# Thread: Can anyone help with this integration thanks

1. ## Can anyone help with this integration thanks

$$\int_{20}^{10} -1/x dx$$

2. ## Re: Can anyone help with this integration thanks

$\displaystyle \int -\dfrac 1 x~dx = -\ln(x)$

$\displaystyle \int_{20}^{10} -\dfrac 1 x~dx = -\ln(10)+\ln(20) = \ln\left(\dfrac{20}{10}\right)=\ln(2)$

3. ## Re: Can anyone help with this integration thanks

thanks I was confused with the + and - signs with the ln cheers

4. ## Re: Can anyone help with this integration thanks

woops actually my book says its -ln(2) is the book wrong?

5. ## Re: Can anyone help with this integration thanks

Originally Posted by bee77
woops actually my book says its -ln(2) is the book wrong?
the way you've written it $\ln(2)$ is correct.

Check the limits. If you have them reversed the answer will be $-\ln(2)$

6. ## Re: Can anyone help with this integration thanks

Originally Posted by romsek
$\displaystyle \int -\dfrac 1 x~dx = -\ln(x)$
$\displaystyle \int -\dfrac 1 x~dx = -\ln|x|$

It was a fluke that the answer worked out.

7. ## Re: Can anyone help with this integration thanks

Originally Posted by bee77
$\int_{20}^{10} -1/x dx$
\begin{align*}\int_{20}^{10} -1/x dx&=\int_{10}^{20} 1/x dx \\&=\left. {\log(x)} \right|_{10}^{20}\\&=\log(2) \end{align*}

Originally Posted by greg1313
$\displaystyle \int -\dfrac 1 x~dx = -\ln|x|$
It was a fluke that the answer worked out.
It was by no means a fluke. Anyone experienced enough would see that the domain of integration is a subset of the positive real numbers, therefore the absolute values is not necessary.

Moreover, when doing an integral as apposed to an antederivative one need not use a C.

8. ## Re: Can anyone help with this integration thanks

$-\int_{20}^{10} -1/x dx= -[\int_{10}^{20} 1/x dx]= \left[ln|x|\right]_{10}^{20}= [ln(20)- ln(10)]= ln(20/10)= ln(2)$

9. ## Re: Can anyone help with this integration thanks

Originally Posted by Plato
\begin{align*}\int_{20}^{10} -1/x dx&=\int_{10}^{20} 1/x dx \\&=\left. {\log(x)} \right|_{10}^{20}\\&=\log(2) \end{align*}

It was by no means a fluke. Anyone experienced enough would see that the domain of integration is a subset of the positive real numbers, therefore the absolute values is not necessary.

Moreover, when doing an integral as apposed to an antederivative one need not use a C.
You don't know what you're talking about. The *correct* antiderivative formula is as I stated with the absolute value bars. That is a separate issue from continuing into the next part
of the problem. Anyone experienced enough knows that. Therefore, *after* noticing that, then the qualification may be stated that parentheses are sufficient where it is placed into
the definite integral. So, yes, it was by means a fluke, because the argument without the absolute value didn't change anything. Please be careful next time with your logic.

Moreover, in my haste to correct him, I left off the "+ C" for the indefinite integral. It should be:

$\displaystyle \int -\dfrac 1 x~dx = -\ln|x| + C$