## 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 }

1. anonymous

This is the question. $\int\limits_{-2}^{1}\frac{ 1 }{ x^4 }dx$

2. Zarkon

what happens when x=0

3. anonymous

it would be undefined?

4. Zarkon

so this integral is an improper integral

5. anonymous

what? I don't quite understand...

6. Zarkon

was all that work above given and you are to tell why it is wrong?

7. Zarkon

8. anonymous

It's not my work, it's from the textbook

9. Zarkon

to use the FTC the integrand (the thing inside the integral) must be continuous on the interval you are integrating over

10. anonymous

okay...

11. zepdrix

still confused? :o

12. anonymous

yes... :(

13. zepdrix

1/x^4 has an asymptote at x=0, ya? :o

14. anonymous

yes

15. zepdrix

so our function is not continuous over the given interval.|dw:1433734647150:dw|See the problem the 0 is creating for us? +_+

16. anonymous

Oooh I see... I think I got now.. Thanks!

17. 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