Question #5003d

2 Answers
Jan 1, 2015

I am not sure if I got the meaning of your question right but I'll try anyway:

#int_0^xt^2dt=2(x-1)#
#t^3/3|_0^x=2(x-1)#
#x^3/3=2(x-1)#

Rearranging:
#x^3-6x+6=0#

To solve this cubic equation I used the Cardano-Tartaglia formula that you can find in most texts or on the internet such as in Wikipedia:

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I supposed that you needed only real roots so I neglected the ones which are complex numbers.
You have only one real root (an irrational number) given as:
#x_1=-2,84732#

You can check it by substituting in:
#x^3/3=2(x-1)#

Probably there are simpler techniques but I do not know. Hope it helped.

There is no value of #x# that solves this problem.

By "greatest integer function of #t#" I assume you mean #|~t~|#, called the ceiling of #t# or the ceiling function.

For every finite interval chosen as it's domain, the ceiling function and it's square #|~t~|^2# are step functions , which means that there is a finite collection of intervals in which the function takes a constant value for each interval. Equivalently, #f# is called a step function if, for a finite number #m# of intervals #I_k#

#f(x) = c_k(b_k - a_k), x in I_k #

where #b_k - a_k# is the length of the interval #I_k#

The integral of any step function is simply the sum of the lengths of all intervals multiplied by their respective constants:

#int_(a_1)^(b_m) f(x) = sum_(k=1)^m c_k (b_k - a_k)#

Now, the function #|~t~|^2# is defined as

#|~t~|^2 = n^2, x in (n-1, n)#

where #n# are integers.

It's integral is, observing that all intervals of the form # (n-1, n) # have length #1# and that the last interval # ( |__ x __|, x ) # has length # x- |__ x __| #, where # |__ x __| # is the floor function:

#int_(0)^(x) |~t~|^2 = sum_(k=1)^(|__ x __|) k^2 + |~x~|^2 (x-|__ x __|)#

Now, your question also gives us the condition that

#int_(0)^(x) |~t~|^2 = 2(x-1)#

Wich means:

#sum_(k=1)^(|__ x __|) k^2 + |~x~|^2 (x-|__ x __|) = 2x-2#

Now, the sum of the first #q# squares is #(2q^3+3q^2+q)/6#.

So,

#sum_(k=1)^(|__ x __|) k^2= (2(|__ x __|)^3+3(|__ x __|)^2+|__ x __|)/6#

With the additional condition given to us by the question:

#(2(|__ x __|)^3+3(|__ x __|)^2+|__ x __|)/6 + |~x~|^2 (x-|__ x __|) = 2x-2#

Finding the solution to this equation is not simple. But noticing that the left side behaves itself similarly to a #3#rd degree polynomial, as the value of #x# gets bigger, it should grow faster than the right hand side.

Consider the left side equation for #x=1#. We have:

#(2(|__ 1 __|)^3+3(|__ 1 __|)^2+|__ 1 __|)/6 = (2 times 1 + 3 times 1 +1)/6 = 1#

The right hand side:

#2 times 1 - 2 =0#

So even where the function at the left side grows at it's slowest, the left side of the equation is greater than the right side. So, for #x>0#, there is no solution.

Another way of thinking about this problem is graphing the integral. For #x<1#, the graph of the integral is a line segment of slope #1# passing through the origin. For #1 geq x < 2#, it's a line segment of slope #4#, and as #x# get's bigger, so does the slope. For #x<1#, #2x-2<0#, so this line does not intersect the graph of the integral in this interval. But for #x>1#, the integral grows much faster then #2x-2#, so the two graphs will never intersect.