Integral of x^n using Reimann sums

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Homework Statement
prove [tex] \int_{a}^{b} x^n dx = \frac{ b^{n+1} - a^{n+1}{n+1} [/tex] 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!
 
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stunner5000pt said:
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.
 
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PeroK said:
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?
 
stunner5000pt said:
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##.
 
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