# Integral of x^n using Reimann sums

• stunner5000pt
In summary: It can't be much more difficult algebraically than the case where ##a = 0##.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
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!

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.

stunner5000pt
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##.

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 $$x_{i+1}^{n+1} - x_i^{n+1} = (x_{i+1}-x_i)(x_{i+1}^n + x_ix_{i+1}^{n-1} + \dots + x_i^kx_{i+1}^{n-k} + \dots + x_i^n)$$

## 1. What is the definition of the integral of x^n using Reimann sums?

The integral of x^n using Reimann sums is a method for approximating the area under a curve of a function with the power of n. It involves dividing the interval of integration into smaller subintervals, calculating the area of each subinterval using rectangles, and then summing up these areas to approximate the total area under the curve.

## 2. How do you calculate the width of each subinterval in Reimann sums?

The width of each subinterval in Reimann sums is calculated by dividing the total interval of integration (b-a) by the number of subintervals (n). This gives us the width, or the change in x, for each rectangle used in the approximation.

## 3. What is the difference between a left, right, and midpoint Reimann sum?

In a left Reimann sum, the height of each rectangle is determined by the function value at the left endpoint of the subinterval. In a right Reimann sum, the height is determined by the function value at the right endpoint. In a midpoint Reimann sum, the height is determined by the function value at the midpoint of the subinterval. These different methods can give slightly different approximations of the integral.

## 4. How does the number of subintervals affect the accuracy of the Reimann sum approximation?

The more subintervals used in the Reimann sum, the more accurate the approximation will be. As the number of subintervals increases, the width of each rectangle decreases, resulting in a closer approximation to the actual area under the curve.

## 5. Can Reimann sums be used to calculate the exact value of the integral of x^n?

No, Reimann sums can only provide an approximation of the integral of x^n. The accuracy of the approximation can be improved by using more subintervals, but it will never be an exact value. To find the exact value of the integral, we need to use other methods such as the Fundamental Theorem of Calculus or integration techniques.

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