Integral of x^n using Reimann sums

[stunner5000pt]

Homework Statement

prove $$\int_{a}^{b} x^n dx = \frac{ b^{n+1} - a^{n+1}{n+1}$$ using left & right side sums for n greater than or equal to 1

Relevant Equations

Reimann upper & lower sums, Faulhaber's formula

We don't need to worry about the n = -1 so we can assume that the function is continuous on any interval [a,b] where a, b are real numbers

if I separate my interval into N partitions, then the right side values in my interval are
$$a + \frac{b-a}{N}, a + 2 \frac{b-a}{N}, ... , a + k \frac{b-a}{N}, ... , a + N\frac{b-a}{N}$$

then the right side sum is
$$\frac{b - a}{N} \left( \left((a + \frac{b-a}{N} \right)^n + \left( a + \frac{b-a}{N} \right)^n + ... b^n \right)$$

$$\frac{b-a}{N^{n+1}} \sum_{k = 1}^{N} \left( a + k(b-a) \right)^n$$

here is where i'm stuck. First of all, is this the right way to go about this?

or perhaps, should we use an interval like [0,a] (which I was able to prove easily) and then use instead?

$$\int_{a}^{b} f(x) dx = \int_{0}^{b} f(x) dx - \int_{0}^{a} f(x) dx$$

Thanks again for all your help!

 

[PeroK]

or perhaps, should we use an interval like [0,a] (which I was able to prove easily) and then use instead?

$$\int_{a}^{b} f(x) dx = \int_{0}^{b} f(x) dx - \int_{0}^{a} f(x) dx$$

That would certainly be a perfectly valid approach.

 

[stunner5000pt]

That would certainly be a perfectly valid approach.

Thanks a lot - it certainly seems much easier

Out of curiosity, is the 'direct' approach that I was taking above even solvable?

 

[PeroK]

Thanks a lot - it certainly seems much easier

Out of curiosity, is the 'direct' approach that I was taking above even solvable?

It can't be much more difficult algebraically than the case where $a = 0$.

 

[pasmith]

Consider $$(n+1)x_i^n < x_{i+1}^n + x_ix_{i+1}^{n-1} + \dots + x_i^kx_{i+1}^{n-k} + \dots + x_i^n< (n+1)x_{i+1}^n$$ since $x_i < x_{i+1}$, and $$x_{i+1}^{n+1} - x_i^{n+1} = (x_{i+1}-x_i)(x_{i+1}^n + x_ix_{i+1}^{n-1} + \dots + x_i^kx_{i+1}^{n-k} + \dots + x_i^n)$$