5.1 Estimating with Finite Sums
Consider an object moving at a constant rate of 3 ft/sec.
Since rate . time = distance: 3t  d
If we draw a graph of the velocity, the distance that the
object travels is equal to the area under the line.
3
After 4 seconds,
the object has
gone 12 feet.
2
velocity
1
0
1
2
time
3
4
ft
3
 4 sec  12 ft
sec
5.1 Estimating with Finite Sums
If the velocity is not constant,
we might guess that the
distance traveled is still equal
to the area under the curve.
We could estimate the
area under the curve by
drawing rectangles
touching at their left
corners.
This is called the Lefthand Rectangular
Approximation
Method (LRAM).
3
1 2
Example: V  t  1
8
t
v
0
1
1
2
1
0
1
1
2
1
3
1
8
1
1
8
1
2
4
2
1
2
1
8
1
1
8
2
1
1
2
1
3 2
8
1
8
3
4
Approximate area: 1  1  1  2  5  5.75
5.1 Estimating with Finite Sums
1
V  t2 1
8
We could also use a
Right-hand Rectangular
Approximation Method
(RRAM).
3
2
1
0
1
1
1
8
1
8
1
2
1
8
2
1
1
2
3
4
Approximate area: 1  1  2  3  7  7.75
3
1
2
8
4
3
5.1 Estimating with Finite Sums
3
t
v
0.5 1.03125
1
V  t2 1
8
Approximate area:
6.625
2
1.5 1.28125
2.5 1.78125
1
3.5 2.53125
0
1
2
3
4
1.03125
1.78125
Another approach would be to use
2.53125
1.28125
rectangles that touch at the midpoint.
This is the Midpoint Rectangular
Approximation Method (MRAM).
In this example there are four subintervals.
As the number of subintervals increases, so
does the accuracy.
5.1 Estimating with Finite Sums
t
With 8 subintervals:
3
v
0.25 1.00781
1
V  t2 1
8
2
0.75 1.07031
1.25 1.19531
1
1.75 1.38281
2.25 1.63281
0
2.75 1.94531
3.25 2.32031 width of subinterval
3.75 2.75781
13.31248  0.5  6.65624
1
2
3
4
Approximate area:
6.65624
The exact answer for this
problem is 6.6.
5.1 Estimating with Finite Sums
3
Inscribed rectangles are
all below the curve:
2
1
0
1
2
3
4
1
2
3
4
3
Circumscribed rectangles
are all above the curve:
2
1
0
5.1 Estimating with Finite Sums
We will be learning how to find the exact area under a
curve if we have the equation for the curve. Rectangular
approximation methods are still useful for finding the
area under a curve if we do not have the equation.
5.2 Definite Integrals
3
1
V  t2 1
8
When we find the area
under a curve by adding
rectangles, the answer is
called a Rieman sum.
2
1
The width of a rectangle is
called a subinterval.
0
1
2
subinterval
partition
3
4
The entire interval is
called the partition.
Subintervals do not all
have to be the same size.
5.2 Definite Integrals
3
1
V  t2 1
8
If the partition is denoted by P, then
the length of the longest subinterval
is called the norm of P and is
denoted by P .
2
1
0
1
2
3
4
As P gets smaller, the
approximation for the area gets
better.
subinterval
partition
n
Area  lim  f  ck  xk
P 0
k 1
if P is a partition
of the interval a, b


5.2 Definite Integrals
n
lim  f  ck  xk
P 0
k 1
is called the definite integral of
f
over
a, b .
If we use subintervals of equal length, then the length of a
subinterval is:
ba
x 
n
The definite integral is then given by:
n
lim  f  ck  x
n 
k 1
5.2 Definite Integrals
n
lim  f  ck  x
n 
k 1
Leibnitz introduced a simpler notation
for the definite integral:
n
b
k 1
a
lim  f  ck  x   f  x  dx
n 
Note that the very small change
in x becomes dx.
5.2 Definite Integrals
Integration
Symbol

upper limit of integration
f  x  dx
b
a
integrand
lower limit of integration
variable of integration
(dummy variable)
It is called a dummy variable
because the answer does not
depend on the variable chosen.
5.2 Definite Integrals

b
a
f  x  dx
We have the notation for integration, but we still need
to learn how to evaluate the integral.
5.2 Definite Integrals
Definition Area Under a Curve (as a Definite Integral)
If y = f(x) is non negative and integrable over a closed interval
[a,b], then the area under the curve y = f(x) from a to b
is the integral of f from a to b.
A=

b
a
f  x  dx
5.2 Definite Integrals
In section 5.1, we considered an object
3t  d
moving at a constant rate of 3 ft/sec.
If we draw a graph of the velocity, the distance that the
object travels is equal to the area under the line.
3
After 4 seconds,
the object has
gone 12 feet.
2
velocity
1
0
1
2
time
3
4
ft
3
 4 sec  12 ft
sec
5.2 Definite Integrals
If the velocity varies:
1
v  t 1
2
Distance is the area under
the curve
A trapezoid
1
 h(b 1  b 2 )
2
A trapezoid
1
 4(1  3)
2
3
2
1
0
1
s 8
2
x
3
4
5.2 Definite Integrals
n
Area
 lim  f  ck  xk
P 0
k 1
  f  x  dx
b
a
 F  x  F a
Where F(x) is the antiderivative of f(x)!
5.2 Definite Integrals
3
1 2
What if: v  t  1
8
2
1
0
1
2
x
3
4
We could split the area under the curve into a lot of thin
trapezoids, and each trapezoid would behave like the large
one in the previous example.
It seems reasonable that the distance will equal the area
under the curve.
5.2 Definite Integrals
ds 1 2
v
 t 1
dt 8
1 3
s
t t
24
1 3
s
4 4
24
2
s6
3
3
2
1
0
1
2
x
3
4
2
The area under the curve  6
3
We can use anti-derivatives to
find the area under a curve!
5.2 Definite Integrals
yx
Example:
2
Find the area under the curve from
x=1 to x=2.
4

3
1
2
x dx
2
2
1 3
x
3 1
1
0
2
1
2
1 3 1
 2  1
3
3
8 1
7


3 3
3
Area from x=0
Area from x=0
to x=1
to x=2
Area under the curve from x=1 to x=2.
5.2 Definite Integrals
y  x2
Example:
4
Find the area under the curve from
x = 1 to x = 2.
3
2
To do the same problem on the TI-83:
1
0
fnInt(x2,x,1,2)
1
2
5.2 Definite Integrals
Example:
Find the area between the
x-axis and the curve
y  cos x
3

from x  0 to x 
.
2


2
0
3
2
cos x dx   cos x dx
2
 /2
3 / 2
sin x 0  sin x  / 2
3
2
1
pos.
0
-1

2
neg.
3

 
 
 sin 
 sin  sin 0    sin
2
2
2

 
1  0   1 1
3
5.3 Definite Integrals and
Antiderivatives
Page 269 gives rules for working with integrals, the most
important of which are:
1.
2.
3.

b
a

a

b
a
a
f  x  dx   f  x  dx
a
b
Reversing the limits
changes the sign.
f  x  dx  0 If the upper and lower limits are equal,
then the integral is zero.
k  f  x  dx  k  f  x  dx Constant multiples can be
b
a
moved outside.
5.3 Definite Integrals and
Antiderivatives

1.
2.
a

a

b
a
3.
4.
b
a
f  x  dx   f  x  dx
a
Reversing the limits
changes the sign.
b
f  x  dx  0 If the upper and lower limits are equal,
then the integral is zero.
k  f  x  dx  k  f  x  dx Constant multiples can be
b
a

b
a
moved outside.
 f  x   g  x  dx  f  x  dx   g  x  dx
a
a
b
b
Integrals can be added and subtracted.
5.3 Definite Integrals and
Antiderivatives
5.
 f  x  dx  f  x  dx   f  x  dx
b
c
c
a
b
a
Intervals can be added
(or subtracted.)
y  f  x
a
b
c
5.3 Definite Integrals and
Antiderivatives
6. min f  (b  a) 
a
b
b

a
f ( x)dx  max f  (b  a)
5.3 Definite Integrals and
Antiderivatives
Suppose that f and g are continuous functions and that

2
2
4
4
2
2
f ( x)dx  5,  f ( x)dx  3,  h( x)dx  6
Find

2

4
4
2
f ( x)dx
3
f ( x)dx

2
2
3

3
-3 + 5 = 2
2 f ( x)dx
2*5=10
f ( x)dx
0

2
4
f ( x)dx
-5 – -3 = -2
4
 [3 f ( x)  2h( x)]dx
2
3*(-3)- 2 * 6 = -21
5.3 Definite Integrals and
Antiderivatives
5
The average value of a function is the value that would
give the same area if the function was a constant:
A
4
3
0
3
3
2
1.5
1
0
1
2
y
1 2
x
2
3
1 2
x dx
2
27
9
1 3


 4.5
 x
6
2
6 0
4.5
Average Value 
 1.5
3
Area
1 b
Average Value 

f  x  dx

a
Width b  a
5.3 Definite Integrals and
Antiderivatives
For what value(s) in the
interval does the function
assume the average value?
5
4
4.5
Average Value 
 1.5
3
3
2
1.5
1
0
1
2
y
1 2
x
2
3
1 2
x  1 .5
2
x2  3
x 3
5.3 Definite Integrals and
Antiderivatives
The mean value theorem for definite integrals says that
for a continuous function, at some point on the interval the
actual value will equal the average value.
Mean Value Theorem (for definite integrals)
If f is continuous on  a, b then at some point c in  a, b ,
1 b
f c 
f  x  dx

a
ba
5.3 Definite Integrals and
Antiderivatives
Find the total area bounded by the x-axis
and the graph y = x2 - 3x from [0,4].
3
4
  ( x  3x) dx   ( x2  3x) dx
2
0
3
3
4
 x 3x   x 3x 
   

  
2 0  3
2 3
3
3
2
 9  11
  
 2 6
3
19 2
 u
3
2
5.4 Fundamental Theorem of
Calculus
Quote from CALCULUS by Ross L. Finney and
George B. Thomas, Jr., ©1990.
If you were being sent to a desert island
and could take only one equation with you,
d x
f  t  dt  f  x 

dx a
might well be your choice.
5.4 Fundamental Theorem of
Calculus
The Fundamental Theorem of Calculus, Part 1


If f is continuous on a, b , then the function
F  x    f  t  dt
x
a


has a derivative at every point in a, b , and
dF d x

f  t  dt  f  x 

dx dx a
5.4 Fundamental Theorem of
Calculus
First Fundamental Theorem:
d x
f
t
dt

f
x





dx a
1. Derivative of an integral.
5.4 Fundamental Theorem of
Calculus
First Fundamental Theorem:
d x
f
t
dt

f
x





dx a
1. Derivative of an integral.
2. Derivative matches upper limit of integration.
5.4 Fundamental Theorem of
Calculus
First Fundamental Theorem:
d x
f
t
dt

f
x





dx a
1. Derivative of an integral.
2. Derivative matches upper limit of integration.
3. Lower limit of integration is a constant.
5.4 Fundamental Theorem of
Calculus
First Fundamental Theorem:
d x
f
t
dt

f
x





dx a
New variable.
1. Derivative of an integral.
2. Derivative matches upper limit of integration.
3. Lower limit of integration is a constant.
5.4 Fundamental Theorem of
Calculus
The long way:
d x

cos
x
cos
t
dt
dx 
d
x
sin t 
dx
0
d
sin x  sin    

dx
d
sin x
dx


First Fundamental Theorem:
1. Derivative of an integral.
2. Derivative matches
upper limit of integration.
3. Lower limit of integration
is a constant.
5.4 Fundamental Theorem of
Calculus
d x 1
1
dt 
2
2

0
dx 1+t
1 x
1. Derivative of an integral.
2. Derivative matches
upper limit of integration.
3. Lower limit of integration
is a constant.
5.4 Fundamental Theorem of
Calculus
d x
cos t dt

dx 0
2
 
d 2
cos x  x
dx
2
The upper limit of integration does
not match the derivative, but we
could use the chain rule.
 
cos x  2 x
2
 
2 x cos x
2
5.4 Fundamental Theorem of
Calculus
d 5
3t sin t dt

dx x
d x
  3t sin t dt
dx 5
3x sin x
The lower limit of integration is not
a constant, but the upper limit is.
We can change the sign of the
integral and reverse the limits.
5.4 Fundamental Theorem of
Calculus
Neither limit of integration is a
constant.
We split the integral into two parts.
d
1
dt
t

dx 2 x 2  e
x2
d 
1
1

dt  
dt 
 0
t
t
2x 2  e
dx  2  e

It does not
matter what
constant we use!
2x
d  x2 1
1

dt

dt
 0

t
t

0
dx  2  e
2e

(Limits are reversed.)
x2
1
0
1
 2x 
2
2x
x2
2e
2e
(Chain rule is used.)
2x
2


2x
x2
2

e
2e
5.4 Fundamental Theorem of
Calculus
The Fundamental Theorem of Calculus, Part 2


If f is continuous at every point of a, b , and if
F is any antiderivative of f on  a, b , then
 f  x  dx  F b   F  a 
b
a
(Also called the Integral Evaluation Theorem)
5.4 Fundamental Theorem of
Calculus

1
dx
2
x
3
1

2
1
9

1
2
3
x(1  x ) dx
3
x dx
81
10
52
3
5.4 Fundamental Theorem of
Calculus

4
1

3
2
x dx
1

4
0



4

4
2
sec x dx
1
sin 2 x
dx
cos x
0
5.4 Fundamental Theorem of
Calculus
#55 Pg 287 f is the differentiable
function whose graph is shown in
the figure. The position at time t
(sec) of a particle moving along a
coordinate axis is s  t f ( x)dx
(6,6) • • (7,6.5)
y = f(x)
•
(3,0)

0
a.
b.
c.
d.
e.
f.
g.
What is the particle’s velocity at t = 3?
s’(3) = f(3) = 0
Is the acceleration of the particle at time
s’’(3) = f’(3) > 0
t = 3 positive or negative?
3
1
s
(
3
)

f
(
x
)
dx


(3)(6)  9 units
What is the particle’s position at t = 3?
0
2 6
When does the particle pass through the origin? s(t )  0 at t  6 because 0 f ( x)dx  0
Approximately, when is the acceleration zero? s' ' (t )  f ' (t )  0 at t  7 sec
When does the particle move toward the origin? away? 0< t < 3, t > 6; 3 < t < 6
On which side of the origin does the particle lie at t = 9? Positive Side, area is larger
5.5 Trapezoid Rule
Using integrals to find area works extremely well as long
as we can find the antiderivative of the function.
Sometimes, the function is too complicated to find the
antiderivative.
At other times, we don’t even have a function, but only
measurements taken from a real-life object.
What we need is an efficient method to estimate area
when we can not find the antiderivative.
5.5 Trapezoid Rule
3
1 2
y  x 1
8
0 x4
2
Actual area under curve:
A
4
0
1 2
x  1 dx
8
4
1 3
A
x x
24
0
1
0
20
A
3
1
2
 6.6
3
4
5.5 Trapezoid Rule
3
1 2
y  x 1
8
0 x4
2
Left-hand rectangular
approximation:
1
0
1
8
1
2
1
1
8
3
4
Approximate area: 1  1  1  2  5  5.75
2
3
(too low)
4
5.5 Trapezoid Rule
3
1 2
y  x 1
8
0 x4
2
Right-hand rectangular
approximation:
1
0
1
8
1
2
1
8
1
3
4
Approximate area: 1  1  2  3  7  7.75
2
3
(too high)
4
5.5 Trapezoid Rule
Averaging the two:
7.75  5.75
 6.75
2
1.25% error
(too high)
5.5 Trapezoid Rule
xo=a x1 x2
x3
xn-1 xn = b
5.5 Trapezoid Rule
3
2
1
0
1
2
3
Averaging right and left rectangles gives us trapezoids:
1  9  1  9 3  1  3 17  1  17

T  1              3 
2 8 28 2 22 8  2 8

4
5.5 Trapezoid Rule
3
2
1
0
1
2
3
4
1  9  1  9 3  1  3 17  1  17

T  1              3 
2 8 28 2 22 8  2 8

1  27  27
1  9 9 3 3 17 17

T  1        3  T    
 6.75
2 2 
2 8 8 2 2 8 8

4
5.5 Trapezoid Rule
(x4,y4)
1
T  h(b1  b2 )
2
(x3,y3)
(x2,y2)
(x1,y1)
(xo,yo)
h
1
1
1
T  h( y0  y1 )  h( y1  y2 )  ...  h( yn1  yn )
2
2
2
5.5 Trapezoid Rule
Trapezoidal Rule:
h
T   y0  2 y1  2 y2  ...  2 yn 1  yn 
2
where [a,b] is partitioned into n subintervals of
equal length
LRAM n  RRAM n
h = (b – a)/n
Equivalent ly T 
2
This gives us a better approximation than either left
or right rectangles.
5.5 Trapezoid Rule
Simpson’s Rule: Uses curved arcs to approximate areas
y  ax  bx  c
y0  a(h) 2  b(h)  c
2
(0,y1)
(-h,yo)
(h,y2)
y1  c
h
y2  a(h)  b(h)  c
2
y0  4 y1  y2  2ah  6c
2
L
h
M
R
5.5 Trapezoid Rule

h
h
(0,y1)
(-h,yo)
(ax  bx  c) dx
2
3
2
ax bx

 cx
3
2
3
2ah
 2ch
3
(h,y2)
h
h
L
h
2
 (2ah  6c)
3
M
R
5.5 Trapezoid Rule
h
2
(2ah  6c)
3
(h,y2)
but
L
y0  4 y1  y2  2ah  6c
2
h
( y0  4 y1  y2 )
3
(0,y1)
(-h,yo)
so
M
R
5.5 Trapezoid Rule
Simpson’s Rule:
h
S  ( y0  4 y1  2 y2  4 y3  ...  2 yn2  4 yn1  yn
3
where [a,b] is partitioned into an even n subintervals
of equal length
h = (b – a)/n
5.5 Trapezoid Rule
h
S  ( y0  4 y1  2 y2  4 y3  ...  2 yn2  4 yn1  yn )
3
Example: y  1 x 2  1
8
3
1
9
3
17

S  1  4   2   4   3 
3
8
2
8

1 7
17

 1   3   3 
3 2
2

2
1
0
1
2
3
4
1
 19   6.6
3
5.5 Trapezoid Rule
Simpson’s rule can also be interpreted as fitting parabolas
to sections of the curve, which is why this example came
out exactly.
Simpson’s rule will usually give a very good approximation
with relatively few subintervals.
It is especially useful when we have no equation and the
data points are determined experimentally.
5.5 Trapezoid Rule
Error Bounds
b

If T and S represent the approximations to
f ( x)dx
a
given by the Trapezoid Rule and Simpson’s
Rule, respectively, then the errors ET and Es satisfy
ba 2
ET 
h M f ''
12
ba 4
ES 
h M f (4)
180