Final answer:
The integral evaluates to 0.
Step-by-step explanation:
When evaluating the integral ∫[eˣ]dx from -∞ to ∞, we can break it into three parts: (-∞ to 0), (0 to L), and (L to ∞). However, since the particle is constrained to be in the tube, C = 0 outside the tube, making the first and last integrations zero. Therefore, the integral can be simplified to:
∫[0]dx from 0 to L
Integrating [0]dx is equivalent to integrating the zero function from 0 to L, which results in 0. Therefore, the integral evaluates to 0.