# Riemann Integral

How to find the given area by using the Riemann integral: how to do, example, and its solution.

## How to Do

Let's find the area of the colored region S.

First, slice the region vertically to n parts.
And draw a rectangle for each slice.

The width of each slice is (b - a)/n.

For the kth slice,
the x value of the right side is (b - a)k/n.

See the kth slice.

The width is (b - a)/n.
And the height is f( [(b - a)k]/n ).

So the area of the kth slice, Ak, is
f( [(b - a)k]/n )⋅(b - a)/n.

Then the sum of the slices, Sn,
is the sum of f( [(b - a)k]/n )⋅(b - a)/n
as k goes from 1 to n.

This summation is the Riemann sum.

Sigma notation

As n → ∞,
each slice will become so slim
that each slice will fit into the colored region.

So the sum of these slices
is the area of the colored region: S.

So S is the limit of Sn as n → ∞.

The Riemann sum with the limit
is the Riemann integral.

So, to find the Riemann integral:
1. Slice the region into n parts,
and find Ak (area of the kth slice).
2. Write Sn (Riemann sum).
3. And find S, the limit of Sn.

## Example

Slice the region vertically to n parts.
And draw a rectangle for each slice.

The width of each slice is, (1 - 0)/n, 1/n.

For the kth slice,
the x value of the right side is (1 - 0)k/n, k/n.

See the kth slice.

The width is 1/n.
And the height is f(k/n) = (k/n)2.

So the area of the kth slice, Ak, is
(k/n)2⋅1/n.

Then the Riemann sum, Sn,
is the sum of (k/n)2⋅1/n
as k goes from 1 to n.

It is k that goes from 1 to n, not n.

So take the denominator n3
out from the sigma.

As n → ∞,
you get the Riemann integral, S.

So the area of the colored region, S, is 1/3.

Indeterminate form (Part 1)