Prove that the function: l1-2xl has a Integral in the section(see in the picture)

thanks.

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- Dec 13th 2010, 09:18 AMZOOZProve the Integral
Prove that the function: l1-2xl has a Integral in the section(see in the picture)

thanks. - Dec 13th 2010, 09:45 AMtonio

Did you mean to prove that the improper integral $\displaystyle \int\limits^\infty_{-\infty}|1-2x|\,dx$ exist? Because this integral

doesexist (i.e., it's divergent), since $\displaystyle |1-2x|=1-2x\Longleftrightarrow x\leq\frac{1}{2}$ , so**not**

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 - Dec 15th 2010, 05:15 AMZOOZ
hi i mean how i prove that this function have a Antiderivative (in the same section)

thanks. - Dec 15th 2010, 05:25 AMHallsofIvy
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".