Definite Integral

In order to find the integral by first principle, see the following steps:

First, we should make partitions on the interval [0,2]. However, in this step, you should consider the length of each partition (i.e. delta x) first. Here we take it be 1/n (i.e. delta x = 1/n). Since the length of the interval is 2 and the length of each partition is 1/n, therefore the number of partitions will be 2n.

Next, we should find what will be the value of (xi-1, xi) in terms of this delta x in the i-th partition, then we can know what will be the xi.
In the i-th partition, the value of xi = i/n.

Therefore, the integral x dx = limit sum [f(xi) delta xi] {limit takes as n tends to infinity and the sum is from i = 1 to i = 2n}

i.e. integral x dx
= limit sum [f(i/n) 1/n] {limit takes as n tends to infinity and the sum is from i = 1 to i = 2n}
= limit sum [i/square(n)]
= limit 1/square(n) (sum i) {i from 1 to 2n}
= limit 1/square(n) [2n(2n+1)/2]
= limit (2n+1)/n {as n tends to infinity}
= limit (2 + 1/n) {as n tends to infinity}
= 2

If you have any problem, don't hesitate to ask me.

Let f(x) = x.
f(0) = 0
f(2) = 2

∫ x dx (Upper limit:2, Lower limit:0)
= Area under the graph y=f(x) from 0 to 2
= 2(2)/2 (Note: it is a triangle)
= 2