Question

The value of $\int_{0}^{3} \frac{e^x + e^{-x}}{[x]!} dx$ is equal to (where $[\, \cdot \,]$ denotes greatest integer function):

• A. $\frac{e^3 + e^2 - e^{-2} - e^{-3}}{2}$
• B. $\frac{e^3 - e^2 - e^{-2} + e^{-3}}{2}$
• C. $e^3 + e^2 - e^{-2} - e^{-3}$
• D. $e^3 - e^2 - e^{-2} + e^{-3}$

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The greatest integer function \( [x] \) changes its value at every integer.
To evaluate the integral, we must split the interval \( [0, 3] \) into sub-intervals where \( [x] \) is constant.
Step 2: Key Formula or Approach:
Split the integral as:
\[ I = \int_{0}^{1} \dots dx + \int_{1}^{2} \dots dx + \int_{2}^{3} \dots dx \]
For \( x \in [0, 1) \), \( [x] = 0 \implies [x]! = 0! = 1 \).
For \( x \in [1, 2) \), \( [x] = 1 \implies [x]! = 1! = 1 \).
For \( x \in [2, 3) \), \( [x] = 2 \implies [x]! = 2! = 2 \).
Step 3: Detailed Explanation:
Substitute the values into the integral:
\[ I = \int_{0}^{1} \frac{e^x + e^{-x}}{1} dx + \int_{1}^{2} \frac{e^x + e^{-x}}{1} dx + \int_{2}^{3} \frac{e^x + e^{-x}}{2} dx \]
Combine the first two integrals:
\[ I = \int_{0}^{2} (e^x + e^{-x}) dx + \frac{1}{2} \int_{2}^{3} (e^x + e^{-x}) dx \]
Integrate term by term:
\[ I = [e^x - e^{-x}]_{0}^{2} + \frac{1}{2} [e^x - e^{-x}]_{2}^{3} \]
\[ I = (e^2 - e^{-2}) - (e^0 - e^0) + \frac{1}{2} [(e^3 - e^{-3}) - (e^2 - e^{-2})] \]
\[ I = e^2 - e^{-2} + \frac{e^3 - e^{-3} - e^2 + e^{-2}}{2} \]
\[ I = \frac{2e^2 - 2e^{-2} + e^3 - e^{-3} - e^2 + e^{-2}}{2} = \frac{e^3 + e^2 - e^{-2} - e^{-3}}{2} \]
Step 4: Final Answer:
The value of the definite integral is \( \frac{e^3 + e^2 - e^{-2} - e^{-3}}{2} \).