Let f be a continuous non-negative function defined on a closed interval [a, b]. Since the value of the function is non-negative, the graph of the function is a curve above X-axis. Let the graph of the curve be as shown in the figure.Let f be a continuous non-negative function defined on a closed interval [a, b]. Since the value of the function is non-negative, the graph of the function is a curve above X-axis. Let the graph of the curve be as shown in the figure.
figure (a)
The question is how we find the area under the curve y = f(x) bounded by the X-axis and the lines x = a and x = b. This region is shaded in the graph.
To understand this problem easily let us consider three special such functions.
Let
This function is continuous, non-negative in the interval [1, 2], which is shown in the figure.
Being a rectangular region, the area of f(x) = 2 bounded by X- axis, x = 1 and x = 2 is given by base X height, the height being equal to 
Base = (2 - 1) = 1 units, height = 2 units
This region is triangular above the axis bounded by x = 0 and x = 1.
The area of this region is given by 
or
The region under the centre bounded by X - axis, x = 1 and x = 3 is a trapezium, where area is given by
(Since the area of the trapezium = base x 
(the sum of the parallel sides))
In all the three cases, we have seen that, the area of the regions are obtained by multiplying the base with average height of the curve.
Using this fact, how can we find the area under the curve in figure (a) above?
The base is the length of the domain interval [a, b] = b - a. Now our problem is to find the average height of the curve. This is indeed the average value of the function in the interval [a, b]
Average Value of a Function in an Interval
We can take the value of f at a (i.e., f(a)) as first estimate for average value of the function.
Divide [a, b] into two equal parts such that 
then the second estimate of the average value of the function can be taken as second estimate of the average value of the function can be taken as
(see the above figure)
Clearly the second estimate of the average value is better than the first estimate.
If we divide the interval into three equal parts such that 
then the improved estimate for the average value of f(a) is
(see the above figure)
In this process, if we divide the closed interval [a, b] into more and more equal parts, and take the average of functional values at these points, we are closer to the average value of the function in closed interval [a, b].
Let us divide the closed intervals to n equal parts, then the average value of the function is
where 
as shown in the figure below
For larger value of n, equation (1) will be appropriate estimate for the average value of the function in the given closed interval. With this discussion, we can define average value of f in [a, b]
Therefore the area under the curve y = f(x) bounded by X-axis, x = a and x = b.
= base x average height
Definite Integral
Let f (x) be a single valued continuous function defined in the interval [a,b] where b > 0 and let the interval [a,b] be divided into n equal parts each of length h, so that nh = b - a; then we define
The method of evaluating 
by using the above definition is called integration from first principles.Definite Integral Through Area of Triangles
The definition,
, can be explained in another way also. We rewrite above definition as
Here the first term is hf (a). It is the area of the rectangle marked as 1 in figure below (because h and f (a) are the adjacent sides of this rectangle). Similarly, the second term hf (a+h) is the area of the rectangle marked as 2 in the figure below.
Thus, hf (a) + hf (a + h) + .......+ 
is the sum of the areas of these n rectangles marked in above figure. The union of these rectangles is approximately the region between the curve and the x-axis. When n is larger, the number of rectangles is more, and the approximation is closer. Therefore if we take the limit as n ® ¥, we obtain that 
as in equation (1) is the area of the region bounded by the curve y = f(x) and the lines y = 0, x = a and x = b.If we take the right end-points instead of the left, then also, we get the same areas as the limit of areas of unions of some other rectangles.
is the area of the same region.Note that any one of the processes, viz., taking the left hand end-points or the right hand end-points will be sufficient for calculating the desired area.
Terminology
We have the following terminology associated with the symbol
Remark
The value of the definite integral of a function over any particular interval depends on the function and the interval, but not on the variable of integration that we choose to represent the independent variable. If the independent variable is denoted by t or u instead of x, 
Hence, the variable of integration is called a dummy variable.Example:
Integrate the following definite as limit of sums:
Solution:
We are given that a = 0, b = 4 
or nh = 4By definition,
E-Store
Let f be a continuous non-negative function defined on a closed interval [a, b]. Since the value of the function is non-negative, the graph of the function is a curve above X-axis. Let the graph of the curve be as shown in the figure.Let f be a continuous non-negative function defined on a closed interval [a, b]. Since the value of the function is non-negative, the graph of the function is a curve above X-axis. Let the graph of the curve be as shown in the figure.
(see the above figure)
(see the above figure)
by using the above definition is called integration from first principles.
, can be explained in another way also. We rewrite above definition as
is the sum of the areas of these n rectangles marked in above figure. The union of these rectangles is approximately the region between the curve and the x-axis. When n is larger, the number of rectangles is more, and the approximation is closer. Therefore if we take the limit as n ® ¥, we obtain that 
as in equation (1) is the area of the region bounded by the curve y = f(x) and the lines y = 0, x = a and x = b.
is the area of the same region.
figure (a)
The question is how we find the area under the curve y = f(x) bounded by the X-axis and the lines x = a and x = b. This region is shaded in the graph.
To understand this problem easily let us consider three special such functions.
Let
This function is continuous, non-negative in the interval [1, 2], which is shown in the figure.
Being a rectangular region, the area of f(x) = 2 bounded by X- axis, x = 1 and x = 2 is given by base X height, the height being equal to
Base = (2 - 1) = 1 units, height = 2 units
This region is triangular above the axis bounded by x = 0 and x = 1.
The area of this region is given by
or
The region under the centre bounded by X - axis, x = 1 and x = 3 is a trapezium, where area is given by
(Since the area of the trapezium = base x
(the sum of the parallel sides))
In all the three cases, we have seen that, the area of the regions are obtained by multiplying the base with average height of the curve.
Using this fact, how can we find the area under the curve in figure (a) above?
The base is the length of the domain interval [a, b] = b - a. Now our problem is to find the average height of the curve. This is indeed the average value of the function in the interval [a, b]
Average Value of a Function in an Interval
We can take the value of f at a (i.e., f(a)) as first estimate for average value of the function.
Divide [a, b] into two equal parts such that
then the second estimate of the average value of the function can be taken as second estimate of the average value of the function can be taken as
Clearly the second estimate of the average value is better than the first estimate.
If we divide the interval into three equal parts such that
then the improved estimate for the average value of f(a) is
In this process, if we divide the closed interval [a, b] into more and more equal parts, and take the average of functional values at these points, we are closer to the average value of the function in closed interval [a, b].
Let us divide the closed intervals to n equal parts, then the average value of the function is
where
as shown in the figure below
For larger value of n, equation (1) will be appropriate estimate for the average value of the function in the given closed interval. With this discussion, we can define average value of f in [a, b]
Therefore the area under the curve y = f(x) bounded by X-axis, x = a and x = b.
= base x average height
Definite Integral
Let f (x) be a single valued continuous function defined in the interval [a,b] where b > 0 and let the interval [a,b] be divided into n equal parts each of length h, so that nh = b - a; then we define
The method of evaluatingDefinite Integral Through Area of Triangles
The definition,
Here the first term is hf (a). It is the area of the rectangle marked as 1 in figure below (because h and f (a) are the adjacent sides of this rectangle). Similarly, the second term hf (a+h) is the area of the rectangle marked as 2 in the figure below.
Thus, hf (a) + hf (a + h) + .......+If we take the right end-points instead of the left, then also, we get the same areas as the limit of areas of unions of some other rectangles.
Note that any one of the processes, viz., taking the left hand end-points or the right hand end-points will be sufficient for calculating the desired area.
Terminology
We have the following terminology associated with the symbol
Remark
The value of the definite integral of a function over any particular interval depends on the function and the interval, but not on the variable of integration that we choose to represent the independent variable. If the independent variable is denoted by t or u instead of x,
Hence, the variable of integration is called a dummy variable.Example:
Integrate the following definite as limit of sums:
Solution:
We are given that a = 0, b = 4
or nh = 4By definition,