Question #5003d

Jan 1, 2015

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

${\int}_{0}^{x} {t}^{2} \mathrm{dt} = 2 \left(x - 1\right)$
${t}^{3} / 3 {|}_{0}^{x} = 2 \left(x - 1\right)$
${x}^{3} / 3 = 2 \left(x - 1\right)$

Rearranging:
${x}^{3} - 6 x + 6 = 0$

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:
${x}^{3} / 3 = 2 \left(x - 1\right)$

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 $\lceiling t \rceiling$, 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 $\lceiling t {\rceiling}^{2}$ 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 ${I}_{k}$

$f \left(x\right) = {c}_{k} \left({b}_{k} - {a}_{k}\right) , x \in {I}_{k}$

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:

${\int}_{{a}_{1}}^{{b}_{m}} f \left(x\right) = {\sum}_{k = 1}^{m} {c}_{k} \left({b}_{k} - {a}_{k}\right)$

Now, the function $\lceiling t {\rceiling}^{2}$ is defined as

$\lceiling t {\rceiling}^{2} = {n}^{2} , x \in \left(n - 1 , n\right)$

where $n$ are integers.

It's integral is, observing that all intervals of the form $\left(n - 1 , n\right)$ have length $1$ and that the last interval $\left(\lfloor x \rfloor , x\right)$ has length $x - \lfloor x \rfloor$, where $\lfloor x \rfloor$ is the floor function:

${\int}_{0}^{x} \lceiling t {\rceiling}^{2} = {\sum}_{k = 1}^{\lfloor x \rfloor} {k}^{2} + \lceiling x {\rceiling}^{2} \left(x - \lfloor x \rfloor\right)$

Now, your question also gives us the condition that

${\int}_{0}^{x} \lceiling t {\rceiling}^{2} = 2 \left(x - 1\right)$

Wich means:

${\sum}_{k = 1}^{\lfloor x \rfloor} {k}^{2} + \lceiling x {\rceiling}^{2} \left(x - \lfloor x \rfloor\right) = 2 x - 2$

Now, the sum of the first $q$ squares is $\frac{2 {q}^{3} + 3 {q}^{2} + q}{6}$.

So,

${\sum}_{k = 1}^{\lfloor x \rfloor} {k}^{2} = \frac{2 {\left(\lfloor x \rfloor\right)}^{3} + 3 {\left(\lfloor x \rfloor\right)}^{2} + \lfloor x \rfloor}{6}$

With the additional condition given to us by the question:

$\frac{2 {\left(\lfloor x \rfloor\right)}^{3} + 3 {\left(\lfloor x \rfloor\right)}^{2} + \lfloor x \rfloor}{6} + \lceiling x {\rceiling}^{2} \left(x - \lfloor x \rfloor\right) = 2 x - 2$

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:

$\frac{2 {\left(\lfloor 1 \rfloor\right)}^{3} + 3 {\left(\lfloor 1 \rfloor\right)}^{2} + \lfloor 1 \rfloor}{6} = \frac{2 \times 1 + 3 \times 1 + 1}{6} = 1$

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 $1 \ge q x < 2$, it's a line segment of slope $4$, and as $x$ get's bigger, so does the slope. For $x < 1$, $2 x - 2 < 0$, so this line does not intersect the graph of the integral in this interval. But for $x > 1$, the integral grows much faster then $2 x - 2$, so the two graphs will never intersect.