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Step-by-Step Solution
Step 1: Rewrite the Given Integral
We are given the integral
$$
\int_{0}^{5} \frac{x + [x]}{e^{\,x - [x]}} \, dx = \alpha e^{-1} + \beta,
$$
where $[x]$ is the greatest integer less than or equal to $x$. Our goal is to find $(\alpha + \beta)^2$ given that $5\alpha + 6\beta = 0.$
Step 2: Break the Integral into Integer Intervals
Since $[x]$ (the greatest integer function) is constant on each interval $[n, n+1)$ (for integer $n$), we break the integral from 0 to 5 as follows:
$$
I = \int_{0}^{5} \frac{x + [x]}{e^{(x - [x])}} \, dx
= \int_{0}^{1} \frac{x + 0}{e^{(x - 0)}} \, dx
+ \int_{1}^{2} \frac{x + 1}{e^{(x - 1)}} \, dx
+ \int_{2}^{3} \frac{x + 2}{e^{(x - 2)}} \, dx
+ \int_{3}^{4} \frac{x + 3}{e^{(x - 3)}} \, dx
+ \int_{4}^{5} \frac{x + 4}{e^{(x - 4)}} \, dx.
$$
Denote these integrals as $I_1, I_2, I_3, I_4,$ and $I_5$ respectively. Hence,
$$
I = I_1 + I_2 + I_3 + I_4 + I_5.
$$
Step 3: Evaluate the First Integral $I_1$
The first integral is
$$
I_1 = \int_{0}^{1} \frac{x}{e^{\,x}} \, dx.
$$
We can integrate by parts or use a known standard integral. One straightforward approach is integration by parts:
Let
$$
u = x, \quad dv = e^{-x} \, dx \quad \bigl(\text{since } \frac{1}{e^x} = e^{-x}\bigr).
$$
Then
$$
du = dx, \quad v = \int e^{-x}\,dx = -e^{-x}.
$$
Therefore,
$$
I_1
= \int_{0}^{1} x e^{-x} \, dx
= \bigl[ -x e^{-x} \bigr]_{0}^{1} + \int_{0}^{1} e^{-x}\,dx
= \Bigl(-1 \cdot e^{-1} - 0\Bigr) + \Bigl[ -e^{-x} \Bigr]_{0}^{1}.
$$
Simplify:
$$
I_1
= -e^{-1} + \bigl(-e^{-1} + 1\bigr)
= 1 - 2e^{-1}.
$$
Step 4: Evaluate $I_2$
The second integral is
$$
I_2 = \int_{1}^{2} \frac{x + 1}{e^{x - 1}} \, dx.
$$
Make the substitution $x = t + 1 \implies dx = dt$, and when $x = 1$, $t = 0$; when $x = 2$, $t = 1$. Thus,
$$
I_2 = \int_{0}^{1} \frac{(t + 1) + 1}{e^{t}} \, dt
= \int_{0}^{1} \frac{t + 2}{e^{t}} \, dt
= \int_{0}^{1} \frac{t}{e^{t}} \, dt + 2 \int_{0}^{1} \frac{1}{e^{t}} \, dt.
$$
The first part $\int_{0}^{1} \frac{t}{e^t} \, dt$ is exactly $I_1$. So
$$
I_2 = I_1 + 2\int_{0}^{1} e^{-t}\, dt = I_1 + 2 \bigl[ -e^{-t} \bigr]_{0}^{1}
= I_1 + 2 \bigl(1 - e^{-1}\bigr).
$$
Step 5: Evaluate $I_3, I_4, I_5$
Following the same pattern for $I_3, I_4,$ and $I_5$, we get:
$I_3 = I_1 + 4 \bigl(1 - e^{-1}\bigr)$
$I_4 = I_1 + 6 \bigl(1 - e^{-1}\bigr)$
$I_5 = I_1 + 8 \bigl(1 - e^{-1}\bigr)$
These forms arise because in general, if we shift $x$ by $n$, the integral will add $2n$ in front of the $e^{-t}$ term when integrated from 0 to 1. The constant term in each bracket follows similarly.
Step 6: Sum All Integrals
Now, sum $I_1 + I_2 + I_3 + I_4 + I_5$:
$$
I = 5I_1 + \bigl[(2 + 4 + 6 + 8)\bigl(1 - e^{-1}\bigr)\bigr].
$$
Since $2+4+6+8 = 20,$ we have
$$
I = 5I_1 + 20\bigl(1 - e^{-1}\bigr).
$$
Recall from Step 3 that $I_1 = 1 - 2e^{-1}.$ Hence,
$$
I = 5(1 - 2e^{-1}) + 20\bigl(1 - e^{-1}\bigr)
= 5 - 10e^{-1} + 20 - 20e^{-1}
= 25 - 30e^{-1}.
$$
Step 7: Identify $\alpha$ and $\beta$
We have $I = \alpha e^{-1} + \beta$. Comparing with $25 - 30e^{-1}$, we deduce
$$
\alpha = -30, \quad \beta = 25.
$$
These values also satisfy the given condition $5\alpha + 6\beta = 0$.
Step 8: Compute $(\alpha + \beta)^2$
$$
\alpha + \beta = -30 + 25 = -5, \quad (\alpha + \beta)^2 = (-5)^2 = 25.
$$
Therefore, the value of $(\alpha + \beta)^2$ is 25.
Final Answer
The correct answer is 25.
