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 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

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