If [t] denotes the greatest integer t, then the value of _0^1[2x-|3x^2-5x+2|+1],dx is:

Question

If $[t]$ denotes the greatest integer $\le t$, then the value of $\displaystyle\int_0^1\big[2x-|3x^2-5x+2|+1\big]\,dx$ is:

Accepted Answer

$3x^2-5x+2=(3x-2)(x-1)$ changes sign at $x=2/3$; write $I=\displaystyle\int_0^1\big[2x-|(3x-2)(x-1)|\big]\,dt+\int_0^1 1\,dt$.

On $[0,2/3]$: $I_1=\displaystyle\int_0^{2/3}(-3x^2+7x-2)\,dt$ (bracket value determined by where $2x-(3x^2-5x+2)$ falls); on $[2/3,1]$: $I_2=\displaystyle\int_{2/3}^1(3x^2-3x+2)\,dt$, where $[3x^2-3x+2]=1$ throughout (since $3x^2-3x+2\in(4/3,2)$ there), contributing $\dfrac13$.

Working through the roots $\alpha,\beta=\dfrac{7\mp\sqrt{37}}6$ (of $-3x^2+7x-2=0$ relevant to the floor breaks) and $\dfrac{7\mp\sqrt{13}}6$-type roots for the other piece, the source arrives at

$$I = \frac13-(\alpha+\beta)+\frac13+1 = \frac53-\frac{7-\sqrt{37}}6+\frac{7-\sqrt{13}}6 = \frac{\sqrt{37}+\sqrt{13}-4}6$$

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