02-28-2024, 07:00 PM
02-29-2024, 08:55 AM
(02-28-2024 07:00 PM)ReinXXL Wrote: [ -> ]integral(ln(x+2)dx
answer: x*ln(x+2)-x+2*ln(x+2)-2
why is there a -2 at the end?
Why not? Indefinite integrals are always +/- any arbitrary constant value, of which "-2" is a special case. I assume this results from Xcas implementation.
02-29-2024, 02:25 PM
I had the same thought. Was wondering how/why XCAS came up with a constant equaling 2 as opposed to something else!
02-29-2024, 02:46 PM
Maybe it is the airspeed velocity of an unladen African sparrow?
02-29-2024, 03:42 PM
(02-29-2024 02:46 PM)KeithB Wrote: [ -> ]Maybe it is the airspeed velocity of an unladen African sparrow?
And because the airspeed velocity is negative (-2), perhaps it is flying backwards :)
02-29-2024, 04:31 PM
Why -2?
You can rewrite the answer x*ln(x+2)-x+2*ln(x+2)-2 as (x+2)*ln(x+2)-(x+2).
You can rewrite the answer x*ln(x+2)-x+2*ln(x+2)-2 as (x+2)*ln(x+2)-(x+2).
02-29-2024, 04:42 PM
(02-28-2024 07:00 PM)ReinXXL Wrote: [ -> ]why is there a -2 at the end?
We can consider the singularity at \(x=-2\) a natural lower bound of the definite integral.
This choice of the integral constant makes it \(0\) at that value:
\(
\begin{align}
F(x)
&= \int_{-2}^{x} \log(t+2) \; \mathrm{d}t \\
\\
&= (t+2) \log(t+2) - t \Big|_{-2}^x \\
\\
&= (x+2) \log(x+2) - x - 2 \\
\end{align}
\)
02-29-2024, 05:47 PM
02-29-2024, 06:23 PM
Not really mysterious, it's a linear change of variable.