Hints for learning activity 3b
Imagine that the speed of the objects is changing stepwise. Lets us divide the interval [0, 3] into
three equal parts where the length of each subinterval [0, 1], [1, 2] and [2, 3] is 1 second.
Figure 3b-I Sum of areas of three rectangles- overestimates - Figure 3b-II Sum of areas of three rectangles
The sum of the areas of the three rectangles equals to S = ((2 x 1) + (5 x1) + (10 x1) = 17 (see
figure 3b-I). This is actually an overestimate of the total area under the curve as some areas
are above the curve.
For a better estimate, we could double the number of rectangles to 6 (see figure 3b-II).
What is the length of each subinterval this time …..?
What is the sum of the six rectangles? ……
The sum of the areas of the six rectangles is now
= 0.625 + 1.0 + 1.625 + 2.5 + 3.625 + 5.0 = 14.375
To get the total area we adding up all the areas of the rectangles where the area of each
rectangle =……
1
Do you realize that the estimates improve as you increase the number of rectangles see table ).
The table below shows the estimates of the area under f (t )  t 2  1 as the number of
rectangles increases.
Number of rectangles
3
6
12
24
48
96
192
384
Area estimates
17
14.375
~13.156
~12.570
~12.283
~12.141
~12.070
~12.035
Table 1: Overestimates
Which number are these estimates heading towards? ….
We could have also obtained underestimates of the area by constructing a stepwise function
with area less than the required area. Similarly, it would be possible to obtain better estimates
by increasing the number of rectangles (see figures 3b-III and 3b-IV).
The sum of the areas of the three rectangles equals to S = (1 x 1) + (2 x1) + (5 x1) = 8.
How was this calculated?.......................................................................................................
And as before, to obtain a better estimate we double the number of rectangles to 6 (see figure
3b-IV ).The sum of the area = 9.875. Can you work this out?
Figure 3b-III Sum of areas of three rectangles-underestimates- Figure 3b-IV Sum of areas of three rectangles
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Again, the table below shows the estimates of the area under f (t )  t 2  1 as the number
of rectangles increases.
Number of rectangles
3
6
12
24
48
96
192
384
Area estimates
8
9.875
~10.906
~11.445
~11.721
~11.860
~11.930
~11.965
Table 2: Underestimates
Which number are these estimates heading towards? ….
The sum of the areas of the rectangles is an estimate of the distance covered. The actual area
lies somewhere between the overestimate and the underestimate. What is interesting is that
either way, the sum seems to be approaching a particular number. In fact, by using quite
involved algebraic manipulations, one can show this sum approaches the exact distance.
*[You will handle this in a formal Calculus course].
So what is the exact distance?
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Calculating the exact distance
Recall that when the derivative of ƒ(x) is given as ƒ'(x). Near, x = a we have
f '(a) 
f ( x)  f (a )
or
xa
derivative of f 
difference in f
difference in x
Therefore, f ( x)  f (a)  f '(a)( x  a) …..[Equation 1]
or difference in f  derivative of f x difference in x
If we assume that the interval between two consecutive values of x is very small, and we
introduce a new function F(x) whose derivative is ƒ(x) implying that:
F '( x)  f ( x)
then using equation 1
difference in F  derivative of F x difference in x
If we add all the differences we obtain:
F(x1) - F(x0) +F(x2) - F(x1) + F(x3) - Fx2) + F(x4) - F(x3) +... F(xn) - F(xn-1) ≈ f(x0)(x1 - x0) + f(x1)(x2x1)+ f(x2)(x3- x2)+ f(x3)(x3- x2)+... f(xn)(xn- xn-1)
*Some terms in the left hand expression cancel so that we remain with only the first and the last
term:
F(xn) - F(x0) ≈ f(x0)(x1 - x0) + f(x1)(x2- x1)+ f(x2)(x3- x2)+ f(x3)(x3- x2)+... f(xn)(xn- xn-1)
The left hand part of the expression is a constant term (a difference). The right hand part of this
expression is a sum of the derivatives multiplied by differences in x.
By increasing the number of points x0, x1, x2, x3....xn infinitely, the ratios become smaller, Δx
tends to zero and the total sum gets closer and closer to the total area under the curve.
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At this point, the relationship changes from an approximation to equality and we have the sum
converging to a limit.
If we let An denote the sum of the areas and A represent the total area under the curve, then
This expression defines and calculates the number A in cases for which the limit of An exists.
b
lim An   f ( x)dx
n 
a
which is the integral from a to b or the definite integral.
In our case F (3) -F (0) which is a constant value ≈ the approximate total sum in the stepwise
function.
Recall that ƒ (t) is the derivative of F (t) or F'(t) = ƒ (t)
In order to find the area we find the integral
3

f (t )dt using the antiderivative F (t ) 
0
t3
t
3
 33
  03

 3    0  9  3  12
3
 3

The total area = exact distance = F (3)  F (0)  
The exact distance = 12m.
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