230 Ordinary Differential Equations
If f(x) is piecewise continuous on a finite interval [ a, b] and is continuous
except possibly at the points a = a 1 < a 2 < · · · < an = b, then f is integrable
on [a, b] and
l
b 1 a2 1 a3 1an
a f(x) dx = a1 f(x) dx + a2 f(x) dx + ... + an-1 f(x) dx.EXAMPLE 7 Laplace Transform of a Piecewise Continuous
FunctionCompute the Laplace transform of the piecewise continuous function{x ,f(x) =
3,0::::; x < 2
SOLUTION
By definition,(4) .C[f(x) ] = l:xo f(x)e-sx dx = 1
2
xe-sx dx + 1= 3e-sx dx.To calculate the first integral on the right-hand side of equation ( 4), we use
integrations by parts. Letting u = x and dv = e-sx and differentiating
and integrating, we find du= dx and v = -(1/s)e-sx. Substituting these
expressions into the integration by parts formula, yields
(5)
12 - x 12
112 -2 { 1 12
xe-sx dx = - e-sx + - e-sx dx = - e-2s - 2e-sx }
0 S 0 So S S 0= -2 -e-2s _ -[^1 e-2s _ l].
s s^2Evaluating the second integral on the right-hand side of equation (4), we find
(6)
100
3e-sx dx = -3 - e-sx 1= =^3 - e-2s
2 s 2 sfor s > 0.
Substituting the results from equations (5) and (6) into ( 4), we get
-2 1 3 1 1
.C[f(x) ] = - e-2s - -[e-2s -1] + - e-2s = - e-2s - -[e-2s -1] for s > 0.
s s^2 s s s^2