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

492 Chapter 16 Further topics in integration
f (t)
2
1 3 t
Figure16.3
Thefunction f (t) =
{ 2 0t<1
t 2 1<t 3.
On the intervals (0, 1) and (1, 3), f (t)iscontinuous, so
∫ 3
0
fdt=
=
∫ 1
0
[
2t
2dt+
] 1
0
∫ 3
1
t 2 dt
[ ] t
3 3
+ = 2 +
3
1
[
9 − 1 ]
= 32 3 3
EXERCISES16.5
1 Given
⎧
⎪⎨
3 −1t<1
f(t)= 2t 1t2
⎪⎩
t 2 2<t3
evaluate
∫ 1 ∫ 1.5
(a) f dt (b) f dt
−1
−0.5
∫ 2.5 ∫ 3
(c) f dt (d) f dt
0
−1
2 Given
⎧
⎨3t
0t<3
g(t) = 15−2t 3t<4
⎩
6 4t6
evaluate
∫ 2 ∫ 4
(a) g(t)dt (b) g(t)dt
0 2
∫ 5
∫ 6
(c) g(t)dt (d) g(t)dt
3
0
∫ 4.5
(e) g(t)dt
3.5
3 Givenu(t)isthe unitstepfunction, evaluate
∫ 4
(a) u(t)dt
0
∫ 2
(b) u(t)dt
−3
∫ 4
(c) 2u(t +1)dt
−2
∫ 2
(d) tu(t)dt
−1
∫ 4
(e) e kt u(t −3)dt k constant
0
Solutions
1 (a) 6 (b) 5.75
(c) 8.5417 (d) 15.3333
2 (a) 6 (b) 15.5 (c) 14
(d) 33.5 (e) 6.75
3 (a)4 (b)2
(c) 10 (d) 2
e 4k −e 3k
(e)
k