A car moving with initial velocity of 5 mph accelerates at
a t  2.4t mph per second for 8 seconds.

(a) How fast is the car going when the 8 seconds are up?
(b) How far did the car travel during these 8 seconds?
Section 7.1b
The integral is a natural tool to calculate net change
and total accumulation of more quantities than just
distance and velocity (ex: growth, decay, consumption,
acceleration).
Whenever you want to find the cumulative effect of a
varying rate of change, integrate it!!!
A car moving with initial velocity of 5 mph accelerates at
a t  2.4t mph per second for 8 seconds.

(a) How fast is the car going when the 8 seconds are up?
Net velocity change:

mph

1.2t

76.8
a
t
dt

2.4
tdt


0
0
0
8
8
2 8
Final velocity after 8 seconds:
Initial velocity + Net velocity change
 5  76.8  81.8 mph
A car moving with initial velocity of 5 mph accelerates at
a t  2.4t mph per second for 8 seconds.

(b) How far did the car travel during these 8 seconds?
Velocity function: 1.2t 2
 5 mph
Distance traveled:

8
0
v  t  dt   1.2t  5 dt   0.4t  5t  0
0
 244.8 mph x sec
8
244.8 mi x sec
1 hr
3
2
x
1 hr
3600 sec
 0.068 mi
8
The graph of the velocity of a particle moving on the x-axis is
given below. The particle starts at x = 2 when t = 0.
(a) Find where the particle is at the end of the trip.
(b) Find the total distance traveled by the particle.
v (m/sec)
Final position:
2   v  t  dt  2  0.5  2  3
10
3
0
t (sec)
10
–3
0.5 1 3  3  3
0.5 1 3  0.5  33
 2.5 meters
The graph of the velocity of a particle moving on the x-axis is
given below. The particle starts at x = 2 when t = 0.
(a) Find where the particle is at the end of the trip.
(b) Find the total distance traveled by the particle.
v (m/sec)
Total distance:

3
10
0
t (sec)
10
–3
v  t  dt  0.5  2 3
0.5 1 3  3  3
0.5 1 3  0.5  33
 19.5 meters
From 1970 to 1980, the rate of potato consumption in a particular
country was C t  2.2  1.1t millions of bushels per year, with
t being years since the beginning of 1970. How many bushels
were consumed from the beginning of 1972 to the end of 1973?

We need the cumulative effect of the consumption rate for
2t 4
Consumption:

4
C  t  dt    2.2  1.1  dt
4
t
2
2
Evaluate numerically:
NINT  2.2  1.1 , t , 2, 4   7.066 million
bushels
t
Suppose that the average rate of electricity consumption for a
certain home is modeled by the given function, where C(t) is
measured in kilowatts and t is the number of hours past midnight.
Find the average daily consumption for this home, measured in
kilowatt-hours.
Consumption:

24
0
C  t   3.9  2.4sin  t 12


28.8
 t 
  t 
 3.9  2.4sin  12   dt  3.9t   cos  12  
  0
 


24
 93.6 kilowatt-hours
Oil flows through a cylindrical pipe of radius 3 in, but friction from
the pipe slows the flow toward the outer edge. The speed at
which the oil flows at a distance r in from the center is
8 10  r 2 inches per second.


(a) In a plane cross section of the pipe, a thin ring with thickness
r at a distance r inches from the center approximates a
rectangular strip when you straighten it out. What is the area
of the strip (and hence the approximate area of the ring)?
Width
 r
Area
Length
 2 rr
 2 r
Oil flows through a cylindrical pipe of radius 3 in, but friction from
the pipe slows the flow toward the outer edge. The speed at
which the oil flows at a distance r in from the center is
8 10  r 2 inches per second.


(b) Explain how we can find the rate at which oil passes through
this ring.
Volume per sec. = Inches per sec. x Cross section area
3
in
in
2
8 10  r 2 
2

r

r
  in = flow in
sec
sec
Oil flows through a cylindrical pipe of radius 3 in, but friction from
the pipe slows the flow toward the outer edge. The speed at
which the oil flows at a distance r in from the center is
8 10  r 2 inches per second.


(c) Set up and evaluate a definite integral that will give the rate
(in cubic inches per sec) at which oil is flowing through the pipe.
 8 10  r   2 r  dr  16  10r  r  dr
3
3
2
0
3
0
3
in
in
 2 1 4
 1244.071
 16 5r  r   396
sec
sec
4 0

3
3
In science, the term work refers to a force acting on a body and
the body’s subsequent displacement.
When a body moves a distance d along a straight line as a result
of the action of a force of constant magnitude F in the direction
of the motion, the work done by the force is
W  Fd
This is called the constant-force formula for work.
Units of work are force x distance. In the metric system, the
unit is Newton-meter (called a Joule); in the U.S. customary
system, the most common unit is the foot-pound.
Hooke’s Law for springs says that the force it takes to stretch or
compress a spring x units from its natural (unstressed) length is
a constant times x. In symbols:
F  kx
where k, measured in force units per unit length, is a
characteristic of the spring called the force constant.
Another important note: Numerically, work is area under
the force graph!!!
It takes a force of 10 N to stretch a spring 2 m beyond its natural
length. How much work is done in stretching the spring 4 m from
its natural length?
First, we need to find the “force equation” for this particular spring:
F  x   kx
F  2  10  k  2  k  5 N m
F  x   5x
for this particular spring
It takes a force of 10 N to stretch a spring 2 m beyond its natural
length. How much work is done in stretching the spring 4 m from
its natural length?
Now, we construct an integral for the work done in applying F
over the interval from x = 0 to x = 4:
4
4
4
2

0
5 
F  x  dx   5xdx  x   40 N m
0
2 0