Every slope is a derivative.

x
Velocity = slope of the tangent
line to a position vs.
time graph
t

v
t

a

 dx
v
dt
Acceleration = slope of the
velocity vs. time
graph

dv
a
dt
2
t
d x
a 2
dt
How then can we move up the stack of graphs?

a

a
t
t1
t2
Area under the curve tt12  A0t2  A0t1


 at2  at1

 a t2  t1 

 at
 
v  at

v

v2

v1
t
t2
t1
Area
Area under the curve tt12  A0t2  A0t1
1  1 
 t2v2  t1v1
2
2
But
  
v1  vo  at1
  
v2  vo  at2
1 
1 
 t2 at2   t1 at1 
2
2

v

v2

v1
t
t2
t1
1 2 1 2
Area under
 at2  at1
2
2


 x2  x1

t
Area under the curve t  x
the curve tt12
2
1
from t1 to t2

v

v
What if the graph is
Fill the area with rectangles
t
t1
t
t2
One method to calculate the
area under the curve
from t1 to t2 would be
t1

v
t2
Increase the # of rectangles
Areatt12  Area0t2  Area0t1
t
t1
t2

v

v
What if the graph is
Fill the area with rectangles
t
t
t1
t1
t2
Area under the curve  lim  Arec
n 

v
t2
Increase the # of rectangles
where n  # of rectangles
Note that the area of a rectangle is yx,
while the slope is y .
x
t
t1
t2

v
t
t1
t2
Definite Integral – the area under the curve between definite limits
Upper limit
t2
t1
f ( x)dx  F (t2 )  F (t1 )
Lower limit
The integral of the function f ( x) with respect to x from t1 to t2
Let a, b, g, e, m, and n be constants.
Differentiation
Integration
Formula
Formula
y  ax
dy d
 (ax)
dx dx
d
ax   a
dx
Power Rule
ya
g
e
y dx  eg a dx
g
e a
y  xn
dy d n
 x
dx dx
d n
x  nx n1
dx
dx  ax eg
y  xn
g
e
g
e
y dx  eg x n dx
n 1 g
x 
x dx  

n

1

e
n
Let a, b, g, e, m, and n be constants.
Differentiation
Integration
Formula
Formula
y  ax n
dy d n
 ax
dx dx
dy
d n
a x
dx
dx
d n
ax  anx n1
dx
y  ax n
g
e
y dx  eg ax n dx
g
n
ax
dx
e
g
n
ax
dx
e
 a eg x n dx
n 1 g
 ax 


 n  1 e
Let a, b, g, e, m, and n be constants.
Differentiation
Integration
Formula
Formula
Sum or Difference
y  axn  bxm
dy d
 (ax n  bx m )
dx dx
dy d
d
n
 (ax )  (bx m )
dx dx
dx
dy
 anx n1  bmx m1
dx
y  axn  bxm
g
e
y dx 
g
n
(
ax
e
 bx m )dx
 eg ax n dx  eg bx m dx
n 1
m 1 g
 ax
bx 



n

1
m

1

e
Example 1

An object has a velocity given by v  (4 m
2 )t
 (1m ).
s
s
a. What is the object' s displacement from 0 s  t  1 s?
 s

vm
Method 1


x  Area under the v vs. t graph
18
16

x  A 01
14
12
1

m
x  (1 )(1 s )  (1 s )(4 m )
s
s
2

x  3 m
10
8
6
4
2
0
1
2
3
4
t s 
Example 1

An object has a velocity given by v  (4 m
2 )t
 (1m ).
s
s
a. What is the object' s displacement from 0 s  t  1 s?
 s

vm
Method 2
 1
x  0 v dt
 1
x  0 (4t  1) dt
 1
x  0 4t dt  011 dt
1
1 01 
  4 11
x  
t 
t 
0  1 0
1  1
1

2
x  2t  t 0
18
16
14
12
10
8
6
4
2
0
1
2
3
4
t s 
 

x  2(1)  1  2(0)
2

x  3 m
2
0

Example 1

An object has a velocity given by v  (4 m
2 )t
 (1m ).
s
s
b. What is the object' s displacement from 2 s  t  4 s?
 s

vm
Method 1


x  Area under the v vs. t graph

x
x  A04  A02
1
 

m
m
x  4 s  1
 4 s  16
s
s 
2

1


m
 2 s  1
 2 s  8 m 
s 2
s

18
16
14
12
10
8
6
4
2
0
1
2
3
4
t s 
 


 



x  26 m
Example 1

An object has a velocity given by v  (4 m
2 )t
 (1m ).
s
s
b. What is the object' s displacement from 2 s  t  4 s?

Method 2
vm
s
 4
18
x  2 v dt
16
 4
14
x  2 (4t  1) dt
12
 


x  2t 2  t
10
8

6
 

x  2(4) 2  4  2(2) 2  2
4
2
0

4
2
1
2
3
4
t s 

x  26 m

Example 2

Calculate the change in velocity (v ) from 1 s  t  2 s for an object whose accelerati on

is given by a  t 2 .
 2
v  1 a dt

a  m 2 
 2 2
 s 
v  1 (t ) dt
21 2
 t 
v  

 2  11
2
 1 3 
v   t 
 3 1
1
2
t s 
 1 3 1 3
v  (2)  (1)
3
3

v  2.3 m
s