Prove that the function: l1-2xl has a Integral in the section(see in the picture)
thanks.
Did you mean to prove that the improper integral $\displaystyle \int\limits^\infty_{-\infty}|1-2x|\,dx$ exist? Because this integral
does not exist (i.e., it's divergent), since $\displaystyle |1-2x|=1-2x\Longleftrightarrow x\leq\frac{1}{2}$ , so
we can write $\displaystyle \int\limits^\infty_{-\infty}|1-2x|\,dx=\int\limits^{1/2}_{-\infty}(1-2x)\,dx+\int\limits^\infty_{1/2}(2x-1)\,dx$ , and both these integrals diverge.
Tonio
Using what basis? That's a continuous function and all continuous functions are integrable.
If you do not know that theorem, then the simplest way to show a function is "integrable" is to find its integral!
If x< 1/2, |1- 2x|= 1- 2x. Integrate that, including the "constant of integration" C. If x> 1/2, |1- 2x|= -(1- 2x)= 2x- 1. Integrate that, including the "constant of integration" D. At x= 1/2, those must be the same, allowing you to write either C in terms of D or D in terms of C so that you have only a single "constant of integration".