# Integrating x^n with geometric sequence strips

- Author:
- Richard Trimble

- Topic:
- Integral Calculus

## Working out one bar

Normally when evaluating an integral by first principles we choose strips of an equal width. There is a major benefit to taking strips whose boundaries follow a geometric sequence.
We need to know a geometric sequence with n terms is:
If the common ratio is between 0 and 1 then the sequence will be decreasing, tending towards zero.
The sum of a geometric series is:
Finite case
.
Infinte case
Firstly observe that as the bars become thinner and thinner but the number of strips created can still be considered as an infinte series. In other words our approximation of the area will become exact in the limit as .
Create bars as shown in the diagram for the curve
Call the first bar (closest to X) bar 1 and the others bar i etc.
Bar i has height
Bar i has width
(remember that is the smaller term!)
The area of a single bar, bar i is thus

## Summing all the bars

We know the area of bar i and only one term actually depends on i.
We switch the order of exponentiation in this term to and now it becomes clear that the areas of the bars themselves form a geometric sequence with first term and common ratio .
As is smaller than 1 we can find the sum of the infinite sequence of bars using the formula above.
So we have the area of all the (infinite number of) bars .
What we need to do now to evaluate the integral is allow .
To do this without dividing zero by zero we need to recognise that is the reciprocal of the sum of a geometric series with first term 1 and common ratio r to n+1 terms!
i.e.
So we re-express the sum of the area of the bars in this expanded form.
as each power of r will tend to 1 as well giving:
as expected!

## Reflections

One of the amazing benefits of this approach is that it allows us to calculate the integral for all n with a single argument. Using equal size bars requires understanding for and for etc. These sums can be found but there is no easy way to generalise them to any power.
There are a few questions still to consider...
Does this proof work for all possible n values?
What adaptations are needed if n is negative?
Are any values of n excluded, if so which ones?