Question

Question #5003d

Answer 1

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:

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.

Answer 2

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.