Answer
See below
Work Step by Step
Obtain: $F(s)=\int^{\infty}_0 e^{-st}f(t)dt\\
=\int^1_0 2e^{-st}dt+\int^{\infty}_1 te^{-st} dt\\
=\frac{2(1-e^{-s})}{s}+\frac{e^{-s}(1+s)}{s^2}\\
=\frac{e^{-s}[1+(2e^s-1)s]}{s^2}$
Differential Equations and Linear Algebra (4th Edition)
by Goode, Stephen W.; Annin, Scott A.
Published by
Pearson
ISBN 10:
0-32196-467-5
ISBN 13:
978-0-32196-467-0
Chapter 10 - The Laplace Transform and Some Elementary Applications - 10.10 Chapter Review - Additional Problems - Page 718: 8
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