Question

Let \([\,]\) denote the greatest integer function. Then the value of} \[ \int_{0}^{3}\left(\frac{e^x+e^{-x}}{[x]!}\right)dx \] is:

• A. \(e^2+e^3-\frac{1}{e^2}-\frac{1}{e^3}\)
• B. \(\frac{1}{2}\left(e^2+e^3-\frac{1}{e^2}-\frac{1}{e^3}\right)\)
• C. \(e^2+e^3-\frac{1}{2e^2}-\frac{1}{2e^3}\)
• D. \(\frac{1}{2}(e^2+e^3)-\frac{1}{e^2}-\frac{1}{e^3}\)

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The Greatest Integer Function \([x]\) remains constant between integer intervals. We must break the integral into sub-intervals \([0, 1), [1, 2), [2, 3)\) to simplify the denominator.
Step 2: Key Formula or Approach:
1. \(\int (e^x + e^{-x}) dx = e^x - e^{-x}\).
2. Break the integral: \(\int_0^3 f(x) dx = \int_0^1 f(x) dx + \int_1^2 f(x) dx + \int_2^3 f(x) dx\).
Step 3: Detailed Explanation:
Break the integral:
- For \(x \in [0, 1), [x] = 0 \implies [x]! = 0! = 1\).
- For \(x \in [1, 2), [x] = 1 \implies [x]! = 1! = 1\).
- For \(x \in [2, 3), [x] = 2 \implies [x]! = 2! = 2\).
Integral \(I = \int_0^1 (e^x + e^{-x}) dx + \int_1^2 (e^x + e^{-x}) dx + \int_2^3 \frac{e^x + e^{-x}}{2} dx\).
Combine the first two terms:
\[ I = \int_0^2 (e^x + e^{-x}) dx + \frac{1}{2} \int_2^3 (e^x + e^{-x}) dx \]
\[ I = [e^x - e^{-x}]_0^2 + \frac{1}{2} [e^x - e^{-x}]_2^3 \]
\[ I = (e^2 - e^{-2}) - (e^0 - e^0) + \frac{1}{2} [(e^3 - e^{-3}) - (e^2 - e^{-2})] \]
\[ I = e^2 - \frac{1}{e^2} + \frac{1}{2} e^3 - \frac{1}{2 e^3} - \frac{1}{2} e^2 + \frac{1}{2 e^2} \]
\[ I = \frac{1}{2} e^2 + \frac{1}{2} e^3 - \frac{1}{2 e^2} - \frac{1}{2 e^3} = \frac{1}{2} (e^2 + e^3 - \frac{1}{e^2} - \frac{1}{e^3}) \]
Step 4: Final Answer:
The integral evaluates to \(\frac{1}{2}(e^2 + e^3 - \frac{1}{e^2} - \frac{1}{e^3})\).