riemann sum
riemann sum can be applied to integrable functions to find area
consider the area below the function
in the interval 
we split the shape into rectangles that are as narrow as possible as the more rectangles we use the more accurate the result
the following image shows the area split into
rectangles
#+RESULTS[34b9c30f69f07672565e241f613511f4b4f7f849]: riemann_sum
you can probably tell that the sum of the areas of these rectangles isnt really close to the area we want to find to be more accurate we increase the number of rectangles to
and we get a nicer result:
#+RESULTS[39af29f7b116f189aff75a9b3bf41a0a82c2e2d9]:
you can probably already tell that since this is a "narrower" coverage of the area, the sum of the rectangles would be closer to the value of the actual area
and now that we know that the more rectangles we have the more accurate we get, we know we should be aiming at
rectangles
the width of a rectangle would be
and the height would be 
where 
is the index of the rectangle starting at 1, the sum of the area 
of the rectangles would be:
we factor out
and since we want 
to be as big as possible, we take 
and since we know that the area in the interval is equal to the definite integral in said interval then:
which is the common pattern for riemann sums
consider the area below the function
in the interval
we split the shape into rectangles that are as narrow as possible as the more rectangles we use the more accurate the result
the following image shows the area split into
rectangles
#+RESULTS[34b9c30f69f07672565e241f613511f4b4f7f849]: riemann_sum
you can probably tell that the sum of the areas of these rectangles isnt really close to the area we want to find to be more accurate we increase the number of rectangles to
and we get a nicer result:
#+RESULTS[39af29f7b116f189aff75a9b3bf41a0a82c2e2d9]:
you can probably already tell that since this is a "narrower" coverage of the area, the sum of the rectangles would be closer to the value of the actual area
and now that we know that the more rectangles we have the more accurate we get, we know we should be aiming at
rectangles
the width of a rectangle would be
and the height would be where is the index of the rectangle starting at 1, the sum of the area of the rectangles would be:
we factor out
and since we want to be as big as possible, we take
and since we know that the area in the interval is equal to the definite integral in said interval then:
which is the common pattern for riemann sums
we find the area of 
in the interval 
using the definite integral method we do:
using riemann sum:
in the interval
using the definite integral method we do:
using riemann sum:
consider the following limit
