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Step-by-Step Solution
Step 1: Understand the integral and the functions involved
We need to evaluate the definite integral
$ \displaystyle I \;=\; \int_{-3}^{101} \Bigl( \bigl[\sin(\pi x)\bigr] \;+\; e^{\bigl[\cos(2\pi x)\bigr]} \Bigr)\,dx,$
where $[t]$ denotes the greatest integer less than or equal to $t.$
The expression has two parts:
$\bigl[\sin(\pi x)\bigr]$
$e^{\bigl[\cos(2\pi x)\bigr]}$
Notice that:
$\sin(\pi x)$ has period 2, so $\bigl[\sin(\pi x)\bigr]$ repeats its values every 2 units of $x$.
$\cos(2\pi x)$ has period 1, so $\bigl[\cos(2\pi x)\bigr]$ repeats every 1 unit of $x$. Hence $e^{\bigl[\cos(2\pi x)\bigr]}$ also has period 1.
Step 2: Break the original interval according to periodicity
The integral's limits are from $x = -3$ to $x = 101$. The total length of this interval is $101 - (-3) = 104.$
Since $\bigl[\sin(\pi x)\bigr]$ repeats every 2 units, we can group the interval into segments of length 2. Specifically, $104 \div 2 = 52$ full periods. Thus we may write:
$$
I \;=\; \int_{-3}^{101} \Bigl( \bigl[\sin(\pi x)\bigr] + e^{\bigl[\cos(2\pi x)\bigr]} \Bigr)\,dx
\;=\; 52 \times \int_{0}^{2} \Bigl(\bigl[\sin(\pi x)\bigr] + e^{\bigl[\cos(2\pi x)\bigr]}\Bigr)\,dx.
$$
(A shift in the starting point by an integer multiple of the periods does not change the value of the integral over one full period.)
Step 3: Evaluate the integral of $[\sin(\pi x)]$ from 0 to 2
Over one period from $x = 0$ to $x = 2$:
For $0 \le x < 1$: $\sin(\pi x)$ is nonnegative and takes values from 0 up to 1. Hence $\bigl[\sin(\pi x)\bigr] = 0$ for all $x \in (0,1)$. (Points where $\sin(\pi x) = 1$ are isolated and do not affect the definite integral.)
For $1 \le x < 2$: $\sin(\pi x)$ is negative, strictly between -1 and 0. Hence $\bigl[\sin(\pi x)\bigr] = -1$ for $x \in (1,2).$
Therefore,
$$
\int_{0}^{2} \bigl[\sin(\pi x)\bigr] \,dx
\;=\; \int_{0}^{1} 0 \,dx \;+\; \int_{1}^{2} (-1)\,dx
\;=\; 0 \;+\; \bigl(-1 \times (2 - 1)\bigr)
\;=\; -1.
$$
Step 4: Evaluate the integral of $e^{[\cos(2\pi x)]}$ from 0 to 2
Since $\cos(2\pi x)$ has period 1, it suffices to evaluate from 0 to 1, then double it for the interval from 0 to 2.
Over one interval $x \in [0,1]$, $\cos(2\pi x)$ completes one full cycle from 1 (at $x=0$) back to 1 (at $x=1$). We break it down:
$0 \le x < \tfrac{1}{4}$: $\cos(2\pi x)$ is in $(0,1]$, so $\bigl[\cos(2\pi x)\bigr] = 0.$ Therefore $e^{[\cos(2\pi x)]} = e^0 = 1.$
$\tfrac{1}{4} \le x < \tfrac{3}{4}$: $\cos(2\pi x)$ goes from 0 down to -1 and back up to 0, so $\bigl[\cos(2\pi x)\bigr] = -1.$ Hence $e^{[\cos(2\pi x)]} = e^{-1} = \tfrac{1}{e}.$
$\tfrac{3}{4} \le x < 1$: $\cos(2\pi x)$ goes from 0 to 1, so $\bigl[\cos(2\pi x)\bigr] = 0.$ Thus $e^{[\cos(2\pi x)]} = 1.$
So, for $x \in [0,1]:$
$$
\int_{0}^{1} e^{[\cos(2\pi x)]} \,dx
\;=\; \int_{0}^{\tfrac{1}{4}} 1\,dx
\;+\; \int_{\tfrac{1}{4}}^{\tfrac{3}{4}} \frac{1}{e}\,dx
\;+\; \int_{\tfrac{3}{4}}^{1} 1\,dx.
$$
Compute each part:
$$
\int_{0}^{\tfrac{1}{4}} 1\,dx = \tfrac{1}{4},
\quad
\int_{\tfrac{1}{4}}^{\tfrac{3}{4}} \frac{1}{e}\,dx = \frac{1}{e} \left(\tfrac{3}{4} - \tfrac{1}{4}\right) = \frac{1}{2e},
\quad
\int_{\tfrac{3}{4}}^{1} 1\,dx = \tfrac{1}{4}.
$$
Sum these:
$$
\int_{0}^{1} e^{[\cos(2\pi x)]} \,dx
= \tfrac{1}{4} + \tfrac{1}{2e} + \tfrac{1}{4}
= \tfrac{1}{2} + \tfrac{1}{2e}.
$$
Since the function repeats for $1 \le x < 2$, the integral from 0 to 2 is twice this value:
$$
\int_{0}^{2} e^{[\cos(2\pi x)]} \,dx
= 2 \left(\tfrac{1}{2} + \tfrac{1}{2e}\right)
= 1 + \tfrac{1}{e}.
$$
Step 5: Combine the results for one period (0 to 2)
Summing both parts over $x \in [0,2]$:
$$
\int_{0}^{2} \Bigl(\bigl[\sin(\pi x)\bigr] + e^{\bigl[\cos(2\pi x)\bigr]}\Bigr)\,dx
\;=\; \underbrace{\int_{0}^{2} \bigl[\sin(\pi x)\bigr] \,dx}_{-1}
\;+\; \underbrace{\int_{0}^{2} e^{[\cos(2\pi x)]} \,dx}_{1 + \tfrac{1}{e}}
\;=\; -1 \;+\; \Bigl(1 + \tfrac{1}{e}\Bigr)
= \tfrac{1}{e}.
$$
Step 6: Multiply by the number of full periods
Since there are 52 full periods of length 2 in the original interval from -3 to 101, the total integral is
$$
I
=\; 52 \times \int_{0}^{2} \Bigl(\bigl[\sin(\pi x)\bigr] + e^{\bigl[\cos(2\pi x)\bigr]}\Bigr)\,dx
=\; 52 \,\times\, \frac{1}{e}
=\; \frac{52}{e}.
$$
Final Answer
The value of the integral is
$$
\boxed{\frac{52}{e}}.
$$
