Question

integral
-2 to 3 of ((1/x^2) +x)dx

Answer 1

\[\int\limits_{-2}^{3}\left( \frac{ 1 }{ x^2}+x \right)dx\]

Answer 2

I might have posted this one before but I don't recall. Calculator won't evaluate this integral. I doublechecked the graph, is it because 0 is included

Answer 3

and 0 is not defined in the domain

Answer 4

you have a problem at 0. break it up into two intervals... -2 to 0 and 0 to 3. you'll have to take the limits at 0.

Answer 5

you gotta do the improper integral

Answer 6

You're using the wrong calculator, try this one:

http://www.wolframalpha.com/input/?i=integral%20from%20-2%20to%203%20(x%5E-2%2Bx)dx&t=crmtb01

Answer 7

\[\int\limits_{-2}^{3}\left( \frac{ 1 }{ x^2 }+x \right)dx=\lim_{t \rightarrow 0}\left[ \int\limits_{-2}^{t}\left( \frac{ 1 }{ x^2 }+x \right)dx+\int\limits_{t}^{3}\left( \frac{ 1 }{ x^2 }+x \right)dx \right]\]

Answer 8

make sense?

Answer 9

yes, but I think this problem is in the incorrect place. I believe improper integrals is not in my course but the next course. Thanks for the set up

Answer 10

first sem calc? should be in there

Answer 11

I didn't see em till calc II

Answer 12

not label as improper integral or is this not improper integral?

Answer 13

I am interested in seeing how to finish solving it using the limit approach. If nothing else, just for fun :)

Answer 14

In calculus, an improper integral is the limit of a definite integral as an endpoint of the interval(s) of integration approaches either a specified real number or ∞ or −∞ or, in some cases, as both endpoints approach limits - Wikipedia
so technically it's not an improper integral as the problem is presented, but if you want to show that it diverges, you need to make it into two improper integrals

Answer 15

you do as normal but you take the limit when evaluating at the trouble point

Answer 16

see I didn't see any diverges or converges problems until Calculus III

Answer 17

i find this site much better for this level...

http://tutorial.math.lamar.edu/Classes/CalcII/ImproperIntegrals.aspx

Answer 18

I just wanted to explain why it wasn't improper as it was originally stated, but we made it improper

Answer 19

np thanks

Answer 20

\[\lim_{t \rightarrow 0}\int\limits_{-2}^{t}\left( \frac{ 1 }{ x^2 }+x \right)dx=\lim_{t \rightarrow 0}\left( -\frac{ 1 }{ x } +\frac{ x^2 }{ 2 }\right)_{-2}^{t}=\]

Answer 21

check out Paul's online notes (previously posted). really quite good.

Answer 22

thanks again.

Answer 23

you're welcome!
btw, i only did part of the integral above, but you probably realized that!

Answer 24

that is ok, gotta run an errand. Be back in a little bit....sorry I can't stay