Question

For integral (-5 to -20) for 1/x dx, why is the answer ln(1/4) ?

Answer 1

What's the derivative of lnx?

Answer 2

I understand that part, but I don't get how you can just change -5 to -20 to 5 to 20

Answer 3

ln doesn't exist on the negative side right?

Answer 4

\[\int\limits_{-5}^{-20} \frac{ 1 }{ x }dx = \int\limits_{5}^{20} \frac{ 1 }{ x }dx\] is this what you're trying to say

Answer 5

Note that \(\displaystyle \int \frac{1}{x}\,dx = \ln \color{red}|x\color{red}|+C\).

Answer 6

You flip integrals to make them negative -> positive and vise versa

Answer 7

the limits of integration

Answer 8

So it's the absolute signs?

Answer 9

Yes, they're important to have here.

Answer 10

\[\int\limits_{a}^{b} f(x) dx = F(b)-F(a) \implies -\int\limits_{a}^{b} f(x) = - (F(b)-F(a))\] but in your case it's absolute value and this is just the fundamental theorem of calculus

Answer 11

Do you know why we need the absolute values?

Answer 12

Not really.

Answer 13

the function 1/x is defined in (-5, -20) so there will be area under/above this curve too. you can find it using integrals or approximating

Answer 14

oh right, I am finding the area under 1/x, not ln|x|

Answer 15

You can't evaluate the logarithm without absolute values, I guess if you remember the piece wise function it would be easier to visualize

Answer 16

you're finding area under/above 1/x and it equals ln|x| as long as you stay away from 0

Answer 17

Yeah I got confused

Answer 18

When \(x>0\), \(\ln x\) is defined and thus \(\dfrac{d}{dx}\ln x = \dfrac{1}{x}.\) This then implies the indefinite integral formula \(\displaystyle \int\frac{1}{x}\,dx =\ln x+C\).

However, when \(x<0\), \(\ln x\) is no longer defined. We then note that if \(x<0\), then \(-x>0\). Hence, \(\ln (-x)\) is defined for \(x<0\) and we have \(\dfrac{d}{dx}\ln(-x)=\dfrac{1}{-x}\cdot (-1) = \dfrac{1}{x}\). This then implies the indefinite integral formula \(\displaystyle \int \frac{1}{x}\,dx = \ln(-x)+C\).

Putting the two pieces together, we see that
\[\int\frac{1}{x}\,dx = \begin{cases}\ln x + C & x>0\\ \ln(-x)+C & x<0\end{cases}\]
Recalling that \(|x|=\begin{cases}x & x\geq 0\\ -x & x< 0\end{cases}\), we can now see that \(\ln|x|=\begin{cases}\ln x & x>0\\ \ln(-x) & x< 0\end{cases}\).

Therefore, \(\displaystyle\int\frac{1}{x}\,dx = \begin{cases}\ln x + C & x>0\\ \ln(-x)+C & x<0\end{cases}=\ln|x|+C\).