My Math Forum Bounds of integration with inequality constraint

 Probability and Statistics Basic Probability and Statistics Math Forum

 November 18th, 2015, 12:57 PM #1 Newbie   Joined: Nov 2015 From: Boynton Beach, FL Posts: 5 Thanks: 0 Bounds of integration with inequality constraint Hi, all. If the integral of a function when y>1 goes from 1 to infinity, then why does an integral of a function for 0
 November 18th, 2015, 01:15 PM #2 Global Moderator   Joined: May 2007 Posts: 6,807 Thanks: 717 It doesn't sound right. Clarify! Actually the integral can be defined to be over a part of the domain.
 November 18th, 2015, 01:46 PM #3 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,675 Thanks: 2655 Math Focus: Mainly analysis and algebra The most likely reason is that the function is zero between 0 and 1.
 November 18th, 2015, 02:03 PM #4 Newbie   Joined: Nov 2015 From: Boynton Beach, FL Posts: 5 Thanks: 0 Sorry, but I don't understand either reply. Let me rephrase. The density function is f(y)=2/y^3 for y>1 and the integral goes from 1 to infinity. Another function is f(x) = 0.5003e^(-x/2) for 0
 November 18th, 2015, 02:40 PM #5 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,675 Thanks: 2655 Math Focus: Mainly analysis and algebra Typo?
 November 18th, 2015, 02:48 PM #6 Newbie   Joined: Nov 2015 From: Boynton Beach, FL Posts: 5 Thanks: 0 No, there is no typo.
November 19th, 2015, 01:51 PM   #7
Global Moderator

Joined: May 2007

Posts: 6,807
Thanks: 717

Quote:
 Originally Posted by Math1stlove Sorry, but I don't understand either reply. Let me rephrase. The density function is f(y)=2/y^3 for y>1 and the integral goes from 1 to infinity. Another function is f(x) = 0.5003e^(-x/2) for 0
Although the function is defined for 0<x<15, the integral doesn't have to be over the entire interval. You are free to integrate over any piece that you want. You need to give us the exact wording of the question.

 November 19th, 2015, 02:00 PM #8 Newbie   Joined: Nov 2015 From: Boynton Beach, FL Posts: 5 Thanks: 0 Okay, please type "soa problems exam p" into the Google search engine and click on the first link. Then, look at problems 53 and 54 and explain why both integrand start at 1. Btw, there are also solutions if you type in the above-mentioned quote and put solutions after what is already quoted.
 November 19th, 2015, 04:17 PM #9 Math Team   Joined: Nov 2014 From: Australia Posts: 689 Thanks: 244 They're probably accounting for the 1000 deductible. The first \$1000 of damage don't matter.
 November 20th, 2015, 03:33 AM #10 Newbie   Joined: Nov 2015 From: Boynton Beach, FL Posts: 5 Thanks: 0 I see! I knew I could count on you!

 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post sk3blue Calculus 2 February 15th, 2015 11:51 PM helloprajna Economics 0 February 17th, 2013 11:04 PM Gekko Calculus 3 June 12th, 2010 09:16 AM Justin Lo Applied Math 0 December 8th, 2009 02:40 AM shack Linear Algebra 3 December 17th, 2007 01:11 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top