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Step-by-Step Solution
Step 1: Understand the Integral
We need to evaluate the definite integral
$ \int_{0}^{2} \Bigl(\bigl|2x^{2} - 3x\bigr| \;+\; \bigl\lfloor x - \tfrac12 \bigr\rfloor \Bigr)\, dx, $
where $|\,\cdot\,|$ denotes the absolute value function and $\lfloor\,\cdot\,\rfloor$ denotes the greatest integer (floor) function.
Step 2: Identify Intervals for the Absolute Value Part
The expression $2x^{2} - 3x$ changes sign where $2x^{2} - 3x = 0.$ Solving,
$2x^{2} - 3x = x(2x - 3) = 0 \implies x = 0 \text{ or } x = \tfrac{3}{2}.$
Hence, we split the $x$-axis for the absolute value part as follows:
$0 \leq x < \tfrac{3}{2}$: Here $2x^{2} - 3x \le 0,$ so $\bigl| 2x^{2} - 3x \bigr| = 3x - 2x^{2}.$
$\tfrac{3}{2} \leq x \le 2$: Here $2x^{2} - 3x \ge 0,$ so $\bigl| 2x^{2} - 3x \bigr| = 2x^{2} - 3x.$
Step 3: Identify Intervals for the Floor Function Part
We examine $\lfloor x - \tfrac12 \rfloor$ over $[0,2]$:
$0 \le x < \tfrac12$: Then $x - \tfrac12 < 0,$ so $\lfloor x - \tfrac12 \rfloor = -1.$
$\tfrac12 \le x < \tfrac{3}{2}$: Then $0 \le x - \tfrac12 < 1,$ so $\lfloor x - \tfrac12 \rfloor = 0.$
$\tfrac{3}{2} \le x \le 2$: Then $1 \le x - \tfrac12 \le \tfrac{3}{2},$ so $\lfloor x - \tfrac12 \rfloor = 1.$
Step 4: Express the Integral in Piecewise Form
Using these intervals, the integral splits into:
$ \int_{0}^{2} \Bigl(\bigl|2x^{2} - 3x\bigr|\Bigr)\, dx
\;+\;
\int_{0}^{2} \Bigl(\lfloor x - \tfrac12 \rfloor\Bigr)\, dx
$
which becomes
$ \underbrace{\int_{0}^{\tfrac{3}{2}}(3x - 2x^{2})\,dx}_{\text{Abs. value, first part}}
\;+\;
\underbrace{\int_{\tfrac{3}{2}}^{2}(2x^{2} - 3x)\,dx}_{\text{Abs. value, second part}}
\;+\;
\underbrace{\int_{0}^{\tfrac12}(-1)\,dx}_{\text{Floor, }= -1}
\;+\;
\underbrace{\int_{\tfrac12}^{\tfrac{3}{2}}(0)\,dx}_{\text{Floor, }= 0}
\;+\;
\underbrace{\int_{\tfrac{3}{2}}^{2}(1)\,dx}_{\text{Floor, }=1}.
Step 5: Compute the Absolute Value Integrals
From $0$ to $\tfrac{3}{2}$:
$$
\int_{0}^{\tfrac{3}{2}} (3x - 2x^{2})\,dx
= \left[ \frac{3x^{2}}{2} - \frac{2x^{3}}{3} \right]_{0}^{\tfrac{3}{2}}.
$$
Evaluate at $x = \tfrac{3}{2}$:
$$
\frac{3\left(\tfrac{3}{2}\right)^{2}}{2} - \frac{2\left(\tfrac{3}{2}\right)^{3}}{3}
= \frac{3 \cdot \tfrac{9}{4}}{2} - \frac{2 \cdot \tfrac{27}{8}}{3}
= \frac{27}{8} \;-\; \frac{54}{24}
= \frac{27}{8} \;-\; \frac{9}{4}
= \frac{27}{8} \;-\; \frac{18}{8}
= \frac{9}{8}.
$$
From $\tfrac{3}{2}$ to $2$:
$$
\int_{\tfrac{3}{2}}^{2} (2x^{2} - 3x)\,dx
= \left[ \frac{2x^{3}}{3} - \frac{3x^{2}}{2} \right]_{\tfrac{3}{2}}^{2}.
$$
Evaluate:
At $x = 2$: $ \tfrac{2 (2^3)}{3} - \tfrac{3 \cdot (2^2)}{2} = \tfrac{16}{3} - 6 = \tfrac{16}{3} - \tfrac{18}{3} = - \tfrac{2}{3}.$
At $x = \tfrac{3}{2}$:
$$
\frac{2\left(\tfrac{3}{2}\right)^{3}}{3} - \frac{3\left(\tfrac{3}{2}\right)^{2}}{2}
= \frac{2 \cdot \tfrac{27}{8}}{3} - \frac{3 \cdot \tfrac{9}{4}}{2}
= \frac{54}{24} - \frac{27}{8}
= \frac{9}{4} - \frac{27}{8}
= \frac{18}{8} - \frac{27}{8}
= -\frac{9}{8}.
$$
Subtracting, we get:
$$
\left(- \tfrac{2}{3}\right) - \left(- \tfrac{9}{8}\right)
= - \tfrac{2}{3} + \tfrac{9}{8}
= - \tfrac{16}{24} + \tfrac{27}{24}
= \tfrac{11}{24}.
$$
Summing these results for the absolute value part:
$ \int_{0}^{\tfrac{3}{2}}(3x - 2x^{2})\,dx + \int_{\tfrac{3}{2}}^{2}(2x^{2} - 3x)\,dx
= \frac{9}{8} + \frac{11}{24}
= \frac{27}{24} + \frac{11}{24}
= \frac{38}{24}
= \frac{19}{12}.
$
Step 6: Compute the Floor Function Integrals
$ \int_{0}^{\tfrac12} (-1)\,dx + \int_{\tfrac12}^{\tfrac{3}{2}} (0)\,dx + \int_{\tfrac{3}{2}}^{2} (1)\,dx.
$
$\int_{0}^{\tfrac12} -1\,dx = -1 \times \bigl(\tfrac12 - 0\bigr) = -\tfrac12.$
$\int_{\tfrac12}^{\tfrac{3}{2}} 0\,dx = 0.$
$\int_{\tfrac{3}{2}}^{2} 1\,dx = 1 \times \bigl(2 - \tfrac{3}{2}\bigr) = \tfrac12.$
Summing these:
$ -\tfrac12 + 0 + \tfrac12 = 0.$
Step 7: Combine Both Parts
The total value of the original integral is the sum of the absolute value part and the floor part:
$ \left(\frac{19}{12}\right) + 0 = \frac{19}{12}.
Final Answer
$\displaystyle \frac{19}{12}
$
