28-Area as Limits VU

Lecture # 28

Area as Limits

• Definition of Area
• Some technical considerations
• Numerical approx of area
• Look at the figure below

In this figure, there is a region bounded below by the x-axis, on the sides by the lines
x= a and x = b, and above by a curve or the graph of a continuous function y = f(x)
which is also non-negative on the interval [a, b].
Earlier we saw that such an area can be computed using anti-derivatives.
Let's make the concept precise.
Recall that the slope of the tangent line was defined in terms of the limit of the slopes of
secant lines.
Similarly, we will define area of a region R as limits of the areas of simpler regions
whose areas are known.
We will break up R into rectangles, and then find the area of each rectangle in R and
then add all these areas up.
The result will be an approximation to the region R. Let's call it
If we let n go to infinity, the resulting will give a better and better approximation to R
as the rectangle in R will get thinner and thinner, and the gaps will be filled in.
Here is the formal idea
Choose an arbitrary positive integer n, and divide the interval [a,b] into n subintervals of width
b−a
by inserting n-1 equally spaced points between a and b say x1 , x2 ,...xn−1
n

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These points of subdivision form a regular partition of [a,b] a, x1 , x2 ,...xn−1 , b

Next draw a vertical line through the points x1* , x2* ,...xn*
This will divide the region R into n strips of uniform width

Now we want to approximate the area of each strip by the area of a rectangle. For this,
choose an arbitrary point in each subinterval ,
Over each subinterval construct a rectangle whose height is the values of the function f
at the selected arbitrary point.

The union of these rectangles form the region Rn that we can regard as a reasonable

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approx to the region R. The area of Rn can be got by adding the areas of all the
rectangles forming it
If now we let n get big, the number of rectangles gets big, and the gaps btw the curve
and the rectangles get filled in.

As n goes to infinity, the approx gets as good as the real thing.

=
So we define A =
area ( R) lim [area ( Rn )]
n→+∞

For all the following work and computation, we will treat n as a POSITIVE integer
For computational purposes, we can write A in a different form as follows

b−a
In the interval [a,b], each of the approximation rectangles has width
n
b−a
We call this delta x or ∆x =
n
The heights of the approximating rectangles are at the points x1* , x2* ,...xn*
So the approximating rectangles making up the region have areas

f ( x1* )∆x, f ( x2* )∆x,..., f ( xn* )∆x

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So the total area of Rn is given by

=
area ( Rn) f ( x1* )∆x + f ( x2* )∆x + ... + f ( xn* )∆x
OR
n
=
area ( Rn ) ∑ f (x
k =1
k
*
)∆x

So now A can be written as

n
=A lim ∑ f ( xk * )∆x
n→+∞
k =1

This we will take to be the PRECISE def of the area of the region R

Some Technical Considerations

The points x1* , x2* ,...xn* were chosen arbitrarily.

What if some other points were chosen? Would the resulting values for f(x) at these
points be different? And if so then the definition of area will not be well-defined?
It is proved in advanced courses that since f is continuous it doesn’t matter what points
are taken in a subinterval.

Usually the point xk* in a subinterval is chosen so that it is the left end point of the
interval, the right end point of the interval or the midpoint of the interval.
Note that we divided the interval [a,b] by the points x1 , x2 ,..., xn−1 with x0 = a and xn = b
into equal parts of width ∆x

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SO

xk * = xk −1 = a + (k − 1)∆x Left end point

xk * = xk = a + k ∆x Right end point
1 1
xk * = ( xk −1 + xk ) =a + (k − )∆x Midpoint
2 2

Example

Use the definition of the Area with xk * as the right end point of each subinterval to find the area
under the line y = x over the interval [1,2].
Subdivide [1,2] into n equal parts, then each part will have length

b − a 2 −1 1
∆=
x = = *
We are taking xk as the right end point so we
n n n

k
Have xk* = a + k ∆x = 1 +
n

Thus, the kth rectangle has area

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 28-Area as Limits VU

 k  k 1
f ( xk * )∆x = xk *∆x = 1 +  ∆x = 1 + 
 n  nn
And the sum of the areas of the n rectangles is
n
3 1  3
=A lim ∑ f ( xk=
*
)∆x lim  += 
n→+∞
k =1 
n→+∞ 2 2n  2
Note that the region we have computed the area of is a trapezoid with height h = 1 and bases b 1 = 1
and b 2 = 2. From basic geometry we have that

1 1 3
Area of trapezoid = =
A h(b1 + b2=
) (1)(1 + 2)
=
2 2 2
Example
Same problem as before but with left end points.[ 1,2].

Use the definition of the Area with xk* as the left end point of each subinterval to find the area under
the line y = x over the interval [1,2].
Subdivide [1,2] into n equal parts, then each part will have length

b − a 2 −1 1
∆=
x = =
n n n
We are taking xk * as the left end point so we

k
Have xk * = a + k ∆x = 1 +
n
Thus, the kth rectangle has area

 k  k 1
f ( xk * )∆x = xk *∆x = 1 +  ∆x = 1 + 
 n  nn
And the sum of the areas of the n rectangles is
n
 k  1  n  1 k   1 n
n
1 n
=
∑
k 1 =
∆x ∑ 1 + =
f ( xk )= *

k 1  n =
∑ + = 
n  k 1  n n 2 =
∑1 + n2 ∑
 n k 1= k 1
k

1 1 1 
= ⋅ n + 2  n(n + 1) 
n n 2 
Note that the region we have computed the area of is a trapezoid with height h = 1 and bases b1 = 1
and b2 = 2 . From basic geometry we have that

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28-Area as Limits VU

1 1 3
Area of trapezoid = A = h(b1 + b=
2) (1)(1 + 2)
=
2 2 2
Example
Use the area definition with right endpoints of each subinterval to find the area under the parabola

y= 9 − x 2 over the interval [0,3].

Do calculations from book on page 272 (end) and page 273 for example.

Numerical Approximations of Area

As we have all seen so far, the computations involved in computing the limits are tedious and long.
In some cases, it is even impossible to carry out the computations by the definition of the area.
In such cases, it is easier to get a GOOD approx for the area using large values of n and using a
computer or a calculator.
Example

Use a computer of a calculator to find the area under the curve y = 9 − x 2 over the interval [0,3]
and n = 10, 20 and 50.

Left end point approximation Right end point approximation Mid point approximation

K xk * 9 − ( xk ) 2 xk * 9 − ( xk ) 2 xk * 9 − ( xk ) 2
1 0.0 9.000000 0.3 8.910000 0.15 8.977500
2 0.3 8.910000 0.6 8.640000 0.45 8.797500
3 0.6 8.640000 0.9 8.190000 0.75 8.437500
4 0.9 8.190000 1.2 7.560000 1.05 7.897500
5 1.2 7.560000 1.5 6.750000 1.35 7.177500
6 1.5 6.750000 1.8 5.760000 1.65 6.277500
7 1.8 5.760000 2.1 4.590000 1.95 5.197500
8 2.1 4.590000 2.4 3.240000 2.25 3.937500
9 2.4 3.240000 2.7 1.710000 2.55 2.497500
10 2.7 1.710000 3.0 0.00000 2.85 0.877500

n =16.605000 =18.022500
∆x ∑ f ( xk ) = 19.30500
∗

k =1

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And for different values of n we have the following approximations.

Left end point Right end point Mid point

n approximation approximation approximation

10 19.305000 16.605000 18.02250

20 18.663750 17.313750 18.005625

50 18.268200 17/728200 18.000900

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