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

23.3 Odd and even functions 731
the interval [−a,0] is the same as the corresponding area in the interval [0,a]. Hence
wecan write
∫ a
−a
f(t)dt =2
∫ a
0
f(t)dt
Example23.11 Evaluate ∫ π
−π tsintdt.
Solution Thefunctionst andsint arebothodd,andhencetheirproductiseven.Therefore,using
the factthatthe integrand iseven we can write
∫ π
−π
tsintdt=2
So,integrating by parts,
2
∫ π
0
∫ π
0
tsintdt
∫ π
)
tsintdt=2
([−tcost] π 0 + costdt
0
= 2((−πcosπ) − (0) +[sint] π 0 )
= 2π
EXERCISES23.3
1 Determine byinspection whether each ofthe (c)f(t) =t 3 ;
(b)f(t) = t 2 +1;
functionsin Figure23.11 isodd, even orneither. (d) f(t) =cost +0.1t 2 .
f(t)
f(t)
2 Byusingthe properties ofodd andeven functions
developed in Example23.9 statewhether the
following are odd, even orneither:
(a) t 3 sin ωt (b) tcos2t
t
t (c) sint sin4t (d) cos ωt sin2ωt
(e) e t sint
(a)
(b)
f(t)
f(t) 3 Evaluate the following integrals usingthe integral
propertiesofodd andeven functionswhere
appropriate:
t
t (a) ∫ 5
−5 t3 dt (b) ∫ 5
−5 t3 cos3tdt
(c)
(d)
(c) ∫ π
−π t2 sintdt (d) ∫ 2
−2 tcosh3tdt
Figure23.11
(a)The function: f (t) = −t;
(e) ∫ 1
−1 |t|dt (f) ∫ 1
−1 t|t|dt