570
Chapter 5 | Integration
5/3
3 | v ( t ) | dt = ⌠ ⌡ 0
5/3 3
∫
− v ( t ) dt + ∫
v ( t ) dt
0
5/3
5/3 3
= ∫
5−3 tdt + ∫
3 t −5 dt
0
⎞ ⎠ | 0
⎠ | 5/3 3
⎛ ⎝ 5 t − 3 t
⎛ ⎝ 3 t 2
t ⎞
5/3
2
=
+
2
2 −5
⎡ ⎣ 5
⎤ ⎦ −0+
⎡ ⎣ 3(5/3) 2
⎤ ⎦
⎛ ⎝ 5 3
⎞ ⎠ − 3(5/3)
⎡ ⎣ 27
⎤ ⎦ −
2
25 3
=
2
2 −15
2 −
25 6
25 6
= 25 3 −
+ 27 2 −15−
+ 25 3
= 41 6
.
So, the total distance traveled is 14 6
m.
5.22
Find the net displacement and total distance traveled in meters given the velocity function f ( t ) = 1 2 e t −2 over the interval [0, 2].
Applying the Net Change Theorem The net change theorem can be applied to the flow and consumption of fluids, as shown in Example 5.26 . Example 5.26 How Many Gallons of Gasoline Are Consumed? If the motor on a motorboat is started at t =0 and the boat consumes gasoline at 5− t 3 gal/hr for the first hour, how much gasoline is used in the first hour? Solution Express the problem as a definite integral, integrate, and evaluate using the Fundamental Theorem of Calculus. The limits of integration are the endpoints of the interval [0, 1]. We have ∫ 0 1 ⎛ ⎝ 5− t 3 ⎞ ⎠ dt = ⎛ ⎝ 5 t − t 4 4 ⎞ ⎠ | 0 1 = ⎡ ⎣ 5(1)− (1) 4 4 ⎤ ⎦ −0
=5− 1 4 =4.75.
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