Transcript Approximating the area under a curve using Riemann Sums

Approximating the area under a curve
using Riemann Sums
Riemann Sums
-Left, Right, Midpoint, Trapezoid
Summations
Definite Integration
• We want to think about the region contained by a
function, the x-axis, and two vertical lines x=a and x=b.
a
b
The goal of this process is to determine the area that is shaded yellow. We will
first use geometry and some basic shapes to estimate the area. We will then
combine the idea of limits with that of integrals to find the exact area.
• There are two methods that we can use to
approximate area. One is creating basic geometry
figures and looking at each individual figure, finding its
area, and then summing those values. This is not
difficult if the number (n) of figures is small.
• If n becomes very large (probably over 10) then this
process can become very tedious and we revert to the
idea of using summations and creating a ‘sigma
notation formula’ that helps evaluate sums when n is
large.
• Since we know we will be using rectangles, and we
know the formula for the area of a rectangle, we
will start with the most basic style of problem.
• When we look at the region under the curve, it will
be bounded by two vertical lines and the x axis, in
order to find the size of the intervals for a desired
number of rectangles we will use this formula :
size of interval 
ba
x
n
• The area of each rectangle will then be the product of the
height and width of that rectangle or subinterval. Then we
will add up all “n” rectangles.
• What do we use as the height of each rectangle? (later on we
will learn that as more and more rectangles are made, this
distinction does not matter) but for computational reasons
we will talk about 4 methods
Right Sums (using right endpoints, right shoulder of rectangle
touches the curve)
Left Sums (using left endpoints, left should of rectangle
touches the curve)
Midpoint Sums (using the midpoints of each subinterval, so
the segment containing the midpoint of the subinterval is
used as the height for that rectangle.
Trapezoid (instead of using rectangles, a better approximation
can be made using irregular trapezoids)
A few pictures to make the connection
We want to divide it into “n” intervals, for this example let n = 6. So
to determine how wide each interval is, we would take (b-a)/n, and
that value is ∆x. Then to find the endpoint of the first interval you
simply add a+∆x, then keep adding ∆x to the previous endpoint to
get the next subinterval’s endpoint until you are done.
Now we’ve simply
Right shoulder of
constructed each
rectangle on
rectangle so that each of
curve, so right
the six rectangles are
sum
touching the curve with
their right shoulders.
The area of the second rectangle is  A=height*width
So, R2= f(x)*∆x
You will then do this for each rectangle. Note to find f(x)’s numerical value
you simply take the x value of that right endpoint of the rectangle and plug
it into f(x) to find the actual numerical height of that rectangle.
f(x)
ba
x
n
• Because all the rectangles go above f(x), the sum of the
areas will be larger than the actual area under the curve, so
if f(x) is increasing this is an overestimate.
• Any ideas on how to make the green areas smaller?
Left Sum (puts left shoulder on f(x))
Left Shoulder is on
the curve, Left Sum
• Find the area of the rectangles in the same fashion as we
did with right sums. The blue areas are under the curve,
but are not included in your rectangle areas, so if f(x) is
increasing the left sum would be an underestimate
Right Sum
Left Sum
By the sandwich theorem or squeeze theorem we know that since
the Right sum is an overestimate, and the left sum is an
underestimate, then the actual area under the curve is stuck
somewhere in between them.
Remember that the right sum will reduce down towards the actual
area if we increase the number of rectangles, likewise the left sum
will increase towards the actual area when we increase n because
f(x ) is increasing in this graph.
Our goal is to eventually get the ∆x to be infinitely small, so that the
overestimate or underestimate is infinitely small.
So if we use the idea of limits then we can approach the actual area
under the curve. That is as n approaches infinity we will get an exact
value for the area.
Examples: Find the area under the
2
curve f ( x)  x  5, between x  0, and x  2.
Using 5 rectangles
ba 20 2
x


n
5
5
Then the first “right endpoint”
is a+∆x
2
5
4
5
6
5
8
5
10
5
f ( x)   x 2  5, between x  0, and x  2.
Find the area under the curve using 10 rectangles.
Right Sum
f ( x )  3x  2 x
2
• Use Riemann Sums – Right Hand & Left Hand
• n=5 [1,4]
1 4
3
f ( x)  x  1.3 x  4
2
• Use Riemann Sums – Right Hand & Left Hand
• 5 rectangles [-1.2,2]
1 3
2
f ( x)   x  4 x  x
3
• Use Riemann Sums – Midpoint
• 5 rectangles
f ( x)  x  6  5
• Use Riemann Sums – Midpoint
• 6 rectangles [6,12]
Right Sum
2
5
4
5
6
5
8
5
10
5
f ( x)   x  5, between x  0, and x  2.
2
Left Sum
2
5
4
5
6
5
8
5
10
5
f ( x)   x2  5, between x  0, and x  2.
Find the area under the curve using 10 rectangles.
Left Sum
• When the value of “n” is increased, the number of
rectangles increases, does the right sum and left
sum squeeze down towards the actual area?
• Which is a better approximation
• Find the area under the curve f ( x)  2x  3, on[0,6]
with n=6,12 Left and Right Sums
2
f ( x)  2  2
x
Find LS and RS with n=6 on [1,5]
Using Midpoints gives a bit of a better
approximation.
• Steps
– Still divide everything up into subintervals.
– Locate the midpoints of each interval
(endpoint + endpoint/2)
– Now use the midpoint as the x value that you plug
into f(x) to find the height of the rectangle.
– Then to find the area of each rectangle you still
take the height*∆x
– Sum all rectangle areas.
This picture describes the concept of finding the heights of each rectangle at the midpoints of
each subinterval. Still broke the interval into congruent sections, but then found the midpoints
of those sections, plug that midpoint into f(x) to find the height of the rectangle.
Notice how in each rectangle, the part of the rectangle above the curve is almost equivalent to
the part of the rectangle not calculated under the curve, so the overestimate and underestimate
almost cancel each other out completely
3
4
f ( x)  x  3x
5
3
Use midpoints with a=1 and b=3 and
n=10. The exact area is
21.20014476021
Is the midpoint method along with
several subintervals very close to this
value?
Use midpoints, left s um, and right sum. Which is a better estimate?
f(x)=sin(x) , [0, pi] use 8 rectangles.
Trapezoids
Notice after creating trapezoids, the area above or below the curve that is over
or under estimated has become so small that coloring it in would be too
difficult to do, this is a very good approximation, but the method is more
involved and the steps are longer because of the formula for finding area of a
trapezoid being more complex than the area of a rectangle.
• The area of the curve using the trapezoid
method is identical to the average of the left
and right Riemann Sums.
Trapezoids give even better approximations. Remember the
formula for finding a trapezoid is 1/2h*(b1+b2)
Examples: Find the area under the curve using trapezoids.
f ( x)   x2  5, between x  0, and x  2.
2
5
4
5
6
5
8
5
10
5
USE Trapezoids. Which is a better estimate?
f(x)=sin(x) , [0, pi] use 8 subintervals.
f ( x)  x  x  .5x
4
3
[0,2] using 6 trapezoids
Riemann Sums when the number of rectangles or subintervals is very large.
n
lim  f ( xi ) x
n 
i 1
Height of the
“ith” rectangle
Width of
the “ith”
rectangle
ba
x
n
xi  a  ( x)i
Counting tool that allows us
to move from the 1st
rectangle to the 2nd rectangle
to the 3rd … to the ith
rectangle
xi  a  ( x)i
Counting tool that allows us
to move from the 1st
rectangle to the 2nd rectangle
to the 3rd … to the ith
rectangle
If we are interested in going between a=1 and b=3 for example with n
rectangles. Δx = (3-1)/n = 2/n
2nd
1st
ith
ith
1+(2/n)4
f(
4th
1+(2/n)3
1+(2/n)2
1+(2/n)
1
3rd
. . .
This is for right
sums. xi
changes when
dealing with left
sums or with
midpoints.
ith
1+(2/n)i
) will be the height of the ith rectangles.
3
So now lets look at the function
f ( x)  x 2  x  1 when a=1 and b=3 with n rectangles
n
lim  f ( xi ) x
n 
x
i 1
b  a 3 1 2


n
n
n
xi  a  ( x)i
n
lim 
n 
i 1
2i
2
 1  i  1
n
n
 2i  2 
f 1   
n  n 

Now in order to evaluate this you need to identify
compute f(1+2i/n) so plug that in for every x in x2-x+1
 2i  2  2i    2 
lim  1    1    1  
n 
n 
n    n 
i 1 

n
• Now simplify the expression and apply your limit
rules to see what the limit approaches and that will
be the area under the curve between a=1 and b=3.
Combining Riemann Sums with
Summation Notation
n
  function height @ a certain height (width)
i 1
left or right height of rectangle
ba
n
We will want to use i as a counting device to move from one
rectangle to the next.
• Right Sum
n
 f (a 
x  i)  x 
i 1
The ∆xi will put us at the right side of the beginning rectangle
• Left Sum
n
 f a  (
x(i  1))  x 
i 1
The ∆x(i-1) will put us at the left side of the beginning
rectangle
f ( x)  x2  5, between x  0, and x  2.
USING RIGHT SUMS
n
 f (a 
xi ) x
i 1
2  2 

f
0

i  


n  n 

i 1
2
n 
 2 
2 
  i   5   

i 1   n 
 n 
2

2 n  2 
  i   5 

n i 1   n 


2 n  4i 2

 2  5
n i 1  n

n

2  4i
  5


2
n  i 1 n
i 1 
n
2
n
2 4 n 2 n 
i   5
2 

n  n i 1
i 1 
20 2
x

n
n
2  4  n(n  1)(2n  1) 


5
n



n  n 2 
6



2  4 2n3  3n 2  n


5
n

n  n 2
6


2  8n3  12n 2  4n
 5n 

2
n
6n

2  8n3  12n 2  4n  30n3 

n 
6n 2

16n3  24n 2  8n  60n3
6n 3
76n3  24n 2  8n
6n3
• Take the previous function, graph it, find the area
using right sums when n=4.
• Now take the previous slide’s expression and plug
in 4 for n.
• What do you notice?
• This should always be the case… What happens
when n=100000? What about n=10000000000?
• Do you see how this allows us to bring in the
idea of limits and allowing n to go to infinity,
with the summation notation we will get the
exact area if n approaches infinity.
• Lets use the same function and compute the
left sum.
n
 f a  (
x(i  1))  x 
i 1
• The following function is using midpoints.
n
 f a  (
i 1
x(i  ))   x 
1
2
f ( x)  x  3x ; [3,5]
2
Complete this using the idea of
Riemann sums and the summation
formulas. What does the area go
towards when n gets larger?
56.6666
f ( x)  x  2x ; [0,3]
3
2
f ( x)  3x  x  4 ; [2,5]
2
f ( x)  4x  2x ; [2,6]
3
2
• We could do the same thing with left hand sums
but as n get really large both the right hand sum
and the left hand sum will converge to the same
value.
• Know that when f(x) is increasing on I the right sum
will be an overestimate and the left sum will be an
underestimate when n is small.
• When f(x) is decreasing on I the right sum will an
underestimate and the left sum will be on
overestimate.
• But the more and more rectangles we create the
better approximations we get, and when n∞ we
get an exact value for the area.
• Stop here, homework
• The idea of using summations and the
corresponding formulas for i, can be very tedious
with just the basic functions we have used so far,
the algebra gets even more complicated when you
raise something to a higher power or take the
square root, it can all still be done using the
summation formulas but we have more efficient
methods. But it must be understood that these
new methods all arose from the foundation of these
summation formulas. The more efficient way is
called definite integration.
Definition of the Definite Integral
n
b
i 1
a
lim  f (ci ) xi   f ( x)dx
|| ||0
This is called the norm of
the partition, it is the width
of the largest subinterval.
If every subinterval is
equal, the partition is
regular and the norm is
represented by
ba
x
n
So, as the norm approaches 0, it
implies that n approaches infinity (as
the subintervals get smaller, there is
room for more rectangles)
The number a is called the lower limit
of integration and the number b is
the upper limit of integration.
Theorem
• Continuity Implies Integrability
– If a function f is continuous on the closed interval [a,b],
then f is integrable on [a,b].
• With definite integration we will do the same
work we did when we integrated in 4.1, but
now we will know an “a” and “b” in which the
graph will be restricted by and we can then
evaluate the integral at those points.
Definitions of Special Definite Integrals
• 1. If f is defined at x=a, then we define

a
a
f ( x)dx  0
• 2. If f is integrable on [a,b], then we define

a
b
b
f ( x)dx   f ( x)dx
a
Additive Interval Property
• If f is integrable on the three closed intervals
determined by a, b, and c, then

b
a
c
b
a
c
f ( x)dx   f ( x)dx   f ( x)dx
a
c
b
Properties of Definite Integrals
• If f and g are integrable on [a,b] and k is a constant,
then the function of kf and f±g are integrable on
[a,b] and
• 1.

b
a
• 2.
b
kf ( x)dx  k  f ( x)dx
a
b
b
a
a
 [ f ( x)  g ( x)]dx  
b
f ( x)dx   g ( x)dx
a
• Homework
• Pg. 278 13-18, 33-42