Question

Evaluate the double integral S(-5)--->(6)S(10)--->12 (1/(x^2-3x+2))dxdy using partial fractions

Answer 1

\[\int\limits_{}^{}\int\limits_{}^{}\frac{1}{x^2-3x+2} dx dy \]
I don't understand S(-5)--->(6)S(10)--->12
I have no clue what that means.

Answer 2

I can help you on the partial fraction part though.

Answer 3

You should be able to factored x^2-3x+2

Answer 4

That is what the integrals go from. The first integrals range (dx) is 10 to 12 and the second integrals range is 5 to 6.

Answer 5

So you mean for those S's to be integral signs.

Answer 6

But anyways let's factor x^2-3x+2
That is the first step

Answer 7

When i did the partial fractions, I got -2(x-1)^-1 + 2(x-2)^-1

Answer 8

Yes they were supposed to be integral signs. haha

Answer 9

\[\frac{1}{x^2-3x+2}=\frac{A}{(x-2)}+\frac{B}{(x-1)}\]
Combine those fractions we get:
\[\frac{1}{x^2-3x+2}=\frac{A(x-1)+B(x-2)}{(x-2)(x-1)}\]
Since we have the bottoms the sign and we want this to be the same fraction then we need to find for what A and B does
1=A(x-1)+B(x-2)
1=(A+B)x+(-A-2B)
There are no x's on that one side so we know A+B=0
We have the constant 1 on that one side so we know we need -A-2B=1
So is this what you did to get A and B?

Answer 10

I'm not getting those 2's you got.

Answer 11

Do you see what I mean?

Answer 12

Thats how I did it, but when I solved for B i got 2.

Answer 13

THe other two was a mistake.

Answer 14

So A+B=0 => -A=B right?

Answer 15

In our second equation we have -A-2B=1

Answer 16

Replace -A with B.

Answer 17

Solve for B.

Answer 18

B-2B=-1

Answer 19

Well B-2B=1

Answer 20

So -1B=1
which implies B=-1

Answer 21

A should be easy to find now since you know that A and B are opposites of each other.

Answer 22

A=1

Answer 23

Ok so we have this now:
\[\int\limits_{-5}^{6}\int\limits_{10}^{12}( \frac{1}{x-2}+\frac{-1}{x-1}) dx dy\]

Answer 24

You can find what I'm about to write by substitution or you can remember this formula for the antiderivative for 1/(x-c).
\[\int\limits_{}^{}\frac{1}{x-c}dx=\ln|x-c|+k \text{ where c is a constant } \]

Answer 25

ok so after the first integral i got (ln 11 - ln 10)-(ln9-ln8)

Answer 26

uhhh.
do i have the intervals backwards or do you ?

Answer 27

no its correct.

Answer 28

So 10 to 12 not 12 to 10 right?

Answer 29

\[[\ln|x-2|-\ln|x-1|]_{10}^{12}=(\ln(10)-\ln(11))-(\ln(8)-\ln(9))\]

Answer 30

ya, andthen after the second integral it would just be equal to that right?

Answer 31

Well no...

Answer 32

\[\int\limits_{-5}^{6}c dy=cy|_{-5}^{6}=c(6)-c(-5)=6c+5c\]

Answer 33

or 11c

Answer 34

oops i meant 5 to 6. Not -5 to 6...