So I wanted to find an area of a simple triangle using basic integration.

I made the equation: f(x)=(1/2) x, and the area from x=0 to x=2

I wanted to use the formula (method) here: ttp://timmurphy.org/examples/integral_lim.jpg

So I know that delta x = 2/n, coming from (2-0)/n,
but I have no idea how to get the xi (height value in terms of n).

I'd really like an explanation using both right end and left end rectangle points if possible!
The only step I could thing of from here is to just plug 2/n into the function f(x)=(1/2)x to get xi

using the method you're suggesting will always get you an approximate value unless the different between x_{i} and x_{i+1} is infinitely small. using the formula method is no good.

you might want to brush up on sigma notation because I think that's where the confusion is. the distance between x_{i} and x_{i + 1} is something you'd "decide" based on how precise of an answer you want.

and the integral would equal the sum of all f(x_{i}) where x_{i-1}, x_{i} and x_{i+1} are equally spaced and the spacing is decided depending on how many calculations you can afford to make.


tl;dr : humans can't use that method/formula you're suggesting unless you're looking for the area of a rectangle-derived shape.

just integrate x/2 from 0 to 2


edit: you can PM me if you have follow-up question [you can post here and tell me to come check or straight up pm]

This post was edited by rx7drifter on Jul 12 2013 05:36pm

The formula you posted is just the technical definition of an integral. Remember that an integral is just a sum of an infinite number of rectangles. The width of each rectangle is given by delta-x, which represents an infinitesimally small change in the x-direction, and that the height of each rectangle is given by f(x). You can use this method to approximate the area underneath the curve by using thinner and thinner rectangles, but an integral will add up an infinite number of infinitely thin rectangles.

To approximate the slope, you could graph your equation, and then make two rectangles with width 1. Then try it with 4 rectangles (width=1/2), or 8 rectangles (width=1/4). You will note that as the number of rectangles increases, you will get closer and closer to the true value of the area.

Check out Simpson's rule for a similar rule that uses trapezoids to approximate the area rather than rectangles.

As for the integration...

If y=(1/2)x and we integrate, then you get
(1/2)*(x)^2 2
= (1/4)*x^2 from 0 to 2.

Plugging in the limits, you get

(1/4)*(2)^2 - (1/4)*(0)^2
= (1/4) * 4
= 1.

A = 1.

great response

Thank you! I work as a math tutor at a local college and I'm studying engineering. I really love these kinds of things.