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Integration of step functions

  • #1
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Homework Statement



This is from Apostol's Calculus Vol. 1. Exercise 1.15, problem 6.(c)

Find all x>0 for which the integral of [t]2 dt from 0 to x = 2(x-1)

Homework Equations



[t] represents the greatest integer function of t.

The Attempt at a Solution


[/B]
Integral of [t]2 dt from 0 to x = {x(x-1)(2x-1)}/6, which on equating with 2(x-1) gives x(2x-1) = 12 which does not have a rational solution. The answers given at the back of the textbook are 1 and 5/2. Can someone please give me a hint as to where I'm going wrong?
 

Answers and Replies

  • #2
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Homework Statement



This is from Apostol's Calculus Vol. 1. Exercise 1.15, problem 6.(c)

Find all x>0 for which the integral of [t]2 dt from 0 to x = 2(x-1)
Your notation really threw me off for a while. I think this is what you're trying to convey:
$$ \int_0^x \lfloor t \rfloor ^2 dt = 2(x - 1)$$
RandomGuy1 said:

Homework Equations



[t] represents the greatest integer function of t.

The Attempt at a Solution


[/B]
Integral of [t]2 dt from 0 to x = {x(x-1)(2x-1)}/6
No, it isn't. You can't just integrate ##\lfloor t ^2 \rfloor## as if it were the same thing as ##t^2##. Draw a sketch of ##y = \lfloor t ^2 \rfloor## and y = 2(t - 1) and compare the cumulative area under the graph of the integrand with the y values on the straight line graph.
RandomGuy1 said:
, which on equating with 2(x-1) gives x(2x-1) = 12 which does not have a rational solution. The answers given at the back of the textbook are 1 and 5/2. Can someone please give me a hint as to where I'm going wrong?
 
  • #3
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0
Yes, that was exactly what I was trying to say :) Sorry, I do not know how to use the definite integral symbol.

But but I'm supposed to be integrating [t]2 and not [t2] as you pointed out. Won't those two be different?
 
  • #4
Ray Vickson
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Yes, that was exactly what I was trying to say :) Sorry, I do not know how to use the definite integral symbol.

But but I'm supposed to be integrating [t]2 and not [t2] as you pointed out. Won't those two be different?
(1) You don't need to know how to do a definite integral sign; you can just say something like int_{t=0..x} [t]^2 dt or int([t]^2 dt, t=0..x). That would be perfectly clear and would avoid the kind of confusion I was victim to when I first tried to read your message. Or you could have said that "the integral of [t]^2 from 0 to x is equal to 2x-1"---that would also have been clear. Saying "...x = 2x-1" is maximally confusing.
(2) Yes, [t^2] and [t]^2 are very different.
 
  • #5
HallsofIvy
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This is, as you say, a step function. It's graph is just a series of horizontal lines and its integral, the area under the graph, is just the sum of areas of rectangles. The first thing you should do is draw the graph. If x is from 0 to 1, [x]= 0 so [x]2= 0. If x is from 1 to 2, [x]= 1 so [x]2= 1. If x is from 2 to 3, [x]= 2 so [x]2= 4, etc.
 
  • #6
Stephen Tashi
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Integral of [t]2 dt from 0 to x = {x(x-1)(2x-1)}/6
In this problem, the formula for summing the squares of integers is only relevant when [itex] x [/itex] is an integer. And when [itex] x [/itex] is an integer, I think you are "off by 1". For example, as I understand the symbol "[t]^2" in this problem, when [itex] 4 < t < 5 [/itex] , [itex] [t]^2 = [4]^2 = 16 [/itex]. So the area involved in [itex] \int_0^5 [t]^2 dt [/itex] does not include a rectangle with height 25. The areas to be summed are [itex] (1)(0^2)+(1)(1^2) + (1)(2^2) + (1)(3^3) + (1)(4^2) [/itex].

If the upper limit of integration is between two integers you must sum the squares of the whole rectangles involved in the are and then add the area of the fraction of the rectangle at the end to the sum.
 
Last edited:
  • #7
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Ray Vickson said:
Saying "...x = 2x-1" is maximally confusing.
Absolutely, as is the one below.
RandomGuy1 said:
Integral of [t]2 dt from 0 to x = {x(x-1)(2x-1)}/6
 

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