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).
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.
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] $