{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

# lec35 - Lecture 35 18.01 Fall 2006 Lecture 35 Improper...

This preview shows pages 1–3. Sign up to view the full content.

± ± ² ² ² ² ² ± ± ² ² ² ² ± ± Lecture 35 18.01 Fall 2006 Lecture 35: Improper Integrals Defnition. An improper integral , defned by ± M f ( x ) dx = lim f ( x ) dx a M →∞ a is said to converge i± the limit exists (diverges i± the limit does not exist). e - kx dx = 1 /k ( k > 0) Example 1. 0 ± M M e - kx dx = ( - 1 /k ) e - kx = (1 /k )(1 - e - kM ) 0 0 Taking the limit as M → ∞ , we fnd e - kM 0 and e - kx dx = 1 /k 0 We rewrite this calculation more in±ormally as ±ollows, 0 e - kx dx = ( - 1 /k ) e - kx 0 = (1 /k )(1 - e - k ) = 1 /k (since k > 0 ) e - kx dx = 1 /k has an easier ±ormula than the Note that the integral over the infnite interval ± M 0 corresponding fnite integral e - kx dx = (1 /k )(1 - e - kM ) . As a practical matter, ±or large M , the 0 term e - kM is negligible, so even the simpler ±ormula 1 /k serves as a good approximation to the fnite integral. Infnite integrals are o±ten easier than fnite ones, just as infnitesimals and derivatives are easier than di²erence quotients. Application: Replace x by t = time in seconds in Example 1. R = rate o± decay = number o± atoms that decay per second at time 0 . At later times t > 0 the decay rate is Re - kt (smaller by an exponential ±actor e - kt ) Eventually (over time 0 t < ) every atom decays. So the total number o± atoms N is calculated using the ±ormula we ±ound in Example 1, Re - kt dt = R/k N = 0 The hal± li±e H o± a radioactive element is the time H at which the decay rate is hal± what it was at the start. Thus e - kH = 1 / 2 = - kH = ln(1 / 2) = k = (ln 2) /H 1

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
± ² ² ² ² ² Lecture 35 18.01 Fall 2006 Hence R = Nk = N (ln 2) /H Let us illustrate with Polonium 210, which has been in the news lately. The half life is 138 days or H = (138 days )(24 hr/day )(60 2 sec/hr ) = (138)(24)(60) 2 seconds Using this value of H
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 6

lec35 - Lecture 35 18.01 Fall 2006 Lecture 35 Improper...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online