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Step-by-Step Solution
Step 1: Rewrite the Integral with the Greatest Integer Function
We need to evaluate
$ \int_0^{10} \frac{[\sin(2\pi x)]}{e^{x - [x]}} \, dx $,
where $[x]$ denotes the greatest integer less than or equal to $x$. Observe that over each integer interval $[n, n+1)$ for $n = 0, 1, 2, \dots, 9$, the fractional part $\{x\} = x - [x]$ varies from 0 to 1. Hence, we can use the property of definite integrals by summing up integrals from 0 to 1, 1 to 2, and so on, which simplifies to:
$ \int_0^{10} \frac{[\sin(2\pi x)]}{e^{x-[x]}} \, dx
= 10 \int_0^1 \frac{[\sin(2\pi x)]}{e^{\{x\}}} \, dx.$
Step 2: Determine the Value of $[\sin(2\pi x)]$ Over One Period
Over one period of the sine function from $x = 0$ to $x = 1$, we have $2\pi x$ going from 0 to $2\pi$. The value of $\sin(2\pi x)$ in the interval $[0, 1]$ is:
From $x = 0$ to $x = \frac{1}{2}$, $\sin(2\pi x)$ ranges from 0 up to 1 and back to 0. Its greatest integer value $[\sin(2\pi x)]$ is 0 in this entire interval (since $\sin(2\pi x)$ ranges between 0 and 1, not reaching 1 except at isolated points).
From $x = \frac{1}{2}$ to $x = 1$, $\sin(2\pi x)$ ranges through negative values (from 0 down to -1 and back to 0). Thus $[\sin(2\pi x)] = -1$ in this interval where $\sin(2\pi x)$ is negative.
Step 3: Split the Integral Accordingly
Based on the value of $[\sin(2\pi x)]$:
$ \int_0^1 \frac{[\sin(2\pi x)]}{e^{\{x\}}} \, dx
= \int_0^{1/2} \frac{0}{e^x} \, dx
+ \int_{1/2}^1 \frac{-1}{e^x} \, dx
= \int_0^{1/2} 0 \, dx
+ \int_{1/2}^1 \frac{-1}{e^x} \, dx. $
Step 4: Integrate over the Relevant Sub-Intervals
Since the first integral is zero:
$ \int_0^{1/2} 0 \, dx = 0,
$
and for the second integral:
$ \int_{1/2}^1 \frac{-1}{e^x} \, dx = -\int_{1/2}^1 e^{-x} \, dx.
$
The antiderivative of $e^{-x}$ is $-e^{-x}$. Therefore:
$ -\int_{1/2}^1 e^{-x} \, dx
= -\left[-e^{-x}\right]_{x=1/2}^{x=1}
= -\left[-e^{-1} + e^{-1/2}\right]
= e^{-1} - e^{-1/2}.
$
Step 5: Combine the Results and Multiply by 10
Putting it all together inside the factor of 10:
$ 10 \int_0^1 \frac{[\sin(2\pi x)]}{e^{\{x\}}} \, dx
= 10 \left( e^{-1} - e^{-1/2} \right).
$
Hence,
$ \int_0^{10} \frac{[\sin(2\pi x)]}{e^{x-[x]}} \, dx
= 10 e^{-1} - 10 e^{-1/2}.
$
Step 6: Match with the Given Form
We compare this result to the form
$ \alpha e^{-1} + \beta e^{-\frac{1}{2}} + \gamma. $
From our result:
$\alpha = 10, \quad \beta = -10, \quad \gamma = 0.$
Therefore,
$ \alpha + \beta + \gamma = 10 + (-10) + 0 = 0.
$
Final Answer
The value of $\alpha + \beta + \gamma$ is $\boxed{0}$.
