Question 2: What is the Fundamental Theorem of Calculus?
We started section 13.2 by estimating change in a function by finding the area of
rectangles. Using the rate of change as the height of the rectangle, we were able to find
the change in revenue or profit when production is changed. In other words, we could
use the rate of change of a quantity to find changes in the quantity itself. There was
nothing special about revenue of profit. We could apply this strategy to any rate of
change to find the corresponding change.
When the number of rectangles is increased, the area of the rectangles gets closer and
closer to the area under the rate of change (or derivative). We used the definite integral
to denote the area under the function. This suggests that there is a connection between
the change in a function and the area under its derivative.
The Fundamental Theorem of Calculus
Suppose f is a continuous function on the interval  a, b  . If F
is an antiderivative of f, then
b
 f  x  dx  F  b   F  a 
a
The left side of this relationship describes the exact area under some function. The
right side is the difference in antiderivative function values. In practice, we may start
with the left side and use it right to calculate the value or vice versa. In using this
theorem, we may use any antiderivative. In practice, it is easiest to use the
antiderivative with C  0 .
Example 2
Find the Definite Integral
In Example 2 of section 13.3, we used geometry formulas to evaluate
the definite integral
5
5
  x  1 dx
0
Find the value of this definite integral using the Fundamental Theorem
of Calculus.
Solution The antiderivative of the integrand is
F  x     x  1 dx
x2
 x
2
Using this antiderivative, the value of the definite integral is
 52
  02

F  5  F  0     5     0 
2
  2

 17.5
This matches the value we calculated geometrically in section 13.3.
The power in the Fundamental Theorem of Calculus is that it allows you to calculate the
definite integral even when the region is not a simple geometric figure. In the next
example, the region depicted by the definite integral is curved. No simple formulas may
be used to find the area. However, the antiderivative is easy to compute so the definite
integral is easily to evaluate.
Example 3
Find the Definite Integral
In Example 5 of section 13.3, we estimated the value of
3

x 2 dx
1
6
using midpoint sums. Find the value of this definite integral using the
Fundamental Theorem of Calculus.
Solution According to the Fundamental Theorem of Calculus, the value
of this definite integral is F (3)  F (1) where F ( x) is the antiderivative of
x2 ,
F  x    x 2 dx
x3

3
We have set the arbitrary constant equal to zero since any
antiderivative may be used in the Fundamental Theorem of Calculus.
The value of the definite integral is
3

x 2 dx  F  3  F 1
1

33
13

3
3

26
3
The midpoint sum for this area was 8.625. This is very close to the
exact area calculated above,
26
3
 8.667 .
The right hand side of the Fundamental Theorem of Calculus, F  b   F  a  , is often
abbreviated as
F  x
b
a
This means you should substitute the values on the ends of the vertical bar in the
function and subtract. Using this notation, we could carry out the steps in the previous
example as
7
3

1
x3
x dx 
3
3
2
1

33 13

3 3

26
3
Writing the work this way allows the antiderivative to be incorporated into the steps.
In some applications, we wish to consider all enclosed areas as positive. In these
cases, we must calculate areas above and below the horizontal axis individually. The
individual negative areas are modified to make them positive.
Example 4
Find the Area of the Enclosed Region
Find the area enclosed between y  e x  1 and the x axis over the
interval  1,1 . Assume all areas above or below the x axis are positive.
Solution Graph the function to see what the enclosed region looks like.
Figure 1 – The region between the x axis and the function over the interval [-1, 1].
If we were to find the area by evaluating the definite integral
8
1
 e
x
 1 dx
1
we would not find the proper amount of area. In this case, area below
the x axis would be negative and area above the x axis would be
positive. To insure the area below the x axis is calculated as positive,
we must find the area over  1, 0 and make it positive. Then we can
add that amount to the area from  0,1 .
The area on the left is
0
 e
x
 1 dx  e x  x
1
0
1
  e0  0    e 1   1 
 e 1
As expected, this is a negative value approximately equal to -0.368.
The area on the right side is
1
 e
x
 1 dx  e x  x
1
0
0
  e1  1   e0  0   e1  2
This is approximately equal to 0.718. The total enclosed area is
Total Enclosed Area  e1  e1  2  1.086
The absolute value insures that the negative value of the definite
integral over  1, 0 is considered a positive value in the total enclosed
area.
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Example 5
Find the Change in Sales
Based on data from 2001 through 2010, the rate at which annual sales
at Apple is changing (in millions of dollars of sales per million dollars of
research) can be modeled by
S   R   0.006 R  34.780
where R is the amount of money spent on research in millions of dollars.
In 2010, Apple spent $1782 million on research. If research spending is
increased to $2000 million, how much will sales increase?
Solution To get at the sales function, we need to apply the Fundamental
Theorem of Calculus
2000
 S   R  dR  S  2000   S 1782 
1782
The antiderivative of the integrand is S ( R)  0.003R 2  34.780 R  C .
Using the antiderivative, we get
2000
 S   R  dR  0.003R
2
 34.780 R
2000
1782
1782

 
 0.003  2000   34.780  2000   0.003 1782   34.780 1782 
2
2
 81560  71504.532
 10055.468
The units on this number are the same units as area under S   R  .
Heights on this graph are in millions of dollars of sales per millions of
dollars of research. Widths are in millions of dollars of research. The
units on the area is found by multiplying these units to give
10

units on area 
millions of dollars of sales
 millions of dollars of research
millions of dollars of research
 millions of dollars of sales
The increase in research spending leads to an increase in sales of
10,055.468 million dollars or $10,055,468,000.
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