Question

integral from -2 to 2 of 1/(x^2-4)
i spit into 2 integrals and do trig sub x=2sec@ @=theta. I get to -1/2 lim t->-2+ of [ln|csc@+cot@|]from t to 0. But when i turn my @ back into x's i get ln[x/(sqrt(x^2-4) + 2/(sqrt(x^2-4)) and now 0 is undefined since it produces a negative under the sqrt... what do i do?

Answer 1

\[\int\limits_{-2}^{2}\frac{ dx }{ x^2-4 }\]

Answer 2

ick

Answer 3

left side=\[\frac{ 1 }{ 2 }\lim_{t \rightarrow -2+} \int\limits_{t}^{0}[-\ln|\csc (\theta)+\cot(\theta)]\]

Answer 4

how did you get that anti derivative?

Answer 5

ah the integral should go away my bad

Answer 6

it was integral of csc@ at some point

Answer 7

but should be [-ln|csc@+cot@] evaluated from t to 0

Answer 8

using partial fractions you get
\[\frac{1}{4}\int \frac{1}{x-2}+\frac{1}{x+2}dx\]

Answer 9

hm i used trig sub was that incorrect?

Answer 10

i dont see any cotangent, but i suck at these

Answer 11

Oh interesting :O
Your steps look fine edcion.
I wasn't sure what the problem was until I threw it into Wolfram.

The integral does not converge over that interval.

Answer 12

i get the log, and no chance for it to converge

Answer 13

i am asked to evauluate however -_-

Answer 14

you cannot

Answer 15

Good luck

Answer 16

the integral (clearly improper as you wrote) does not exist, i.e. that limit does not exist

Answer 17

my problem is when i turn my thetas back into x's i get this : for left side only: \[-\frac{ 1 }{ 2 }\lim_{t \rightarrow -2+}[\ln|\frac{ x }{ \sqrt(x^2-4) }+ \frac{ 2 }{ \sqrt(x^2-4) } t \to 0\]

Answer 18

but can the limit be -inf or positive inf?

Answer 19

that last thing should be t to 0 in which case i have negative sqrt in the denominators of both when evaluating at 0

Answer 20

i am confused about the method to get your integral, but i am not saying it is wrong
i got
\[\frac{1}{4}(\log(x-2)-\log(x+2))\] but in either form it should be clear that the limit as x approaches -2 does not exist
also as x approaches 2
the integral does not converge

Answer 21

Sub x=i2tan(theta)
Idm try it lol

Answer 22

ug i don't even know how to use imaginary numbers in this shiz yet -______-

Answer 23

ill try partial fractions

Answer 24

it is really snappy with partial fractions
integral still does not converge though

Answer 25

right but i need to know then if it is positive or negative infinity (what the prof wants)

Answer 26

evaluating the antiderivative satellite gave you will tell you that it diverges to -infinity

Answer 27

lol so will a picture!

Answer 28

not really

Answer 29

lol first i can't draw the picture, second still need to do it algebraically

Answer 30

if you knew it diverged first then a picture will tell you that it must be -infinity

Answer 31

im trying now with partial fractions

Answer 32

yes that is what i meant
it is pretty clearly negative on that interval

Answer 33

yes it is

Answer 34

you know how to do partial fractions the snap way?

Answer 35

hm a/(x+2) + b/(x-2)

Answer 36

yeah and to find \(a\) do this: since in \(\frac{a}{x+2}\) \(x\) cannot be \(-2\) put your finger over the factor \(x+2\) is
\[\frac{1}{(x+2)(x-2)}\]
and replace \(x\) by \(-2\)
you get
\[\frac{1}{\cancel{(x+2})(-2-2)}=-\frac{1}{4}\]

Answer 37

for \(b\) do the same thing, only put your finger over \((x-2)\) and replace \(x\) by \(2\)
you get
\[\frac{1}{(2+2)\cancel{(x-2)}}=\frac{1}{4}\]

Answer 38

the left side so far is convergent

Answer 39

since the ln's have absolute value

Answer 40

no

Answer 41

\[\lim_{x\to -2^+}\ln(x+2)\] does not exist

Answer 42

integral of 1/x is ln|x|

Answer 43

what about the absolute values?

Answer 44

nvm lol

Answer 45

i was thinking that ln|x+2| as x approaches -2+ is zero

Answer 46

@satellite73 if your thing was correct. 1/4(ln|x-2|-ln|x+2|) then it would be positive inf right? since the - in front of the ln makes -inf turn into positive inf

Answer 47

because as x-> -2+ ln|x+2| -> -inf

Answer 48

or ln|approaches 0| -> -inf