Question

\[ \int_0^3 \left(|x-1|-x[x]\right)dx = \underline{} \] rounded off to one decimal place.

Hint:

For integrals involving the greatest integer function \([x]\), split the interval at integer points and evaluate separately.

Answer 1

Correct Answer: -4

Step 1: Break the range at the jumps.
The floor $[x]$ changes at $x=1$ and $x=2$, so split as $\int_0^1+\int_1^2+\int_2^3$.

Step 2: On $[0,1)$.
Here $[x]=0$ and $|x-1|=1-x$, so the part is $\int_0^1(1-x)\,dx=\tfrac12$.

Step 3: On $[1,2)$.
Here $[x]=1$ and $|x-1|=x-1$, so the integrand is $(x-1)-x=-1$, giving $\int_1^2(-1)\,dx=-1$.

Step 4: On $[2,3]$.
Here $[x]=2$ and $|x-1|=x-1$, so the integrand is $(x-1)-2x=-x-1$.

Step 5: Do that last piece.
\[ \int_2^3(-x-1)\,dx=\Big[-\tfrac{x^2}{2}-x\Big]_2^3=-\tfrac72 \]

Step 6: Add them up.
\[ \tfrac12-1-\tfrac72=-4 \]
\[ \boxed{-4.0} \]