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Ch. 8 - Integration Techniques
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Non è quello che usi tu?Cambia libro di testo
Capitolo 8, Problema 8.9.63

63. Average Lifetime The average time until a computer chip fails (see Exercise 62) is 0.00005 ∫(from 0 to ∞) t e^(-0.00005t) dt. Find this value.

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Recognize that the problem asks for the average lifetime, which is given by the integral \(0.00005 \int_0^{\infty} t e^{-0.00005 t} \, dt\). This is an expected value calculation for a continuous random variable with probability density function proportional to \(e^{-\lambda t}\), where \(\lambda = 0.00005\).
Recall the formula for the expected value of an exponential distribution: if \(X\) has PDF \(f(t) = \lambda e^{-\lambda t}\) for \(t \geq 0\), then \(E[X] = \int_0^{\infty} t \lambda e^{-\lambda t} \, dt = \frac{1}{\lambda}\).
Identify that the integral inside the problem matches the form \(\int_0^{\infty} t e^{-\lambda t} \, dt\), but the problem includes the factor \(0.00005\) outside the integral, which corresponds to \(\lambda\).
Use integration by parts to verify the integral \(\int_0^{\infty} t e^{-\lambda t} \, dt\): let \(u = t\) and \(dv = e^{-\lambda t} dt\), then compute \(du\) and \(v\), and apply the integration by parts formula \(\int u \, dv = uv - \int v \, du\).
After evaluating the integral, multiply the result by \(0.00005\) as given, and simplify to find the average lifetime value.

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