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Step-by-Step Solution
Step 1: Understand the Function
We are given the function
$f(x) = \max\{x^2,\,1 + [x]\}$, where $[x]$ denotes the greatest integer less than or equal to $x$.
Step 2: Analyze the Greatest Integer Part
For $0 \le x < 1$, $[x] = 0$, so $1 + [x] = 1$.
For $1 \le x < 2$, $[x] = 1$, so $1 + [x] = 2$.
Step 3: Compare $x^2$ and $1 + [x]$ on Different Intervals
Interval $0 \le x < 1$:
Here, $1 + [x] = 1$. Compare $x^2$ and 1:
Since $x^2 \le 1$ for $0 \le x \le 1$, the maximum is $1$. Thus, for $0 \le x < 1$, $f(x) = 1$.
Interval $1 \le x < 2$:
Here, $1 + [x] = 2$. Compare $x^2$ and 2:
Find the point where $x^2 = 2$, which is $x = \sqrt{2}$.
• For $1 \le x \le \sqrt{2}$, $x^2 \le 2$, so $f(x) = 2$.
• For $\sqrt{2} < x < 2$, $x^2 > 2$, so $f(x) = x^2$.
Step 4: Set Up the Integral
We divide the integral $\displaystyle \int_{0}^{2} f(x)\,dx$ into three parts:
$\displaystyle \int_{0}^{2} f(x)\,dx
= \int_{0}^{1} f(x)\,dx
+ \int_{1}^{\sqrt{2}} f(x)\,dx
+ \int_{\sqrt{2}}^{2} f(x)\,dx.$
Based on the comparisons:
For $0 \le x < 1$, $f(x) = 1$.
For $1 \le x \le \sqrt{2}$, $f(x) = 2$.
For $\sqrt{2} \le x \le 2$, $f(x) = x^2$.
Thus,
$\displaystyle \int_{0}^{2} f(x)\,dx
= \int_{0}^{1} 1 \,dx
+ \int_{1}^{\sqrt{2}} 2 \,dx
+ \int_{\sqrt{2}}^{2} x^2 \,dx.$
Step 5: Evaluate Each Part
1. $\displaystyle \int_{0}^{1} 1\,dx = [\,x\,]_{0}^{1} = 1.$
2. $\displaystyle \int_{1}^{\sqrt{2}} 2\,dx = 2\bigl[x\bigr]_{1}^{\sqrt{2}}
= 2\bigl(\sqrt{2} - 1\bigr).$
3. $\displaystyle \int_{\sqrt{2}}^{2} x^2\,dx = \left[\frac{x^3}{3}\right]_{\sqrt{2}}^{2}
= \frac{2^3}{3} - \frac{(\sqrt{2})^3}{3}
= \frac{8}{3} - \frac{2\sqrt{2}}{3}.$
Step 6: Combine the Results
Summing up the three parts gives:
$
\bigl(1\bigr)
+ 2\bigl(\sqrt{2} - 1\bigr)
+ \left(\frac{8}{3} - \frac{2\sqrt{2}}{3}\right).
$
First combine the numeric terms:
\[
1 + \frac{8}{3} = \frac{3}{3} + \frac{8}{3} = \frac{11}{3}.
\]
Then combine the terms involving $\sqrt{2}$:
\[
2\sqrt{2} - 2 + \left(-\frac{2\sqrt{2}}{3}\right).
\]
Notice we have $-2 = -2 \cdot \frac{3}{3} = -\frac{6}{3}$, so collecting everything carefully:
\[
\left(\frac{11}{3} - \frac{6}{3}\right) + 2\sqrt{2} - \frac{2\sqrt{2}}{3}
= \frac{5}{3} + \left(2\sqrt{2} - \frac{2\sqrt{2}}{3}\right).
\]
Factor out $2\sqrt{2}$:
\[
2\sqrt{2}\left(1 - \frac{1}{3}\right) = 2\sqrt{2}\cdot \frac{2}{3} = \frac{4\sqrt{2}}{3}.
\]
Finally,
\[
\frac{5}{3} + \frac{4\sqrt{2}}{3} = \frac{5 + 4\sqrt{2}}{3}.
\]
Therefore,
$$
\int_{0}^{2} f(x)\,dx
= \frac{5 + 4\sqrt{2}}{3}.
$$
Graphical References (Unaltered Images)
Combining both the graph we get,
Final Answer
$ \displaystyle \boxed{\frac{5 + 4\sqrt{2}}{3}} $
