Question

\[\int\limits_{-1}^{0}{2 \over 3x-5}dx\]

I think I know how to integrate this, but my substitution isn't working?

Answer 1

Remember the general form, on the picture

Answer 2

Yes, isn't the integral form 2/3 * ln(3x-5)?

Answer 3

yes

Answer 4

int[-1,0,(2)/(3x-5),x]
let
u=3x-5

dx=(du)/(3)

(2ln(|u|))/(3)

(2ln(|(3(0)-5)|))/(3)-[(2ln(|(3(-1)-5)|))/(3)]

ln((5^(2)/(3)*(1)/(4)

Answer 5

But how do you substitute the 0 and -1?

Answer 6

Integrate with respect to u, then substitute back u=3x-5

Answer 7

Then only use 0 and -1

Answer 8

since you have your integration, you will want to plug in the upper bound x value which is 0 to the integration.
And whatever you get subtract from it the value you get by pluging in the lower bound -1 to the integration.

Answer 9

Yes, I know that, but I end up with 2/3 * In(-5) - In(-8) and I don't know how to carry on from there..?

Answer 10

If it comes out to that then probably there is no solution

Answer 11

but the answer is 2/3In5 - In4 ~~~in my textbook

Answer 12

are you sure it is ln(3x - 5)

Answer 13

Yes

Answer 14

Thats wierd

Answer 15

i will redo it and see

Answer 16

\[\huge \int\limits_{-1}^{0}\frac{2}{3x-5}dx\]
Let u=3x-5
du=3dx
\[\frac{du}{3}=dx\]
Substitute du/3 to dx
\[\huge \int\limits\limits_{-1}^{0}\frac{2}{u}\frac{du}{3}\]
\[\huge \frac{2}{3}\int\limits\limits\limits_{-1}^{0}\frac{1}{u}du\]
\[\huge \frac{2}{3}[\ln(u)]^{0}_{-1}\]
u=3x-5
\[\huge \frac{2}{3}[\ln(3x-5)]^{0}_{-1}\]

\[\huge \frac{2}{3}[\ln(-5)-\ln(-8)]\]

Answer 17

I am still getting the same result, maybe that was a typo

Answer 18

Thanls Sam. That's what I got... but the answer in the textbook was different. Hmmm. Perhaps..?

Answer 19

Or the textbook is wrong

Answer 20

No typo. I know. I have the textbook right infront of me.

Answer 21

Are you learning on your own or is this a class

Answer 22

Wait for the natural log, inside is absolute

Answer 23

I'm learning on my own :(

Answer 24

That is amazing, keep it up. :)
If you dont mind how old are you

Answer 25

I can't say my age... but I'm trying to do 2 years of A-level in one for 4 subjects... And the school doesn't combine 2 years teaching in one, so I have to do 1 of the two years on my own...

Answer 26

\[\huge \int\limits_{-1}^{0}\frac{2}{3x-5}dx\]
Let u=3x-5
du=3dx
\[\frac{du}{3}=dx\]
Substitute du/3 to dx
\[\huge \int\limits\limits_{-1}^{0}\frac{2}{u}\frac{du}{3}\]
\[\huge \frac{2}{3}\int\limits\limits\limits_{-1}^{0}\frac{1}{u}du\]
\[\huge \frac{2}{3}[\ln(u)]^{0}_{-1}\]
u=3x-5
\[\huge \frac{2}{3}[\ln|3x-5|]^{0}_{-1}\]

\[\huge \frac{2}{3}[\ln5-\ln8]\]

\[\huge \frac{2}{3}[\ln\frac{5}{8}]\]

Answer 27

Sam, thanks so much for trying, but that doesn't seem to work as the answer either :/

Answer 28

maybe the book made a mistake?

Answer 29

whats the answer

Answer 30

Keep it up, i will also suggest that you use www.khanacademy.org he has free calculus tutorials.

Answer 31

2/3In5 - In4

Answer 32

Thanks deny!

Answer 33

\[\huge \frac{2}{3}[\ln5-\ln8]\]
\[\huge \frac{2}{3}\ln5-\frac{2}{3}\ln8\]
\[\frac{2}{3}\ln5-\ln8^{\frac{2}{3}}\]
\[\frac{2}{3}\ln5-\ln4\]

Answer 34

Ok, that's the only explanation :) Thanks!

Answer 35

no problem :)

Answer 36

@.Sam. good job, I didn't even think about that