Question

i am having problems with this question, integral 4 to 8 ydy/y^2-2y-3, by expressing the integrand as a sum of partial fractions and evaluate the integrals, with the integration of rational functions by partial fractions

Answer 1

\(\large\color{slate}{\displaystyle\int\limits_{4}^{8}\frac{y}{y^2-2y-3}~dy}\)

Answer 2

factor the denominator

Answer 3

well i know you need to break the (y^2-2y-3) into the different things

Answer 4

you just need to tell me (for now) how to factor the \(\large\color{slate}{\displaystyle y^2-2y-3}\).

Answer 5

what are it's factors ?

Answer 6

wait you cant can you

Answer 7

yes, you can

Answer 8

\(\large\color{slate}{\displaystyle (-3)+(1)=-2
\\ (-3) \times (1)=-3}\)

Answer 9

is it (y-1)(y-1)?

Answer 10

no it is not

Answer 11

I gave you the hint

Answer 12

yeah i see what you did, im just not thinking very well

Answer 13

alright, when you factor it, tell me what factors you get.

Answer 14

ok its (y-3)(y+1)

Answer 15

yes.

Answer 16

\(\large\color{slate}{\displaystyle\int\limits_{4}^{8}\frac{y}{(y+1)(y-3)}~dy}\)

Answer 17

now, you need to solve the following:

\(\large\color{slate}{\displaystyle\frac{y}{(y+1)(y-3)}=\frac{A}{y+1}+\frac{B}{y-3}}\)

Answer 18

(note: this equation is true for any value of y)

Answer 19

yes i have that

Answer 20

now, multiply each term times (y+1)(y-3), and rewrite each term.

Answer 21

ok when i get to that part i need help with the intrgral part from 4 to 8
\

Answer 22

you are trying to skip steps/

Answer 23

well i meant, when i finished the part before the integral

Answer 24

lets just do that part right now, and then worry about the steps ahead

Answer 25

when you multiply the above (the last thing I wrote in latex) each term, times (y+1)(y-3) what do you get /

Answer 26

well i dont think they are righ but i have A = 3/4 and B = -3/2

Answer 27

\(\large\color{slate}{ \displaystyle \frac{y}{(y+1)(y-3)} = \frac{A}{y+1} +\frac{B}{y-3} }\)


\(\large\color{slate}{ \displaystyle \left(\frac{y}{(y+1)(y-3)} = \frac{A}{y+1} +\frac{B}{y-3} \right)\times [~(y+1)(y-3) ~] }\)


\(\large\color{slate}{ \displaystyle y=A(y-3)+B(y+1) }\)

Answer 28

ok i got that now

Answer 29

when y=-1
\(\large\color{slate}{ \displaystyle -1=A(-1-3)+B(-1+1) }\)
\(\large\color{slate}{ \displaystyle -1=A(-4)+B(0) }\)
\(\large\color{slate}{ \displaystyle -1=A(-4) }\)

\(\large\color{slate}{ \displaystyle \frac{1}{4}=A }\)

when y=3
\(\large\color{slate}{ \displaystyle y=A(y-3)+B(y+1) }\)
\(\large\color{slate}{ \displaystyle 3=A(3-3)+B(3+1) }\)
\(\large\color{slate}{ \displaystyle 3=A(0)+B(4) }\)
\(\large\color{slate}{ \displaystyle 3=B(4) }\)

\(\large\color{slate}{ \displaystyle \frac{3}{4}=B }\)

Answer 30

ok that is what i am getting at wasnt using -1, used 1

Answer 31

\(\large\color{slate}{\displaystyle\int\limits_{4}^{8}\frac{y}{(y+1)(y-3)}~dy}\)

\(\large\color{slate}{\displaystyle\int\limits_{4}^{8}\frac{A}{(y+1)}+\frac{B}{(y-3)}~dy}\)

\(\large\color{slate}{\displaystyle\int\limits_{4}^{8}\frac{(1/4)}{(y+1)}+\frac{(3/4)}{(y-3)}~dy}\)


then there a couple way to say it, which are all essentially same

(1)

\(\large\color{slate}{\displaystyle\frac{1}{4}\int\limits_{4}^{8}\left(\frac{1}{y+1}+\frac{3}{(y-3)}\right)~dy}\)



(2)

\(\large\color{slate}{\displaystyle\int\limits_{4}^{8}\frac{1}{4(y+1)}+\frac{3}{4(y-3)}~dy}\)

Answer 32

then as you can probably see, the answers will be natural logarithms.

Answer 33

yeah, so then just plug in 8 and 4 and you will get the answer?

Answer 34

you will need to integrate first.

Answer 35

by using u substitution?

Answer 36

well, you can use a substitution from y+1 and y-1 in the denominators, but in the cases of y+1 and y-1, if you try to set u=y+1 and/or u=y-1, then dy=du

when dy=du, that indicates no real need for substitution.

if you know what an integral of 1/y is, then you should be also able to tell me what an integral of 1/(y+1) and the integral of 1/(y-1) is

Answer 37

i need help with that, all i can think of is 1/(2y+y) and 1/(2y-2)

Answer 38

lets take it bit by bit.

Answer 39

What is an integral of 1/y ?

Answer 40

y^2/2?

Answer 41

that is integral of y, but what is integral of 1/y ?

Answer 42

1/2y^2

Answer 43

2/2y^2 actually

Answer 44

\(\large\color{black}{ \displaystyle \int \frac{1}{y} ~dy~~=\ln \left| y\right| \color{gray}{+C}}\)

Answer 45

and, thus
\(\large\color{black}{ \displaystyle \int \frac{1}{y\pm 1} ~dy~~=\ln \left| y\pm 1\right| \color{gray}{+C}}\)

Answer 46

ok so its ln y

Answer 47

yes

Answer 48

\(\large\color{black}{ \displaystyle \int \frac{1}{4(y+1)} +\frac{3}{4(y-3)} dy~~=\frac{1}{4}\ln \left| y+1\right|+\frac{3}{4}\ln \left| y-3\right| \color{gray}{+C}}\)

Answer 49

\(\normalsize\color{black}{ \displaystyle \int_{4}^{8} \frac{1}{4(y+1)} +\frac{3}{4(y-3)} dy~~=\left(~\frac{1}{4}\ln \left| y+1\right|+\frac{3}{4}\ln \left| y-3\right| ~\right){{\huge |}_{4}^{8}}}\)

Answer 50

now plug in the limits

Answer 51

ok thank you for all you help

Answer 52

you welcome!

Answer 53

ok i need help

Answer 54

nvm