More Applications of Definite Integrals 315Total Distance Traveled∫ 21v(t)dt+∫ 32−v(t)dt+∫ 43v(t)dt.Enter∫
(y1(x),x,1,2)and obtain1
4
.
Enter∫
(−y1(x),x,2,3)and obtain1
4
.
Enter∫
(y1(x),x,3,4)and obtain9
4
.
Thus, total distance traveled is(
1
4+
1
4
+
9
4
)
=11
4
.
Example 4
The acceleration function of a moving particle on a coordinate line isa(t)=−4 andv 0 = 12
for 0≤t≤8. Find the total distance traveled by the particle during 0≤t≤8.
a(t)=− 4v(t)=∫
a(t)dt=∫
− 4 dt=− 4 t+CSincev 0 = 12 ⇒−4(0)+C=12 orC= 12.
Thus,v(t)=− 4 t+ 12.Total Distance Traveled=∫ 80∣∣
− 4 t+ 12∣∣
dt.Let− 4 t+ 12 = 0 ⇒t= 3.|− 4 t+ 12 |={
− 4 t+ 12 if 0≤t≤ 3
−(− 4 t+12) ift> 3
∫ 80∣∣
− 4 t+ 12∣∣
dt=∫ 30∣∣
− 4 t+ 12∣∣
dt+∫ 83−(− 4 t+12)dt=
[
− 12 t^2 + 12 t] 3
0 +[
2 t^2 + 12 t] 8
3
= 18 + 50 = 68.
Total distance traveled by the particle is 68.Example 5
The velocity function of a moving particle on a coordinate line isv(t)=3 cos(2t) for
0 ≤t≤ 2 π. Using a calculator:
(a) Determine when the particle is moving to the right.(b) Determine when the particle stops.
(c) The total distance traveled by the particle during 0≤t≤ 2 π.