1. ## Finding Riemann sum

Let f(x) = 3x(4 - x)

Suppose we partition the interval [1; 3] into n subintervals of equal length, and then use the right-hand endpoints of these subintervals as sample points. Find an expression for the corresponding Riemann sum for f(x).

2. Originally Posted by TsAmE
Let f(x) = 3x(4 - x)

Suppose we partition the interval [1; 3] into n subintervals of equal length, and then use the right-hand endpoints of these subintervals as sample points. Find an expression for the corresponding Riemann sum for f(x).
The interval has length $\displaystyle 2$ units. If you divide it into $\displaystyle n$ subintervals, then the length of each subinterval is $\displaystyle \frac{2}{n}$. This is the length of each rectangle.

The width of each rectangle will be $\displaystyle f(x_i) = 3x_i(4 - x_i)$ for $\displaystyle i = 1, 2, 3, \dots, n$ (since you are taking the right hand endpoints you don't use $\displaystyle x_0$).

So that means each rectangle's area is

$\displaystyle \frac{2}{n}\left[3x_i(4 - x_i)\right]$

$\displaystyle = \frac{6x_i(4 - x_i)}{n}$.

The total area therefore is

$\displaystyle \sum_{i = 1}^{n}\frac{6x_i(4 - x_i)}{n}$

By making $\displaystyle n \to \infty$, the area of the rectangles converges on the area underneath the curve.

3. Originally Posted by TsAmE
Let f(x) = 3x(4 - x)

Suppose we partition the interval [1; 3] into n subintervals of equal length, and then use the right-hand endpoints of these subintervals as sample points. Find an expression for the corresponding Riemann sum for f(x).
Each subinterval has length

$\displaystyle \Delta x = \frac{3-1}{n}=\frac{2}{n}$

The right endpoint choices for $\displaystyle x_1^*,...,x_n^*$ is given by:

$\displaystyle x^*_k=x_k=a+k \Delta x$

$\displaystyle x_k^*= 1+k \frac{2}{n} =\frac{2n+4k}{n^2}$

Thus

$\displaystyle \sum ^n_{k=1} f(x^*_k)\Delta x = \sum ^n_{k=1} 3x_k^*(4 - x^*_k) \Delta x$

$\displaystyle = \sum ^n_{k=1} 3( \frac{2n+4k}{n^2} ) (4- \frac{2n+4k}{n^2}) \frac{2}{n}$

This is the required expression for the Riemann sum:

$\displaystyle A= \lim_{(\Delta x \to \infty)} \sum ^n_{k=1} \left[ 3( \frac{2n+4k}{n^2} ) (4- \frac{2n+4k}{n^2}) \frac{2}{n} \right]$

4. $\displaystyle x_k^*= 1+k \frac{2}{n} =\frac{2n+4k}{n^2}$

How did you get: $\displaystyle \frac{2n+4k}{n^2}$ ?