# definite integration of log

• Oct 29th 2009, 10:28 PM
fvaras89
definite integration of log
Hi i've got this question of definite integration (1/5x-6) dx where x=0 from x=1

Ive got [1/5 log (5x+6)] x=1 x=0

= [1/5 log (5+6)] - [1/5log(0+6)]
= 0.208 - 0.1556
=0.0526

Am i supposed to have it as a decimal.. and how do i avoid using a decimal cos my friend got an answer of -log(6)/5

And sorry not sure how to do the integration sign on the computer haha
• Oct 30th 2009, 01:34 AM
fvaras89
Is there some sort of formula to which we keep log in the equation? as the final answer consists of it.

Or can i simply put it into the calculator leaving it as a decimal.
• Oct 30th 2009, 02:12 AM
e^(i*pi)
Quote:

Originally Posted by fvaras89
Hi i've got this question of definite integration (1/5x-6) dx where x=0 from x=1

Ive got [1/5 log (5x+6)] x=1 x=0

= [1/5 log (5+6)] - [1/5log(0+6)]
= 0.208 - 0.1556
=0.0526

Am i supposed to have it as a decimal.. and how do i avoid using a decimal cos my friend got an answer of -log(6)/5

And sorry not sure how to do the integration sign on the computer haha

I assume you got $\displaystyle F(x) = \frac{1}{5}log(5x+6)$ between 0 and 1.

$\displaystyle \frac{1}{5}log(5+6) - \frac{1}{5}log(0x+6) = \frac{1}{5}(log(11)-log(6))$

1/5 is a factor so we can factor that out.
Leave the logs in unless specifically asked to round to a decimal as the exact answer leaves the logs in. You can use the laws of logs to simplify

$\displaystyle log_c(a) - log_c(b) = log_c \left(\frac{a}{b}\right)$

I've no idea how your friend got their answer based on that simplification. What was the original question because it's also unusual to see logs that are not base e in integration
• Oct 30th 2009, 02:28 AM
fvaras89
Quote:

Originally Posted by e^(i*pi)
I assume you got $\displaystyle F(x) = \frac{1}{5}log(5x+6)$ between 0 and 1.

$\displaystyle \frac{1}{5}log(5+6) - \frac{1}{5}log(0x+6) = \frac{1}{5}(log(11)-log(6))$

1/5 is a factor so we can factor that out.
Leave the logs in unless specifically asked to round to a decimal as the exact answer leaves the logs in. You can use the laws of logs to simplify

$\displaystyle log_c(a) - log_c(b) = log_c \left(\frac{a}{b}\right)$

I've no idea how your friend got their answer based on that simplification. What was the original question because it's also unusual to see logs that are not base e in integration

the original question is definite integral of (1/5x-6) dx where x=0 from x=1.

So from your log laws. i got log(11/6) is that right?
• Oct 30th 2009, 03:18 AM
e^(i*pi)
Quote:

Originally Posted by fvaras89
the original question is definite integral of (1/5x-6) dx where x=0 from x=1.

So from your log laws. i got log(11/6) is that right?

Assuming the equation in post 1 is correct then it should be.

------------------------------------------------------------

I note you've put -6 in the original question and +6 in the OP, which is correct? I've took -6 but I can edit if it's +6

$\displaystyle \int^1_0\,\frac{1}{5x-6}\,dx$

I would use a u substitution to make it easier

$\displaystyle u = 5x-6 \: \rightarrow \: du = 5dx \: \rightarrow \: dx = \frac{1}{5}du$

Re-evaluate the limits in terms of u

$\displaystyle u = 5(1)-6 = -1$
$\displaystyle u = 5(0) - 6 = -6$

So we get
$\displaystyle \frac{1}{5} \int^{-1}_{-6}\, \frac{1}{u}\, du$

$\displaystyle = \frac{1}{5}[ln|u|]$ between -1 and -6

$\displaystyle \frac{1}{5}(ln|-1| - ln|-6|) = -\frac{ln6}{5}$