- #1

stunner5000pt

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- 2

- 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

[tex] a + \frac{b-a}{N}, a + 2 \frac{b-a}{N}, ... , a + k \frac{b-a}{N}, ... , a + N\frac{b-a}{N} [/tex]

then the right side sum is

[tex] \frac{b - a}{N} \left( \left((a + \frac{b-a}{N} \right)^n + \left( a + \frac{b-a}{N} \right)^n + ... b^n \right) [/tex]

[tex] \frac{b-a}{N^{n+1}} \sum_{k = 1}^{N} \left( a + k(b-a) \right)^n [/tex]

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?

[tex] \int_{a}^{b} f(x) dx = \int_{0}^{b} f(x) dx - \int_{0}^{a} f(x) dx [/tex]

Thanks again for all your help!

if I separate my interval into N partitions, then the right side values in my interval are

[tex] a + \frac{b-a}{N}, a + 2 \frac{b-a}{N}, ... , a + k \frac{b-a}{N}, ... , a + N\frac{b-a}{N} [/tex]

then the right side sum is

[tex] \frac{b - a}{N} \left( \left((a + \frac{b-a}{N} \right)^n + \left( a + \frac{b-a}{N} \right)^n + ... b^n \right) [/tex]

[tex] \frac{b-a}{N^{n+1}} \sum_{k = 1}^{N} \left( a + k(b-a) \right)^n [/tex]

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?

[tex] \int_{a}^{b} f(x) dx = \int_{0}^{b} f(x) dx - \int_{0}^{a} f(x) dx [/tex]

Thanks again for all your help!