anonymous
  • anonymous
What is wrong with the following calculation? int_{-2}^{1}\frac{ 1 }{ x^4 }dx=\frac{ -1 }{ 3x^3 }]^1(-2)below=\frac{ -1 }{ 3 }-\frac{ 1 }{ 24 }=\frac{ -9 }{ 24 }
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
This is the question. \[\int\limits_{-2}^{1}\frac{ 1 }{ x^4 }dx\]
Zarkon
  • Zarkon
what happens when x=0
anonymous
  • anonymous
it would be undefined?

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Zarkon
  • Zarkon
so this integral is an improper integral
anonymous
  • anonymous
what? I don't quite understand...
Zarkon
  • Zarkon
was all that work above given and you are to tell why it is wrong?
Zarkon
  • Zarkon
or is it your work?
anonymous
  • anonymous
It's not my work, it's from the textbook
Zarkon
  • Zarkon
to use the FTC the integrand (the thing inside the integral) must be continuous on the interval you are integrating over
anonymous
  • anonymous
okay...
zepdrix
  • zepdrix
still confused? :o
anonymous
  • anonymous
yes... :(
zepdrix
  • zepdrix
1/x^4 has an asymptote at x=0, ya? :o
anonymous
  • anonymous
yes
zepdrix
  • zepdrix
so our function is not continuous over the given interval.|dw:1433734647150:dw|See the problem the 0 is creating for us? +_+
anonymous
  • anonymous
Oooh I see... I think I got now.. Thanks!
zepdrix
  • zepdrix
If you want to be more "mathy" about it, you can break up the integral at that 0 point,\[\Large\rm \int\limits_{-2}^1 \frac{1}{x^4}dx=\int\limits_{-2}^0 \frac{1}{x^4}dx+\int\limits_{0}^1 \frac{1}{x^4}dx\]But it's still improper since our function isn't defined at zero, so we introduce some limits,\[\Large\rm =\lim_{b\to 0}\int\limits\limits_{-2}^b \frac{1}{x^4}dx+\lim_{b\to0}\int\limits\limits_{b}^1 \frac{1}{x^4}dx\]And then uhhh, in your calculuations you should end up with some stuff... \(\Large\rm \infty\), that kinda stuff. Which will show you that it doesn't converge. I dunno, I like thinking about the graph better :P

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