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 }

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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
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This is the question. \[\int\limits_{-2}^{1}\frac{ 1 }{ x^4 }dx\]
what happens when x=0
it would be undefined?

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Other answers:

so this integral is an improper integral
what? I don't quite understand...
was all that work above given and you are to tell why it is wrong?
or is it your work?
It's not my work, it's from the textbook
to use the FTC the integrand (the thing inside the integral) must be continuous on the interval you are integrating over
okay...
still confused? :o
yes... :(
1/x^4 has an asymptote at x=0, ya? :o
yes
so our function is not continuous over the given interval.|dw:1433734647150:dw|See the problem the 0 is creating for us? +_+
Oooh I see... I think I got now.. Thanks!
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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