# realguy's question at Yahoo! Answers regarding a Riemann sum

#### MarkFL

Staff member
Here is the question:

Using limit process to solve?

Use the limit process to find the area of the region between the function: f(x) = 4 – x2
And the x-axis over the interval [-2, 2]
I have posted a link there to this topic so the OP can see my work.

#### MarkFL

Staff member
Hello realguy,

If we observe that the given function is even, then we may find the area on the interval $[0,2]$ and then double the result to get the answer. We will divide this interval into $n$ equal subdivisions and use a left sum. The area of the $k$th rectangle is:

$$\displaystyle A_k=\frac{2-0}{n}\left(4-x_k^2 \right)$$

where $$\displaystyle 0\le k\le n-1\in\mathbb{Z}$$ and $$\displaystyle x_k=\frac{2k}{n}$$.

Hence:

$$\displaystyle A_k=\frac{2}{n}\left(4-\left(\frac{2k}{n} \right)^2 \right)=\frac{8}{n^3}\left(n^2-k^2 \right)$$

Thus, the total area is approximated by (recall we need to double the sum):

$$\displaystyle A_n=2\left(\frac{8}{n^3}\sum_{k=0}^{n-1}\left(n^2-k^2 \right) \right)=\frac{16}{n^3}\sum_{k=0}^{n-1}\left(n^2-k^2 \right)$$

Using the formulas:

$$\displaystyle \sum_{k=0}^{n-1}(1)=n$$

$$\displaystyle \sum_{k=0}^{n-1}\left(k^2 \right)=\frac{n(n-1)(2n-1)}{6}$$

We obtain then:

$$\displaystyle A_n=\frac{16}{n^3}\left(n^3-\frac{n(n-1)(2n-1)}{6} \right)=\frac{16}{n^3}\cdot\frac{4n^3+3n^2-n}{6}=\frac{32n^2+24n-8}{3n^2}$$

A form which is more easily evaluated as a limit to infinity is:

$$\displaystyle A_n=\frac{32}{3}+\frac{8}{n}-\frac{8}{3n^2}$$

And so the exact area is given by:

$$\displaystyle A=\lim_{n\to\infty}A_n=\frac{32}{3}$$