modulus (please delete this post)

[burgerandcheese]

What am I missing? the answer key says k = 1/2 (e²+1) but that would mean that they did ln(2k-1) instead of ln|2k-1| ? why?
My teacher said it's because "the answer of ln can only be positive despite its input being positive or negative" what did she mean by that?

edit: nevermind, I just realised why after graphing 1/(2x-1).

 

[yoscar04]

Hi, if you try to compute the integral including the value x=0.5, the integral will diverge, as in the case where k=0.5(1-e^2).

 

[Deleted member 4993]

View attachment 20204

What am I missing? the answer key says k = 1/2 (e²+1) but that would mean that they did ln(2k-1) instead of ln|2k-1| ? why?
My teacher said it's because "the answer of ln can only be positive despite its input being positive or negative" what did she mean by that?

You say:

"the answer of ln can only be positive despite its input being positive or negative"

That is incorrect.

The explanation actually is that we cannot calculate "ln" of a negative number.

So the "argument" of the ln function - the 'a' in ln(a) - is always greater than zero.

If you take: k = 1/2 * (-e^2 +1) \(\displaystyle \to \) k = - 3.195 \(\displaystyle \to \) 2k -1 = - 7.389,

which makes argument of ln(2k-1) negative \(\displaystyle \to \) thus inadmissible.

 

[burgerandcheese]

You say:

"the answer of ln can only be positive despite its input being positive or negative"

That is incorrect.

The explanation actually is that we cannot calculate "ln" of a negative number.

So the "argument" of the ln function - the 'a' in ln(a) - is always greater than zero.

If you take: k = 1/2 * (-e^2 +1) \(\displaystyle \to \) k = - 3.195 \(\displaystyle \to \) 2k -1 = - 7.389,

which makes argument of ln(2k-1) negative \(\displaystyle \to \) thus inadmissible.

I see. But then why use ln(2k-1)? Why not ln|2k-1| ? I thought the integral of 1/x is ln|x|.

 

[Deleted member 4993]

I see. But then why use ln(2k-1)? Why not ln|2k-1| ? I thought the integral of 1/x is ln|x|.

Okay - now I understand the confusion.

1/(2x - 1) has a discontinuity at x = 1/2. Thus while integrating from 1 to 'k' - we cannot include the point x = 1/2.

Thus we cannot integrate this function from 1 to k = -e² (= ~-3.195)