My Math Forum Integrating a discontinuous function

 Calculus Calculus Math Forum

 November 20th, 2017, 11:32 PM #1 Newbie   Joined: Apr 2016 From: Wonderland Posts: 14 Thanks: 0 Integrating a discontinuous function For this piecewise function, f(x) = -x for x > -1 -1 for x < -1 Let's say F(x) is the integral of f(x). At x = -1, is F(x) a discontinuous point, corner, or a tangent? I believe F(x) is a discontinuous point as I integrated f(x) respectively. As a result, I got -x^2/2 for x > -1 and -x for x < -1 So if I substitute -1 in, I get 2 different values, so there should be a discontinuity. However, I am told that it's a corner. Why?
 November 21st, 2017, 12:48 AM #2 Global Moderator   Joined: Dec 2006 Posts: 18,697 Thanks: 1525 You need to choose the constants of integration in such a way that the indefinite integral is continuous. It then has a corner at x = -1. Thanks from Country Boy
 November 21st, 2017, 04:58 PM #3 Newbie   Joined: Apr 2016 From: Wonderland Posts: 14 Thanks: 0 Hi Skipjack, Do you mean something like this? https://imgur.com/a/5dZyo
 November 22nd, 2017, 01:13 PM #4 Global Moderator   Joined: Dec 2006 Posts: 18,697 Thanks: 1525 Yes... for x = -1, -x²/2 + C$_1$ = -x + C$_2$, so C$_1$ = 1.5 + C$_2$ is needed.

 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post Lucas Webb Calculus 4 April 1st, 2016 04:37 AM ungeheuer Calculus 7 January 30th, 2013 05:59 AM ungeheuer Calculus 2 January 24th, 2013 07:31 AM sibtain125 Calculus 7 January 16th, 2012 06:56 PM stuffy Calculus 0 May 26th, 2008 06:42 AM

 Contact - Home - Forums - Cryptocurrency Forum - Top