Thread: How to solve using Riemann Sums

1. How to solve using Riemann Sums

∫ (1-x^2) dx from -1 to 1

I know how to solve using the first fundamental theorem, but I don't understand Riemann Sums.

Can anyone help?

2. A Reimann sum is an illustration of what an integral is. It is the area of an

infinite number of rectangles under the graph. Thus, giving the area.

If we come off of the right side of the rectangles, we can use

$a+k{\Delta}x$

Where ${\Delta}x=\frac{b-(a)}{n}$.

In this case, each subinterval has length $\frac{1-(-1)}{n}=\frac{2}{n}$

So, we get $-1+\frac{2k}{n}$

Thereby, $f(x_{k})=1-(-1+\frac{2k}{n})^{2}$

Giving the area of rectangle k as $\left(1-(-1+\frac{2k}{n})^{2}\right)\cdot\frac{2}{n}$

The sum of the area of all these rectangles is

$\sum_{k=1}^{n}f(x_{k}){\Delta}x=\sum_{k=1}^{n}\lef t(1-(-1+\frac{2k}{n})^{2}\right)\cdot\frac{2}{n}$

$\frac{8}{n^{2}}\sum_{k=1}^{n}k-\frac{8}{n^{3}}\sum_{k=1}^{n}k^{2}$

But we know from some identities that $\sum_{k=1}^{n}k=\frac{n(n+1)}{2}, \;\ \sum_{k=1}^{n}k^{2}=\frac{n(n+1)(2n+1)}{6}$

We sub these is and take the limit as n-->infinity. This means the number of rectangles becomes larger and larger and we approach the area under the curve.

Doing all this algebra we get:

$\lim_{n\to \infty}\left[\frac{4}{3}-\frac{4}{3n^{2}}\right]=\frac{4}{3}$

Which is the value of the integral $\int_{-1}^{1}(1-x^{2})dx$

3. Originally Posted by operaphantom2003
∫ (1-x^2) dx from -1 to 1

I know how to solve using the first fundamental theorem, but I don't understand Riemann Sums.

Can anyone help?

Subdivide the interval [-1,1] in n equal subintervals $\left\{x_0=-1,x_1=1+\frac{2}{n},x_2=-1+\frac{4}{n},\ldots,x_k=-1+\frac{2k}{n},\ldots,x_n=-1+\frac{2n}{n}=1\right\}$ .

In each subinterval choose the leftmost point to evaluate $f(x)=1-x^2$ : $f\left(x_{i-1}\right)=f\left(-1+\frac{2i}{n}\right)=1-\left(-1+\left(\frac{2i}{n}\right)\right)^2$ $=\frac{4i}{n}-\frac{4i^2}{n^2}$ , and now

form your Riemann sums with the above (we can CHOOSE as we did since f is continuous in [-1,1] so we already know it is Riemann integrable there and thus the Riemann sums will converge to the integral's value no matter what sundivision of the interval and what points in it we choose):

$\lim_{n\rightarrow\infty}\sum\limits_{k=1}^kf(x_{i-1})(x_i-x_{i-1})$ $=\lim_{n\to\infty}\frac{2}{n}\sum\limits_{k=0}^{k-1}\left(\frac{4k}{n}-\frac{4k^2}{n^2}\right)$ ...

Tonio