Question

find the integral of 1/((x^2)-4) from 1 to 4

Answer 1

\[\int\limits_{1}^{4}\frac{ dx }{ x ^{2}-4 }\]
should be minus 4

Answer 2

+4 or -4?

Answer 3

sorry not +4 its minus 4

Answer 4

I need to split the integral and find with limits.

Answer 5

I didn't even notice it was improper :/

Answer 6

yeah that got me too

Answer 7

i really just need to see if it diverges or converges. If it converges to what value

Answer 8

In general you have:
\[\int\limits_a^b \frac{dx}{x-c} = \lim_{\beta \rightarrow c^+} \int\limits_a^{\beta}\frac{dx}{x-c}+\lim_{\beta \rightarrow c^-}\int\limits_{\beta}^b \frac{dx}{x-c}; c \in [a,b]\]

Answer 9

If I remember my calc 2 correctly

Answer 10

Ok I think this looks right

Answer 11

I just don't know what to do after I setup the limits

Answer 12

So you would still do partial fraction decomp: i.e.,
\[\frac{1}{x^2-4}=\frac{1}{4(x-2)}-\frac{1}{4(x+2)}\]
And then do the integral, then take the limits. And you'll see the limits are infinite because you'll get a ln|x-2| but when x goes to 2 you get an "infinity" ("negative one" at that).

Answer 13

Which would imply this integral doesn't converge, this confirms that:
http://www.wolframalpha.com/input/?i=integral+of+(x%5E2-4)%5E(-1)+from+1+to+4

Answer 14

\[\lim_{1 \rightarrow a} \int\limits_{1}^{2}\frac{ dx }{ x ^{2}-4 }+\lim_{a \rightarrow 4}\int\limits_{2}^{4}\frac{ dx }{ x ^{2}-4 }\]

Answer 15

where did the 4 in the denominators come from?

Answer 16

oh nvm A and B both equal 1/4 right?

Answer 17

\[\lim_{n \rightarrow 2} \int\limits_{1}^{n} \frac{ 1 }{ x^2-4 } +\lim_{n \rightarrow 2} \int\limits_{n}^{4} \frac{ 1 }{ x^2-4 } \]

Answer 18

A is +1/4 and B is -1/4.
If you have:
\[\frac{1}{(x-a_1)(x-a_2)...(x-a_n)}=\frac{B_1}{x-a_1}+\frac{B_2}{x-a_2}+...+\frac{B_n}{x-a_n}\]
This implies:
\[B_1=\frac{1}{(a_1-a_2)(a_1-a_3)...(a_1-a_n)}; B_2=\frac{1}{(a_2-a_1)(a_2-a_3)...(a_2-a_n)}\]
Etc...
So:
\[\frac{1}{x^2-4}=\frac{B_1}{x-2}+\frac{B_2}{x+2} \implies B_1=\frac{1}{2+2}=1/4; B_2=\frac{1}{-2-2}=-1/4\]

Answer 19

Got it the answer thanks