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 Calculus Calculus Math Forum

 August 17th, 2019, 06:35 AM #1 Member   Joined: Jul 2017 From: KOLKATA Posts: 50 Thanks: 3 Definite integral of box function Evaluate integral if [ t + 1]^3 dt from t = 0 to t=x where [.] denotes the box function Please let me know how to proceed Thanks from idontknow August 17th, 2019, 08:02 AM #2 Global Moderator   Joined: Dec 2006 Posts: 20,972 Thanks: 2222 You might find it useful to sketch the graph of the function. Thanks from topsquark August 17th, 2019, 01:47 PM #3 Global Moderator   Joined: May 2007 Posts: 6,823 Thanks: 723 You are integrating a function which is piecewise constant. Thanks from topsquark August 17th, 2019, 10:47 PM #4 Member   Joined: Jul 2017 From: KOLKATA Posts: 50 Thanks: 3 I proceeded as below: 1. break down interval [ 0 , x] in x sub-intervals [ 0,1 ] , [ 1,2] .... [x-1,x] then integrate 2. It comes to 1^3 + 2^3 ... [x]^3 = (([x]*([x) +1))/2)^2 But the answer is not matching. Need to know - where I am going wrong? Last edited by skipjack; August 18th, 2019 at 02:09 PM. August 18th, 2019, 01:35 PM #5 Global Moderator   Joined: May 2007 Posts: 6,823 Thanks: 723 I assume x is an integer. If so, your answer looks correct. What is the given answer? If x is not an integer, the last term is incorrect. It should be $[x]^3(x-[x])$. August 18th, 2019, 02:47 PM #6 Global Moderator   Joined: Dec 2006 Posts: 20,972 Thanks: 2222 If x isn't an integer, it comes to 1³ + 2³ + . . . + [x]³ + ([[x] + 1)³(x - [x]). August 18th, 2019, 11:41 PM #7 Member   Joined: Jul 2017 From: KOLKATA Posts: 50 Thanks: 3 Many thanks for your help. Answer from skipjack is matching. Yes, nowhere is it mentioned that x is an integer. However, could you please explain how the the last term is coming ([[x] + 1)³(x - [x])? Last edited by skipjack; August 19th, 2019 at 09:30 AM. August 19th, 2019, 05:34 AM #8 Global Moderator   Joined: Dec 2006 Posts: 20,972 Thanks: 2222 For t between [x] and x, [t + 1]³ = ([x] + 1)³. August 19th, 2019, 08:37 AM #9 Member   Joined: Jul 2017 From: KOLKATA Posts: 50 Thanks: 3 Many thanks. But I still have a question - how to prove analytically integral ([t] + 1)^3 from [x] to x is ([x] + 1)^3 (x - [x]). Last edited by skipjack; August 19th, 2019 at 09:18 AM. August 19th, 2019, 09:27 AM #10 Global Moderator   Joined: Dec 2006 Posts: 20,972 Thanks: 2222 For [x] < t < x, [t] = [x] (a constant), so ([t] + 1)³ is also a constant. Thanks from idontknow Tags box, definite, function, integral Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post idontknow Calculus 2 October 5th, 2018 07:14 PM panky Calculus 4 June 22nd, 2016 10:54 PM jiasyuen Calculus 4 February 17th, 2015 01:17 PM zaidalyafey Calculus 1 August 14th, 2012 02:36 AM Ansh Agrawal Calculus 0 February 11th, 2012 12:05 AM

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