# Thread: definite integration of log

1. ## 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

2. 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.

3. 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

4. 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?

5. 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.

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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

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