Question

Need help evaluating an improper integral :/

Answer 1

I tried to integrate and then take the limit. I got 4/3, but that is not right according to the computer... This is the first time I've encountered this kind of question... A little stumped.

Answer 2

i suggest you split into partial fractions for the function

Answer 3

\[4/(2x-3)^{2} = A/(2x-3) + (B/(2x-3)\]

Answer 4

using u-subt it will work

Answer 5

yep... u-sub

Answer 6

4/(2x-3)^2=( Ax+B)/(2x-3)^2

Answer 7

\[4/((2x-3)(2x-3)) = (A/(2x-3))+(B/(2x-3))\]

Answer 8

I'm just a little confused about how I approach this question. I thought maybe I was supposed to do this:
lim t->infinity \[\int\limits_{t}^{1} 4/(2x-3)^2\]

Answer 9

I mean, it's probably wrong, but I was watching a whole tutorial and he did something like that.... Why do I split it into a partial fraction?

Answer 10

take out the function first to simplify it, coz subst. can't be used here

Answer 11

don't do partial fractions, use u- subs , what u get ?
then we'll worry about limits

Answer 12

put 2x-3=t and solve

Answer 13

@earthcitizen, Not to undervalue your help, but I feel like u-substitution would work fine, wouldn't it? because u=2x-3 simplifies nicely to 1/2du=dx.... Anyway, I integrated and got -4/(2x-3)

Answer 14

wait, scratch that.... forgot something lol

Answer 15

yes thats right..Now put your limits

Answer 16

kk, so if you subst. you get (-2/u)+c
partial fractions might be too complex

Answer 17

But its -2/2x-3

Answer 18

yeah I forgot about the 1/2..

Answer 19

@antomic, kk good point the 2 cancels out the 4 to give =(-2/2x-3)+c

Answer 20

ok so now what? do I take the limit?

Answer 21

yup take limits

Answer 22

ok I may need a little refresher here... so for -2/(2x-3) when x approaches negative infinity, the 2x is going to basically become zero right?

Answer 23

yep

Answer 24

so my answer is -2/-3 = 2/3?

Answer 25

uhm, when you divide by infinity you get zero, when you multiply you get infinity

Answer 26

yes thats

Answer 27

it