper integral? Technically, the integrand that you're given is not defined on the entire interval of integration. For that reason alone, I wouldn't consider it to be integrable.
 
  • #4
SammyS
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Homework Statement



∫ [(x + 2)(x - 2)] / (x - 2) dx from -3 to 3

Homework Equations



The Attempt at a Solution


∫ [(x + 2)(x - 2)] / (x - 2) dx from -3 to 3
= [ (x^2)/2 + 2x ] from -3 to 3

I know that this function is integrable, but I could not apply the 2nd part of fundamental theorem of calculus since the function has a removable discontinuity at x=2. So how do I proceed?
Look at
[itex]\displaystyle \int_{-3}^{2 - |a|} \frac{(x + 2)(x - 2)}{x-2}\, dx[/itex]​
and
[itex]\displaystyle \int_{2 + |b|}^{3} \frac{(x + 2)(x - 2)}{x-2}\, dx \ .[/itex]​
 

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