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: Breakdown the integral based on the property of the Greatest Integer Function $[x]$.
The value of $[x]$ changes at integer values of $x$. We split the integration interval $[0, 3]$ into $[0, 1)$, $[1, 2)$, and $[2, 3)$.

For $0 ≤ x < 1$, $[x] = 0$ and $[x]! = 0! = 1$.
For $1 ≤ x < 2$, $[x] = 1$ and $[x]! = 1! = 1$.
For $2 ≤ x < 3$, $[x] = 2$ and $[x]! = 2! = 2$.


Step 2: Rewrite the integral as a sum.
$I = \int_{0}^{1} \frac{e^x + e^{-x}}{1} dx + \int_{1}^{2} \frac{e^x + e^{-x}}{1} dx + \int_{2}^{3} \frac{e^x + e^{-x}}{2} dx$
Combining the first two integrals:
$I = \int_{0}^{2} (e^x + e^{-x}) dx + \frac{1}{2} \int_{2}^{3} (e^x + e^{-x}) dx$


Step 3: Integrate.
$I = \left[ e^x - e^{-x} \right]_{0}^{2} + \frac{1}{2} \left[ e^x - e^{-x} \right]_{2}^{3}$
$I = (e^2 - e^{-2}) - (e^0 - e^0) + \frac{1}{2} ((e^3 - e^{-3}) - (e^2 - e^{-2}))$
$I = e^2 - e^{-2} + \frac{1}{2} e^3 - \frac{1}{2} e^{-3} - \frac{1}{2} e^2 + \frac{1}{2} e^{-2}$


Step 4: Simplify the expression.
$I = \frac{1}{2} e^2 - \frac{1}{2} e^{-2} + \frac{1}{2} e^3 - \frac{1}{2} e^{-3}$
$I = \frac{e^3 + e^2 - e^{-2} - e^{-3}}{2}$
This matches option (1).