Integral calculus
The definite integral Calculating a definite integral from the definition
Riemann sum
Until now, in the definition of the sums, S and s we've used the maximum and the minimum values, Mi and mi of a given continuous function f, so that mi <  f (x) < Mi for x in [xi - 1, xi] i = 1, 2 , . . . , n
Now, if we arbitrarily choose a point xi'  in every interval Dxi and make products  f (xi' ) Dxi, then
mi Dxi <  f (xi' ) Dx < Mi Dxi i = 1, 2 , . . . , n The left-hand and right-hand sides of the above inequality are the sums, s and S respectively that, because of continuity of  f, tend to the same limit value I when the number of subintervals n increases to infinity, such that the length of every interval Dxi tends to zero, for any partition of the interval [a, b] and arbitrarily chosen points xi' in the subintervals [xi - 1, xi]. Hence, the middle term called a Rieman sum, will tend to the same limit value.
Therefore, if  f is a positive continuous function on the interval [a, b] then, the definite integral of the function from a to b is defined to be the limit Note that the limit value of the sum changes as the number of subintervals n increases to infinity while the length (Dxi) of each tends to zero, for any partition of the interval [a, b].
Calculating a definite integral from the definition
As the sequence of inscribed rectangles s tends to the definite integral increasingly while the sequence of circumscribed rectangles S tends to the same value decreasingly then Thus, we can approximate the area under the graph of a function over the interval [a, b] to any desired level of accuracy using the Riemann sum of inscribed or circumscribed rectangles.
The area of the ith rectangle  f (xi' ) Dxi, denoted as height times base, represents the ith term of Riemann sum and is called the element of area.
When we use the partition of the interval [a, b] into n equal subintervals (regular partition) then be the length of the intervals [xi - 1, xi] i = 1, 2 , . . . , n.
Calculating a definite integral from the definition, examples
 Example:   Evaluate where f (x) = 1, using the definition of the definite integral.
Solution:   Since the graph of the constant  f (x) = 1 is the line passing through the point (0, 1) parallel to the x-axis, the region under the graph is the rectangle of the base = (b - a) with the height h = 1.
 Thus, the area A = (b - a) · 1 = b - a, as shows the right figure. Therefore,  Example:   Evaluate where f (x) = x, using the definition of the definite integral.
Solution:   Since the graph of  f (x) = x is the line through the origin, coordinates of every its point  y = x, so the region under the graph is the trapezium with the height b - a and whose parallel sides are a and b.
Let's use the partition of the interval [a, b] into n equal subintervals, so that  Dx = (b - a) / n and calculate
 the lower sum s of inscribed rectangles, as shows the right figure. If we choose the point xi'  to be the left-hand end of each subinterval, then x1' = a                   and        f (x1' ) = a x2' = a + Dx,                      f (x2' ) = a + Dx x3' = a + 2Dx,                    f (x3' ) = a + 2Dx · · ·                                        · · · xn' = a + (n - 1)Dx,           f (xn' ) = a + (n - 1)Dx. We use the Riemann sum to calculate the sum of inscribed rectangles with bases of the same length, therefore,          s = a Dx + (a + Dx)Dx + (a + 2Dx)Dx · · ·  + [a + (n - 1)Dx]Dx
s = Dx[n a + Dx(1 + 2 · · ·  + (n - 1))
to calculate the sum of natural numbers inside of square brackets we use the formula
Sn = [2a1 + (n - 1)d] for the sum of the arithmetic sequence whose first term a1 = 1 and difference d = 1,
so we get the sum equals (n - 1)n / 2, and since  Dx = (b - a) / n then, Thus the area under the graph of  f (x) = x over the interval [a, b]    Calculus contents E 