47. For the probability density P(x) = 0.5 e−0.5x the integral
∫0∞ P(x) dx = __________.
Probability density functions (PDFs) must integrate to 1 over their allowable range. This question tests that concept using a classic exponential distribution:
P(x) = 0.5 e-0.5x, x ≥ 0
We evaluate:
∫₀∞ P(x) dx
Step-by-Step Solution
Given:
P(x) = 0.5 e-0.5x
Required integral:
∫₀∞ 0.5e-0.5x dx
Factor out constants:
0.5 ∫₀∞ e-0.5x dx
Integrate the exponential
Use:
∫ e-ax dx = -(1/a)e-ax
Here a = 0.5:
∫ e-0.5x dx = -2 e-0.5x
Apply limits (0 to ∞):
0.5[-2e-0.5x]\_0∞
Evaluate:
- as x → ∞ → e-0.5x → 0
- at x = 0 → e0 = 1
= 0.5 (0 – (-2)(1)) = 0.5 × 2 = 1
Final Answer
∫₀∞ 0.5e-0.5x dx = 1
Why the Answer Must Be 1
PDFs describe continuous probability; total probability = 1.
This is a classic exponential distribution with parameter λ = 0.5:
∫₀∞ λe-λx dx = 1
Conclusion
The given function is a properly normalized probability density. Results like this appear in biostatistics, decay models, queue theory, and GATE/CSIR probability questions.
