John K. answered • 10/06/14

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Effective Specialist in Never-Been-Good-at-Math situations

Since a > b > 0, in the interval [0, infinity), the graph of e

^{−bx}will always be above the graph of e^{−ax}. So the set-up of the area integral is ∫(e^{−bx}− e^{−ax})dx from 0 to ∞. This can be rewritten as the difference of two integrals:(∫e

^{−bx}dx from 0 to ∞) − (∫e^{−ax}dx from 0 to ∞).For the first, we can write lim N→∞ [(−1/b)e

^{−bx}] from 0 to N and for the second lim N→∞ [(−1/a)e^{−ax}] from 0 to N.Applying the Fundamental Theorem of Calculus:

lim N→∞{[(−1/b)(e

^{−Nb}− e^{0})] − [(−1/a)(e^{−Na}− e^{0})]} the terms with −Nk go to 0, since e^{−Nk}= 1/e^{Nk}→ 0 as N→∞, where k is a constant.We have (−1/b)(−1) − (−1/a)(−1) = (1/b) − (1/a) = (a − b)/ab.