082-Engineering-Mathematics-Anthony-Croft-Robert-Davison-Martin-Hargreaves-James-Flint-Edisi-5-2017

16.5 Integration of piecewise continuous functions 491
∫ ∞ (1+t)δ(t −1)
(e) dt
−∞ 2
∫ ∞
(f) e −kt δ(t)dt
−∞
∫ ∞
(g) e −kt δ(t −a)dt
−∞
∫ ∞
(h) e −k(t−a) δ(t −a)dt
−∞
2 Evaluate the following integrals:
∫ ∞
(a) (sint)δ(t −2)dt
−∞
∫ ∞
(b) e −t δ(t +1)dt
−∞
∫ 0
(c) e −t δ(t +3)dt
−∞
∫ 10
(d) x 3 δ(x−2)dx
0
∫ 1
(e) x 2 δ(x+2)dx
−1
3 Evaluate the following:
∫ ∞
(a) t(sin2t)δ(t −3)dt
−∞
∫ ∞
(b) δ(t+1)−δ(t−1)dt
0
∫ ∞
(c) δ(t −d)dt
0 ∫ a
(d) δ(t −d)dt
−a
Solutions
1 (a) 1 (b) 54.5982
(c) 4.9787 ×10 −2 (d) 36
(e) 1 (f) 1
(g) e −ak (h) 1
2 (a) 0.9093
(b) 2.7183
(c) 20.0855
(d) 8
(e) 0
3 (a) −0.8382
(b) −1
(c) 1 ifd 0,0otherwise
(d) 1 if −a d a,0otherwise
16.5 INTEGRATIONOFPIECEWISECONTINUOUSFUNCTIONS
Integration of piecewise continuous functions is illustrated in Example 16.14. If a discontinuity
occurs within the limits of integration then the interval is divided into subintervals
so thatthe integrand iscontinuouson eachsub-interval.
Example16.14 Given
f(t)=
{
2 0 t<1
t 2
evaluate ∫ 3
0 f(t)dt.
1<t3
Solution The function, f (t), is piecewise continuous with a discontinuity att = 1. The function
is illustrated in Figure 16.3. The discontinuity occurs within the limits of integration.
We split the interval ofintegration atthe discontinuitythus:
∫ 3
0
f(t)dt =
∫ 1
0
f(t)dt+
∫ 3
1
f(t)dt