Talk:Additive Interval Property of Definite Integrals

Regarding "There's a far simpler way of proving this than going to FTC", I often overlook obvious things, please show me the easier way when you have a chance. --GFauxPas 09:01, 22 January 2012 (EST)

If $a < b < c$ then $\displaystyle \int_a^b f\left({x}\right) \ \mathrm dx + \int_b^c f\left({x}\right) \ \mathrm dx = \int_a^c f\left({x}\right) \ \mathrm dx$ from the main result then:

$\displaystyle \int_a^b f\left({x}\right) \ \mathrm dx = \int_a^c f\left({x}\right) \ \mathrm dx - \int_b^c f\left({x}\right) \ \mathrm dx$

and so

$\displaystyle \int_a^b f\left({x}\right) \ \mathrm dx = \int_a^c f\left({x}\right) \ \mathrm dx + \int_c^b f\left({x}\right) \ \mathrm dx$

and the result holds. Saves having to get all the heavy machinery out. --prime mover 09:09, 22 January 2012 (EST)

Oh, I don't know how I missed that. Delete this page then, I added it to the other page. --GFauxPas 09:29, 22 January 2012 (EST)

Done. Made it a redirect. --prime mover 09:44, 22 January 2012 (EST)