# Riemann Integration

Can some please draw a comparison between Riemann Integration and normal definite integration in terms of accuracy.

Hi shreyarora, and welcome to PF

What do you mean with "normal definite integration", to my knowledge that simply is the Riemann integral?
And what do you mean with accuracy? Riemann integration is 100% accurate, it uses no approximations what-so-ever.

We'll be in a better position to answer you if you let is know where you're coming from...

HallsofIvy
Homework Helper
I suspect that by "Riemann integration" he is referring to actually using a finite number of rectangles to approximate the integral- that is, Riemann sums.

Shreyarora, the formula for the accuracy is given in any Calculus book. For example, on page 487 of Salas, Hille, and Etgen's Calculus, the error in using n rectangles to approximate
$$\int_a^b f(t)dt$$
is given as less than
$$(f(b)- f(a))\frac{b- a}{n}$$

So, for example, if you use 10 rectangles to integrate
$$\int_0^1 x^2 dx$$
Your error would be less than (1-0)((1- 0)/10) or 1/10.

Of course, the trapezoidal method and Simpson's rule give better accuracy.

Thanks HallofIvy, I think you got most of what I meant say.

By normal integration I meant that, if you integrate x^2 from a to b, you substitute limits to x^3/3.
whereas, the computation differs while evaluating integral using Riemann Integration.

I am actually writing an article wherein I have to justify that using Riemann Integration yields accurate results to a real life problem over the "normal definite integration" that I have defined above.

Or is it actually possible to compare the two methods?

Its probably better (and less confusing) to say "Riemann sum" instead of Riemann integration. 'Ordinary integration' is Riemann integration. That is,

$$\int_1^2 x^2 dx=\frac{1}{3}x^3|_1^2=\frac{7}{3}$$

I just performed Riemann integration. But I could I have approximated the Riemann integral with a Riemann sum using say, the left endpoint:

$$\sum_{k=0}^{N-1} (1+k\Delta x)^2 \Delta x$$

where $$\Delta x =(2-1)/N$$