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anonymous
 one year ago
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 }
anonymous
 one year ago
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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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0This is the question. \[\int\limits_{2}^{1}\frac{ 1 }{ x^4 }dx\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0it would be undefined?

Zarkon
 one year ago
Best ResponseYou've already chosen the best response.3so this integral is an improper integral

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0what? I don't quite understand...

Zarkon
 one year ago
Best ResponseYou've already chosen the best response.3was all that work above given and you are to tell why it is wrong?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0It's not my work, it's from the textbook

Zarkon
 one year ago
Best ResponseYou've already chosen the best response.3to use the FTC the integrand (the thing inside the integral) must be continuous on the interval you are integrating over

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.21/x^4 has an asymptote at x=0, ya? :o

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2so our function is not continuous over the given interval.dw:1433734647150:dwSee the problem the 0 is creating for us? +_+

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oooh I see... I think I got now.. Thanks!

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2If 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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