You're approximating
with a Riemann sum, which comes in the form
where
are sample points chosen according to some decided-upon rule, and 
is the distance between adjacent sample points in the interval.
The simplest way of approximating the definite integral is by partitioning the interval into equally-spaced subintervals, in which case
, and since ![[a,b]=[1,5]](https://tex.z-dn.net/?f=%5Ba%2Cb%5D%3D%5B1%2C5%5D)
, we have
Using the right-endpoint method, we approximate the area under
with rectangles whose heights are determined by their right endpoints. These endpoints are chosen by successively adding the subinterval length to the starting point of the interval of integration.
So if we had
subintervals, we'd split up the interval of integration as
![[1,5]=[1,2]\cup[2,3]\cup[3,4]\cup[4,5]](https://tex.z-dn.net/?f=%5B1%2C5%5D%3D%5B1%2C2%5D%5Ccup%5B2%2C3%5D%5Ccup%5B3%2C4%5D%5Ccup%5B4%2C5%5D)
Note that the right endpoints follow a precise pattern of
The height of each rectangle is then given by the values above getting squared (since
). So continuing with the example of , the Riemann sum would be
For
,
and so on, so that the definite integral is given exactly by the infinite sum
with a Riemann sum, which comes in the form
where
The simplest way of approximating the definite integral is by partitioning the interval into equally-spaced subintervals, in which case
Using the right-endpoint method, we approximate the area under
So if we had
Note that the right endpoints follow a precise pattern of
The height of each rectangle is then given by the values above getting squared (since
For
and so on, so that the definite integral is given exactly by the infinite sum