Question

(a) Compute the definite integral IN = ∫N0 e^(-st)e^(5t) dt. (b) Take the limit N → ∞ in the expression above to obtain the Laplace Transform F(s) = L[e^(5t)]. (c) Find the domain DF where F(s) is defined.

          (a) Compute the definite integral IN = ∫N0 e^(-st)e^(5t) dt.
(b) Take the limit N → ∞ in the expression above to obtain the Laplace Transform F(s) = L[e^(5t)].
(c) Find the domain DF where F(s) is defined.
        

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Calculus: Early Transcendentals
Calculus: Early Transcendentals
James Stewart 8th Edition
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(a) Compute the definite integral IN = ∫N0 e^(-st)e^(5t) dt. (b) Take the limit N → ∞ in the expression above to obtain the Laplace Transform F(s) = L[e^(5t)]. (c) Find the domain DF where F(s) is defined.
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Transcript

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00:01 I given i n is equal to integral 0 to n e rise to the power minus s t into e rise to the power 3 t into d t we need to compute this integral this implies i n is given by integral 0 to n e rise to the power 3 minus s into t into d t now integrating with respect to t we have e rise to the par 3 minus s into t divided by the coefficient 3 minus s within the limits 0 to n.
00:49 Therefore, substituting the limit and simplifying finally we get in is equal to e rise to the power 3 minus s into n minus 1 divided by 3 minus s.
01:06 Next, we need to take the limit n tending to infinity to the obtained expression, that is, limit intending to infinity, e rise to the power 3 minus s into n minus 1 divided by 3 minus s.
01:28 This can be written as limit intending to infinity, e rise to the power, s minus 3 minus 3 ,000 into minus 3.
01:38 Minus n minus 1 divided by 3 minus s.
01:46 Now applying the limit we have e rise to the power minus infinity is 0.
01:52 Therefore we have minus 1 divided by 3 minus s.
01:57 This can also be written as 1 divided by s minus 3...
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