1. Evaluate the following integrals involving 8(t): e4s cos(s)8(8-2) ds. (a) fe (b) S (8³+ s)8(8-2) ds. (c) S s²ln(1 + s)8(s 1) ds. (Your answer will depend on t.) -

Calculus For The Life Sciences
2nd Edition
ISBN:9780321964038
Author:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Publisher:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Chapter4: Calculating The Derivative
Section4.2: Derivatives Of Products And Quotients
Problem 36E
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Note: Recall that the delta function 8(t) is the first derivative of the step function H(t). The
key property of 8(t) is the way it behaves in integrals: for any number c and any function f,
[ f(s)8(8 — c) ds =
i.e. the integral "picks out" the value of the function f at the location c where 8(t - c) has a
spike. The integral is zero if the spike lies outside the domain of integration.
For integrals from 0 to t, when c> 0, we can rewrite the rule more concisely with a step
function:
(b)
1. Evaluate the following integrals involving 8(t):
(a) fe els cos(s)8(s-2) ds.
S
(c)
[f(c), a<c<b,
otherwise,
[ f(s)5(s – c) ds = f(c)H(t - c).
(s³ + 8)8(82) ds.
Ső
s² ln(1 + s)8(s 1) ds. (Your answer will depend on t.)
Transcribed Image Text:Note: Recall that the delta function 8(t) is the first derivative of the step function H(t). The key property of 8(t) is the way it behaves in integrals: for any number c and any function f, [ f(s)8(8 — c) ds = i.e. the integral "picks out" the value of the function f at the location c where 8(t - c) has a spike. The integral is zero if the spike lies outside the domain of integration. For integrals from 0 to t, when c> 0, we can rewrite the rule more concisely with a step function: (b) 1. Evaluate the following integrals involving 8(t): (a) fe els cos(s)8(s-2) ds. S (c) [f(c), a<c<b, otherwise, [ f(s)5(s – c) ds = f(c)H(t - c). (s³ + 8)8(82) ds. Ső s² ln(1 + s)8(s 1) ds. (Your answer will depend on t.)
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