Definite Integral as the Limit of a Sum

Let f be a continuous function defined on close interval [a, b]. Assume that all the values taken by the function are non negative, so the graph of the function is a curve above the x-axis.
The definite integral `int_a^b`f(x) dx  is the area bounded by the curve y = f(x), the ordinates x = a, x = b and the x-axis. To evaluate this area, consider the region PRSQP between this curve, x-axis and the ordinates x = a and x = b. Fig.

Divide the interval [a, b] into n equal subintervals denoted by `[x_0, x_1]`, `[x_1, x_2]` ,..., `[x_(r – 1), x_r], ..., [x_(n – 1), x_n],` where  `x_0 = a, x_1 = a + h, x_2 = a + 2h, ... , x_r` = a + rh and `x_n` = b = a + nh or `n =(b-a)/h` We note that as n → ∞, h → 0.  From the above fig. we have 
area of the rectangle (ABLC) < area of the region (ABDCA) < area of the rectangle (ABDM)        ... (1) 
Same  as `x_r – x_(r–1) → 0, i.e., h → 0 all the three areas shown in (1) become nearly equal to each other. Now we form the following sums.

`s_n = h[f(x_0) + ...+f(x_(n-1))] = h summation f(x_r)`   ...(2)
and `S_n =h[f(x_1) + f(x_2) + ...+ f(x^n)] = h summation f(x_r) `   ..(3)
Here, `s_n` and `S_n` denote the sum of areas of all lower rectangles and upper rectangles raised over subintervals `[x_(r–1), x_r] for r = 1, 2, 3, …, n, respectively. In view of the inequality (1) for an arbitrary subinterval `[x_(r–1), x_r]`, we have `s_n < area of the region PRSQP < S_n`                ... (4)
As n→∞ strips become narrower and narrower, it is assumed that the limiting values of (2) and (3) are the same in both cases and the common limiting value is the required area under the curve.  
Symbolically, we write
`lim_(n ->∞ ) S_n = lim _(n-> ∞) s_n` =  area of the region PRSQP = `int _a^b f(x)`
dx         ...(5)
It follows that this area is also the limiting value of any area which is between that of the rectangles below the curve and that of the rectangles above the curve. 
`int _a^b f(x)dx =lim_(h->0) h [f(a) + f(a+h) + ...+f(a+(n-1))h]`
or `int _a^b f(x) dx = (b-a) lim_(n ->∞) 1/ n [f(a) + f(a+h) + ...+ f(a+(n-1)h)]` ...(6) 
where `h = (b-a)/n -> 0 as n -> ∞`  

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, we simply write the integral as `int _a^b f(t)dt` or `int _a^b f(u)du instead of  `int _a^b f(x)dx. Hence, the variable of integration is called a dummy variable.
Video link : https://youtu.be/CMG6ucGUQiU