Question:

<p><strong>Step 1: Analyze the floor function on interval [n-1, n]</strong></p><p>For x ∈ [n-1, n), we have [x] = n-1 (the floor function is constant).</p><p>Therefore, on the interval [n-1, n]:</p><p>$$e^{x-[x]} = e^{x-(n-1)} = e^{x-n+1}$$</p><p><strong>Step 2: Evaluate a single integral</strong></p><p>$$\int_{n-1}^{n} e^{x-[x]} dx = \int_{n-1}^{n} e^{x-n+1} dx$$</p><p>Let u = x - n + 1, then du = dx</p><p>When x = n-1: u = 0</p><p>When x = n: u = 1</p><p>$$\int_{n-1}^{n} e^{x-n+1} dx = \int_{0}^{1} e^{u} du = [e^u]_{0}^{1} = e^1 - e^0 = e - 1$$</p><p><strong>Step 3: Sum from n = 1 to 100</strong></p><p>$$\sum_{n=1}^{100} \int_{n-1}^{n} e^{x-[x]} dx = \sum_{n=1}^{100} (e-1)$$</p><p>Since each term equals (e-1) and there are 100 terms:</p><p>$$= 100(e-1)$$</p><p><strong>∴ Answer:</strong> a</p>