Let $$[t]$$ denote the greatest integer $$\leq t$$. Then the value of $$8 \cdot \int_{-\frac{1}{2}}^{1} \left([2x] + |x|\right) dx$$ is _________.

We have to evaluate the quantity $$8\;\cdot\;\displaystyle\int_{-\dfrac12}^{1}\Big([2x]+|x|\Big)\,dx,$$ where $$[t]$$ denotes the greatest integer less than or equal to $$t$$. After finding the definite integral, we will multiply the result by $$8$$.

Because both the greatest-integer function $$[2x]$$ and the absolute-value function $$|x|$$ are piece-wise, it is convenient to break the interval $$\left[-\dfrac12,\,1\right]$$ into smaller parts on which the integrand is simple and constant in form.

First, recall the two standard definitions we shall use again and again:

1. For any real number $$x$$, $$|x|=\begin{cases}-x,& &x<0,\\[4pt] x,& &x\ge 0.\end{cases}$$

2. For any real number $$x$$, $$[2x]$$ is the greatest integer that is &le 2x. To know where $$[2x]$$ changes its value, we solve $$2x=n$$ for integers $$n$$; this gives the critical points $$x=\dfrac n2$$.

Within our interval $$-\,\dfrac12\le x\le1$$ the relevant half-integers are $$x=-\dfrac12,\;0,\;\dfrac12,\;1.$$ Hence we split the entire interval into three sub-intervals:

  (i) $$\left[-\dfrac12,\,0\right)$$   (ii) $$\left[0,\,\dfrac12\right)$$   (iii) $$\left[\dfrac12,\,1\right]$$

We now write the integrand $$[2x]+|x|$$ explicitly on each sub-interval.

Interval (i): If $$-\dfrac12\le x<0,$$ then $$2x$$ lies in $$[-1,0),$$ so $$[2x]=-1.$$ Because $$x<0,$$ we have $$|x|=-x.$$ Thus the integrand becomes $$[2x]+|x|=-1+(-x)=-1-x.$$

Interval (ii): If $$0\le x<\dfrac12,$$ then $$0\le2x<1,$$ so $$[2x]=0.$$ Now $$x\ge0,$$ hence $$|x|=x.$$ Therefore the integrand is $$[2x]+|x|=0+x=x.$$

Interval (iii): If $$\dfrac12\le x\le1,$$ then $$1\le2x\le2,$$ and except for the single point $$x=1$$ (of zero width in integration) we have $$[2x]=1.$$ Again $$x\ge0,$$ so $$|x|=x.$$ Consequently the integrand is $$[2x]+|x|=1+x.$$

With the integrand fully described, we write the original integral as the sum of three simpler definite integrals:

$$\displaystyle\int_{-\frac12}^{1}\big([2x]+|x|\big)\,dx =\int_{-\frac12}^{0}(-1-x)\,dx+\int_{0}^{\frac12}x\,dx+\int_{\frac12}^{1}(1+x)\,dx.$$

Let us evaluate each piece one after another, showing every algebraic step.

First piece

$$\int_{-\frac12}^{0}(-1-x)\,dx =\Bigl[\,-x-\frac{x^{2}}{2}\Bigr]_{-1/2}^{0}.$$

Substituting the upper limit $$x=0$$ gives $$-0-\dfrac{0^{2}}{2}=0.$$

Substituting the lower limit $$x=-\dfrac12$$ gives $$-\!\Bigl(-\dfrac12\Bigr)-\frac{\left(-\dfrac12\right)^{2}}{2} =\dfrac12-\frac{\dfrac14}{2} =\dfrac12-\dfrac18 =\dfrac38.$$

Therefore $$\int_{-\frac12}^{0}(-1-x)\,dx =0-\dfrac38=-\dfrac38.$$

Second piece

$$\int_{0}^{\frac12}x\,dx =\Bigl[\frac{x^{2}}{2}\Bigr]_{0}^{1/2} =\frac{(1/2)^{2}}{2}-0 =\frac{1/4}{2} =\dfrac18.$$

Third piece

$$\int_{\frac12}^{1}(1+x)\,dx =\Bigl[x+\frac{x^{2}}{2}\Bigr]_{1/2}^{1}.$$

Upper limit $$x=1$$ gives $$1+\dfrac{1^{2}}{2}=1+\dfrac12=\dfrac32.$$

Lower limit $$x=\dfrac12$$ gives $$\dfrac12+\frac{(1/2)^{2}}{2} =\dfrac12+\frac{1/4}{2} =\dfrac12+\dfrac18 =\dfrac58.$$

Hence $$\int_{\frac12}^{1}(1+x)\,dx =\dfrac32-\dfrac58 =\frac{12}{8}-\frac58 =\dfrac78.$$

Now we combine all three results:

$$\int_{-\frac12}^{1}\big([2x]+|x|\big)\,dx =-\dfrac38+\dfrac18+\dfrac78 =\frac{-3+1+7}{8} =\dfrac58.$$

Finally we multiply by the external factor $$8$$:

$$8\;\times\;\dfrac58 = 5.$$

So, the answer is $$5$$.