Area Problem:
Given a function f , for simplicity, assume that f is continuous and non-negative on the closed interval [ a, b ], we want to find the area of the region bounded by the graph of f , the x − axis, and the vertical lines x = a and x = b
Using the concept of limit, We can approximate this area by dividing the region into rectangles, and as the width of the rectangles approaches 0 (or as the number of rectangles approaches infinity), the approximation becomes exact. This is easy to describe using pictures intuitively, how do we actually work out the algebra?
Let’s use an actual example to demonstrate this:
Example: Find the area of the region above the x − axis, below the graph of f ( x ) = x
2
+ 1, between x = 0 and x = 2,
We will estimate the area of the region by dividing the interval [0 , 2] into 4

subintervals of equal length. This is called a partition of the interval. What is the length of each of these intervals?
In general, if we want to partition the interval [ a, b ] into n subintervals of equal b − a length, then the length of each interval is given by ∆ x = , where ∆ x n represents the length of each interval.
In this example, the length of each interval is: ∆ x =
2 − 0
=
4
1
2
= 0 .
5.
We will form rectangles whose width is equal to the width of each interval to estimate the area. What about the height of the rectangles? In this example, we use the value of the function f at the right-endpoint of each interval as the height of each rectangle. In other words, since the right end-points of each interval is x = 0 .
5 , x = 1 , x = 1 .
5 , x = 2, the height of each of the rectangles using the value of f at the right end points will be given by: f (0 .
5) = 0 .
5 2
2 2 + 1 = 5
+ 1 = 1 .
25 , f (1) = 1 2 + 1 = 2 , f (1 .
5) = 1 .
5 2 + 1 = 3 .
25 , f (2) =
The area of each of these rectangles can be calculated by:
A
1
= (0 .
5)( f (0 .
5)) = (0 .
5)(1 .
25) = 0 .
625,
A
2
= (0 .
5)( f (1)) = (0 .
5)(2) = 1,

A
3
= (0 .
5)( f (1 .
5)) = (0 .
5)(3 .
25) = 1 .
625,
A
4
= (0 .
5)( f (2)) = (0 .
5)(5) = 2 .
5
The area of the region can be approximated as the sum of these four rectangles, and is equal to:
A r
= 0 .
625 + 1 + 1 .
625 + 2 .
5 = 5 .
75
We use a table to summarize the above calculation: interval width right-end-pt
[0 , 0 .
5] 0 .
5 0 .
5
[0 .
5 , 1] 0 .
5
[1 , 1 .
5] 0 .
5
[1 .
5 , 2] 0 .
5
1
1
.
2
5 height area = width × height f (0 .
5) = 1 .
25 (0 .
5)( f (0 .
5)) = 0 .
625 f (1) = 2 (0 .
5)( f (1)) = 1 f (1 .
5) = 3 .
25 (0 .
5)( f (1 .
5)) = 1 .
625 f (2) = 5 (0 .
5)( f (2)) = 2 .
5
Notice that in the above example, sine f is an increasing function, by using the right-endpoint, we ended up with an over-estimate of the area of the region.
We can also estimate the area by using the value of the function at the left-end-point of the interval as the height of the rectangles:
The left end-points of each of the intervals is: x = 0 , x = 0 .
5 , x = 1 , x = 1 .
5,
The height of each of the rectangles using the value of f at the left-endpoints is given by:

f (0) = 0 2 + 1 = 1 , f (0 .
5) = 0 .
5 2 + 1 = 1 .
25 , f (1) = 1 2 + 1 = 2 , f (1 .
5) = 1 .
5 2 + 1 =
3 .
25
The area of each of the rectangles is given by:
A
1
= (0 .
5)( f (0)) = (0 .
5)(1) = 0 .
5,
A
2
= (0 .
5)( f (0 .
5)) = (0 .
5)(1 .
25) = 0 .
625,
A
3
= (0 .
5)( f (1)) = (0 .
5)(2) = 1,
A
4
= (0 .
5)( f (1 .
5)) = (0 .
5)(3 .
25) = 1 .
625,
The area of the region can be approximated as the sum of the area of these four rectangles:
A l
= 0 .
5 + 0 .
625 + 1 + 1 .
625 = 3 .
75
We also summarize the information with a table: interval width left-end-pt
[0 , 0 .
5] 0 .
5 0
[0 .
5 , 1] 0 .
5
[1 , 1 .
5] 0 .
5
[1 .
5 , 2] 0 .
5
0
1
.
1
.
5
5 height f (0) = 1 area = width × height
(0 .
5)( f (0)) = 0 .
5 f (0 .
5) = 1 .
25 (0 .
5)( f (0 .
5)) = 0 .
625 f (1) = 2 (0 .
5)( f (1)) = 1 f (1 .
5) = 3 .
25 (0 .
5)( f (1 .
5)) = 1 .
625
This time, since f is a decreasing function, using the left end-point gives us an under-estimate.
Yet another approache is to use the value of the function at the mid-point of the interval as the height of the rectangles:

The x − coordinates of the mid-points of the intervals are: x = 0 .
25 , x = 0 .
75 , x = 1 .
25 , x = 1 .
75,
The height of each of the rectangles using the value of f at the mid-points is given by: f (0 .
25) = 0 .
25
2
+ 1 = 1 .
0625 , f (0 .
75) = 0 .
75
2
2 .
5625 , f (1 .
75) = 1 .
75
2
+ 1 = 4 .
0625
+ 1 = 1 .
5625 , f (1 .
25) = 1 .
25
2
+ 1 =
The area of each of the rectangles is given by:
A
1
= (0 .
5)( f (0 .
25)) = (0 .
5)(1 .
0625) = 0 .
53125,
A
2
= (0 .
5)( f (0 .
75)) = (0 .
5)(1 .
5625) = 0 .
78125,
A
3
= (0 .
5)( f (1 .
25)) = (0 .
5)(2 .
5625) = 1 .
28125,
A
4
= (0 .
5)( f (1 .
5)) = (0 .
5)(4 .
0625) = 2 .
03125,
The area of the region can be approximated as the sum of the area of these four rectangles:
A m
= 0 .
53125 + 0 .
78125 + 1 .
28125 + 2 .
03125 = 4 .
625
We similarly use a table to summarize the information: interval width mid-pt
[0 , 0 .
5] 0 .
5 0 .
25 f (0 .
height
25) = 1 .
area =
0625 (0 .
5)( f (0 width
.
×
25)) = 0 height
.
53125
[0 .
5 , 1] 0 .
5
[1 , 1 .
5] 0 .
5
[1 .
5 , 2] 0 .
5
0
1
1
.
.
.
75
25
75 f f f
(0
(1
(1
.
.
.
75) = 1
25) = 2
75) = 4
.
.
.
5625 (0
5625 (0
0625 (0
.
.
.
5)(
5)(
5)( f f f
(0
(1
(1
.
.
.
75)) = 0
25)) = 1
75)) = 2
.
.
.
78125
28125
03125
Out of these three estimates, which one is better? The exact value of the area
14 of this region is ≈ 4 .
667 (we will see later how to find this value), so using
3 the mid-point gives a better approximation for this particular function. While on average the mid-point approach usually gives a more accurate estimate, this is not always the case, so do not make any assumption about the accuracy of each approach.
We could have chosen yet some other points inside the interval and use the value of the function on those points as the height of the rectangle, but other than the right-end-point, left-end-point, and mid-point, all other points will require more complicated mathematics with not much to gain, so in order to estimate the area of the region, we usually stick with one of these three approaches.
What if we want a better estimation? Our intuition tells us that if we use more rectangles, our approximation should be better. We do the same problem with

eight rectangles instead of four, this time not showing the detail of the work, but just summarize the info in a table:
Right End-Point: interval width right-end-pt
[0 , 0 .
25] 0 .
25
[0 .
25 , 0 .
5] 0 .
25
0
0
.
.
25
5
[0 .
5 , 0 .
75] 0 .
25
[0 .
75 , 1] 0 .
25
[1 , 1 .
25] 0 .
25
[1 .
25 , 1 .
5] 0 .
25
[1 .
5 , 1 .
75] 0 .
25
[1 .
75 , 2] 0 .
25
0
1
1
1
.
.
.
.
75
1
25
5
75
2 height area = width × height f (0 .
25) = 1 .
0625 (0 .
25)( f (0 .
25)) = 0 .
265625 f (0 .
5) = 1 .
25 (0 .
25)( f (0 .
5)) = 0 .
3125 f (0 .
75) = 1 .
5625 (0 .
25)( f (0 .
75)) = 0 .
390625 f (1) = 2 (0 .
25)( f (1)) = 0 .
5 f (1 .
25) = 2 .
5625 (0 .
25)( f (1 .
25)) = 0 .
640625 f (1 .
5) = 3 .
25 (0 .
25)( f (1 .
5)) = 0 .
8125 f (1 .
75) = 4 .
0625 (0 .
25)( f (1 .
75)) = 1 .
015625 f (2) = 5 (0 .
25)( f (2)) = 1 .
25
The sum of the area of these eight rectangles is equal to:
A r
= 0
5 .
1875
.
265625 + 0 .
3125 + 0 .
390625 + 0 .
5 + 0 .
640625 + 0 .
8125 + 1 .
015625 + 1 .
25 =

Left End-Point: interval width left-end-pt
[0 , 0 .
25] 0 .
25 0
[0 .
25 , 0 .
5] 0 .
25
[0 .
5 , 0 .
75] 0 .
25
0
0
.
.
25
5
[0 .
75 , 1] 0 .
25
[1 , 1 .
25] 0 .
25
[1 .
25 , 1 .
5] 0 .
25
[1 .
5 , 1 .
75] 0 .
25
[1 .
75 , 2] 0 .
25
0
1
1
1
.
.
.
.
75
1
25
5
75 height f (0) = 1 area = width × height
(0 .
25)( f (0)) = 0 .
25 f (0 .
25) = 1 .
0625 (0 .
25)( f (0 .
25)) = 0 .
265625 f (0 .
5) = 1 .
25 (0 .
25)( f (0 .
5)) = 0 .
3125 f (0 .
75) = 1 .
5625 (0 .
25)( f (0 .
75)) = 0 .
390625 f (1) = 2 (0 .
25)( f (1)) = 0 .
5 f (1 .
25) = 2 .
5625 (0 .
25)( f (1 .
25)) = 0 .
640625 f (1 .
5) = 3 .
25 (0 .
25)( f (1 .
5)) = 0 .
8125 f (1 .
75) = 4 .
0625 (0 .
25)( f (1 .
75)) = 1 .
015625
The sum of the area of these eight rectangles is equal to:
A l
= 0
4 .
1875
.
25 + 0 .
265625 + 0 .
3125 + 0 .
390625 + 0 .
5 + 0 .
640625 + 0 .
8125 + 1 .
015625 =

Mid-Point: interval width mid-pt height area = width × height
[0 , 0 .
25] 0 .
25 0 .
125 f (0 .
125) = 1 .
015625 (0 .
25)( f (0 .
125)) = 0 .
25390625
[0 .
25 , 0 .
5] 0 .
25 0 .
375 f (0 .
375) = 1 .
140625 (0 .
25)( f (0 .
375)) = 0 .
28515625
[0 .
5 , 0 .
75] 0 .
25 0 .
625 f (0 .
625) = 1 .
390625 (0 .
25)( f (0 .
625)) = 0 .
34765625
[0 .
75 , 1] 0 .
25 0 .
875 f (0 .
875) = 1 .
765625 (0 .
25)( f (0 .
875)) = 0 .
44140625
[1 , 1 .
25] 0 .
25 1 .
125 f (1 .
125) = 2 .
265625 (0 .
25)( f (1 .
125)) = 0 .
56640625
[1 .
25 , 1 .
5] 0 .
25 1 .
375 f (1 .
375) = 2 .
890625 (0 .
25)( f (1 .
375)) = 0 .
72265625
[1 .
5 , 1 .
75] 0 .
25 1 .
625 f (1 .
625) = 3 .
640625 (0 .
25)( f (1 .
625)) = 0 .
91015625
[1 .
75 , 2] 0 .
25 1 .
875 f (1 .
875) = 4 .
515625 (0 .
25)( f (1 .
875)) = 1 .
12890625
The sum of the area of these eight rectangles is equal to:
A m
= 0 .
25390625+0 .
28515625+0 .
34765625+0
0 .
91015625 + 1 .
12890625 = 4 .
65625
.
44140625+0 .
56640625+0 .
72265625+
We see that as the number of rectangles increases, the approximation becomes better.

We now come up with a general formula for approximating the area of the region bounded between the x − axis, the function f and between x = a and x = b :
We divide the interval [ a, b ] into n subintervals of equal length, let ∆ x = b − a n be the length of each subinterval. We have the following subintervals:
[ a = x
0
, x
1
] , [ x
1
, x
2
] , [ x
2
, x
3
] , . . .
[ x i
, x i +1
] . . .
[ x n − 1
, x n
= b ]
Let x
∗
1 be any number inside the interval [ x
0
, x
1
], and x
∗
2 interval [ x
1
, x
2
], and in general, let x
∗ i be any number inside the be any number inside the interval [ x i − 1
, x i
].
The sum of the area of the n rectangles with width ∆ x = b − a n is then given by: and height f ( x
∗ i
)
A n
= f ( x
∗
1
)∆ x + f ( x
∗
2
)∆ x + f ( x
∗
3
)∆ x + · · · + f ( x
∗ i
)∆ x + · · · f ( x
∗ n − 1
)∆ x + f ( x
∗ n
)∆ x
If we use the sigma Σ notation, the above formula can be written in more compact form:
A n n
X
= f ( x
∗
1
)∆ x + f ( x
∗
2
)∆ x + f ( x
∗
3
)∆ x + · · · + f ( x
∗ i
)∆ x + · · · f ( x
∗ n − 1
)∆ x + f ( x
∗ n
)∆ x = f ( x
∗ i
)∆ x i =1
Our intuition tells us that, as the number of rectangles, n , becomes larger, the area will be a better approximation. We believe that our approximation will be exact if we allow n go out to infinity. Therefore, we define the area of the region by the following formula:
Defintion: Suppose f is a continuous, non-negative function inside the interval

[ a, b ], then the area bounded above the x − axis, below f , between x = a and x = b , is given by:
A = lim n →∞
A n
= lim n →∞ f ( x
∗
1
)∆ x + f ( x
∗
2
)∆ x + · · · + f ( x
∗ i
)∆ x + · · · + f ( x
∗ n
)∆ x = lim n →∞ n
X i =1 f ( x
∗ i
)∆ x
The sum n
X f ( x
∗ i
)∆ x that occurs in this definition is called a Riemann sum .
i =1
In other words, the area of a region is defined as the limit of a Riemann sum.
Notice that the coordinate of each of the boundary point x i can be obtained by adding the proper multiple of ∆ x to the left end point, a . For example, x
0
= a , x
1
= a + ∆ x = a + (1)∆ x , x
2
= a + (2)∆ x ,
...
x
3
= a + (3)∆ x , x i
= a + ( i )∆ x , x n
= a + ( n )∆ x = a + ( n ) b − a n
= b
In the definition, the number x
∗ i can be any number inside the interval [ x i − 1
, x i
], and the definition will give the same result (if the limit exists). However, in actual practice, if we really wanted to use this definition to find the area of the region, usually either the left-end-point or right-end-point is used for simplicity.
If we use the left-end-point, then x
∗ i x
∗ i
= x i
= x i − 1
; if we use the right end-point, then
It can be shown that if f is a continuous function inside [ a, b ], then the above limit always exists. However, if f is not continuous inside [ a, b ], the limit may not exist.
The above definition can also be used to find the Riemann sum of a function f which is not necessarily always positive in [ a, b ]. How to interpret the result if part or all of the graph of f is below the x − axis?

In this case, we can interpret the Riemann sum as a net area between the (positive) area of the region above the x − axis and the (negative) area of the region below the x − axis.
There are other interpretation of the Riemann sum. Let’s say you are driving and you wanted to estimate how much distance you covered by using your velocity at different time intervals. You do a reading of your velocity every 5 seconds over a
30 seconds period, and come up with the following: time (s) 0 5 10 15 20 25 30 velocity (meters/sec) 8 12 18 20 17 11 5
So how much distance you covered in this 30 seconds?
One way we can estimate this is by assuming that, at the moment you make the reading of your velocity, your velocity will be the same for the next five seconds before you make the next reading. If we make this assumption, since distance = velocity × time , we can calculate the distance we travelled for each of the five second-period in the table: time (s) velocity (meters/sec)
0
8
5 10
12 18
15
20
20 25 30
17 11 5 distance travelled (meters) 40 60 90 100 75 55 25
The total distance travelled is: 40 + 60 + 90 + 100 + 75 + 55 + 25 = 445 meters
We graph the above table, with time t on the x − axis and velocity on the y − axis, and connect the points using a smooth curve.

If we try to estimate the area of the region under the curve using the same method we used before, by dividing into 6 subintervals and using the left end points as the height of the rectangles, we see that the area of the first rectangle is 5 × 8 = 40, which is the same as the distance travelled for the first five seconds in our assumption. The area of the second rectangle is 5 × 12 = 60, the same as the distance travelled in the next five seconds. In fact, the total area of the rectangles is equal to 445, the same as the distance travelled.
We see in this example that, if velocity (as a function of time) is graphed on the y − axis, and time is graphed on the x − axis, then the area under the curve represents the (total) distance travelled by the object with the velocity function over that time period. In fact, if the function f ( x ) is used to represent other quantities, the area under the curve can be interpreted as something of interest as well.
Also notice in this example that, velocity, v ( t ), is the derivative of distance, s ( t ).
This example we just saw indicates that the area under the curve of the velocity function is related to the distance travelled (net change of the distance function).
This is not an isolated case. If the value of f
0
( x ) is graphed as values on the y − axis, then the area under f
0
( x ) and the net change of the value of f ( x ) is intricately related, as we shall see later.

Definite integral :
Suppose f is a function defined on [ a, b ] ( f is not necessarily positive), and the interval [ a, b ] is divided into n subintervals of equal lengths ∆ x = be any point in the subinterval [ x i − 1
, x i b − a
. Let x n
], we define the definite integral of f
∗ i from a to b by:
Z b a f ( x ) dx = lim n →∞ n
X i =1 f ( x
∗ i
)∆ x
If this limit exists, we say that f is integrable on [ a, b ]
Notice that the sum on the right hand side of the above definition is just the
Riemann sum we used to define the area of a region under f if f is always positive inside [ a, b ]. By definition, a definite integral is just a Riemann sum.
Also remember that in our definition, the function f does not have to be always positive inside the interval [ a, b ]. As discussed, if f is not always positive on [ a, b ], then the Riemann sum, hence the definite integral, represents the net area of the region.
By definition, a definite integral is the limit of a Riemann sum. As with any other limits, a limit does not have to exist. However, there is a condition that guarantees the existance of the limit:
Theorem:
If f is continuous on [ a, b ], then f is integrable on [ a, b ] (in other words, if f
P n is continuous on the closed interval [ a, b ], then the limit: lim n →∞ i =1 f ( x i
)∆ x exists).

Properties of Definite Integral
Suppose f and g are continuous functions on [ a, b ] and k is a constant, then:
Z b a f ( x ) dx = −
Z a f ( x ) dx b
Z a f ( x ) dx = 0 a
Z b k dx = k ( b − a ) a
Z b a f ( x ) ± g ( x ) dx =
Z b a f ( x ) dx ±
Z b a g ( x ) dx
Z b a kf ( x ) dx = k
Z a f ( x ) dx b
If c is any real number a ≤ c ≤ b , then
Z c
Z b
Z b f ( x ) dx + f ( x ) dx = f ( x ) dx a c a
If f ( x ) is an odd function, then
Z a
− a f ( x ) dx = 0
If f ( x ) is an even function, then
Z a
− a f ( x ) dx = 2
Z a
0 f ( x ) dx
Z b f ( x ) dx ≤ a
Z b a
| f ( x ) | dx
If f ( x ) ≤ g ( x ) for all x in [ a, b ], then
Z b f ( x ) dx ≤ a
Z b a g ( x ) dx

Fundamental Theorem of Calculus, first form:
If f is a continuous function on [ a, b ]. Define
F ( x ) =
Z x f ( t ) dt for a ≤ x ≤ b , a f ( t ) then F is continuous on [ a, b ] and differentiable on ( a, b ), and F
0
( x ) = f ( x )
Using the above, one can easily establish a more well-known and more frequently applied form of FTC:
Fundamental Theorem of Calculus, second form:
If f is continuous on [ a, b ] and F is any anti-derivative of f , then
Z b f ( x ) dx = F ( b ) − F ( a ) a
The fundamental theorem of calculus provides a (comparatively) easier way of finding definite integral (hence area, volumn, arc-length...etc), which is by finding anti-derivatives.
Theoretically, it tells us that differentiation and integration
(finding limits of infinite sum) are inverse processes of each other. Because of the FTC, one usually use the term integration to mean anti-differentiation, or refer to an anti-derivative of a function and an integral of that function. As long as we know what we are talking about, this is an acceptable way of using the terminology and is widely used.
It is, however, important to note that, by definition , anti-differentiation (finding antiderivative of a function) and integration (finding sum/area in terms of

a Riemann sum) are two different processes and need not have anything to do with each other. The FTC tells us that, as a method to finding definite integral
(a sum), one can use antiderivative. However, there are other ways of finding a definite integral (for example, actually evaluate the Riemann sum, numerical approximation) which has nothing to do with finding anti-derivatives. The sym-
Z b bol f ( x ) dx represents, by definition, a Riemann sum. The fact that it can a be calculated by finding antiderivative is a discovered result. Do not mistaken a definite integral (Riemann sum) from an indefinite integral (anti-derivative).
Example:
Evaluate the definite integral by using the fundamental theorem of calculus:
Z
2 x
2
+ 1 dx
1
Since F ( x ) = x
3
+ x is an antiderivative of x
2
+ 1, the FTC tells us that:
3
Z
2 x
2
+ 1 dx = F (2) − F (1) =
1
2 3
+ 2 −
3
1 3
+ 1 =
3
10
3 x
3
You should note that F ( x ) = + x is not the only antiderivative of x 2 + 1.
3
The FTC says that any antiderivative would work, so we use the simplest one.
Any other antiderivative of x 2 + 1 would differ from F ( x ) by a constant, and the constant would cancel each other in the subtraction of F ( b ) − F ( a ) b
We use the notation F ( x ) to mean F ( b ) − F ( a ), so the above can also be written a as:
Z
2 x
2
+ 1 dx =
1 x
3
+ x
3
2
1
=
2
3
+ 2 −
3
1
3
+ 1 =
3
10
3
Example:
Define g ( x ) =
Z x
3 t + 1 dt Find g
0
( x ).
2
Ans:
Understand that for each value of x , g ( x ) has a particular value by definition of g ( x ). For example,

g (2) =
Z
2
2 t + 1 dt = 0,
2 g (3) =
Z
3
2 t + 1 dt = 6,
2
Geometrically, g ( x ) represents the area under the function y = 2 t + 1 between t = 2 and t = x . The first part of the FTC tells us that g ( x ) is a differentiable function of x , and g
0
( x ) = 2 x + 1.
Also note that when applying the fundamental theorem of calculus, you must make sure that the function f ( x ) inside the integral is continuous over the interval
[ a, b ], otherwise you may not use the FTC. Here’s an example where the FTC can not be directly applied:
Z
2
1
− 1 x 3 dx
If you try to use FTC on this, you will get:
Z
2
1
− 1 x 3
1 dx = −
2 x 2
#
2
− 1
= −
1
8
− −
1
2
=
3
8

1
But this is an incorret answer, since the function is discontinuous at x = 0, x 3 and 0 is inside [ − 1 , 2], the FTC cannot be applied in this example. The definite integral in this example does not exist.