We need to evaluate $8\int_{-1/2}^{1} \left([2x]+|x|\right)dx,$ where $[t]$ denotes the greatest integer less than or equal to $t$.

1. Break the interval according to $[2x]$ and $|x|$

The expression changes when:

1. $x=0$ because of $|x|$,
2. $2x$ crosses integers, i.e. at $x=-\tfrac12,0,\tfrac12,1$.

So split the interval into: $\left[-\frac12,0\right),\quad \left[0,\frac12\right),\quad \left[\frac12,1\right).$

2. Find $[2x]$ on each interval

(i) For $-\frac12 \le x<0$

Then $-1\le 2x<0,$ so $[2x]=-1.$ Also, since $x<0$, $|x|=-x.$ Thus integrand is $[2x]+|x|=-1-x.$

(ii) For $0\le x<\frac12$

Then $0\le 2x<1,$ so $[2x]=0.$ Also $|x|=x$. Hence integrand is $x.$

(iii) For $\frac12\le x<1$

Then $1\le 2x<2,$ so $[2x]=1.$ Also $|x|=x$. Hence integrand is $1+x.$

At the single point $x=1$, $[2]=2$, but changing value at one point does not affect the integral.

3. Compute the integral piecewise

So $I=\int_{-1/2}^{1}([2x]+|x|)dx$ becomes $I=\int_{-1/2}^{0}(-1-x)dx+\int_{0}^{1/2}x\,dx+\int_{1/2}^{1}(1+x)dx.$

First part

$\int_{-1/2}^{0}(-1-x)dx=\left[-x-\frac{x^2}{2}\right]_{-1/2}^{0}.$ At $x=0$, value is $0$. At $x=-\tfrac12$, $-\left(-\frac12\right)-\frac{( -1/2)^2}{2}=\frac12-\frac{1/4}{2}=\frac12-\frac18=\frac38.$ So, $\int_{-1/2}^{0}(-1-x)dx=0-\frac38=-\frac38.$

Second part

$\int_{0}^{1/2}x\,dx=\left[\frac{x^2}{2}\right]_{0}^{1/2}=\frac{1/4}{2}=\frac18.$

Third part

$\int_{1/2}^{1}(1+x)dx=\left[x+\frac{x^2}{2}\right]_{1/2}^{1}.$ At $x=1$: $1+\frac12=\frac32.$ At $x=\tfrac12$: $\frac12+\frac{1/4}{2}=\frac12+\frac18=\frac58.$ So, $\int_{1/2}^{1}(1+x)dx=\frac32-\frac58=\frac{12-5}{8}=\frac78.$

4. Add all parts

$I=-\frac38+\frac18+\frac78=\frac58.$

Therefore, $8I=8\cdot \frac58=5.$

5. Compare with stored answer

Derived answer = $5$. Stored correct answer = $5$. They match.