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Understanding the area under a curv and trapizum rule 
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#1
Jan1813, 12:39 PM

P: 106

I am having a bit of difficulty understanding the intuition behind it. So here my understanding so far, I have attached some photos of the book I am using to understand integration.
This is from the book I have posted, I am just going up to the part I do not understand. S basically I have to consider a rectangle ( which is the one at the bottom of the first photograph ). This rectangle has a width of δx. An area of δA which is the shaded area. The area of rectangle ABEF is yδx: which is pretty self explanatory. The area of rectangle ABCD is (y+δy)δx Therfore we have: yδx<δA <(y+δy)δx: So all this is doing is comparing the area to the 2 rectangles Now the confusion begins, it say beceause δx >0 You should divided through by. Why is this? Letting δx [itex]\rightarrow[/itex]0 gives [itex]\frac{δA}{δx}\rightarrow\frac{dA}{d} [/itex] and δy[itex]\rightarrow[/itex]0 therefore [itex]\frac{dA}{dx}=y[/itex] this to me is saying that x is the independent variable and δA dependent variable; correct? So as more rectangle are add then obviously the area will get smaller and the more rectangles add δy will reach zero. But how dose [itex]\frac{dA}{dx}=y[/itex] This is the trouble I am having up to a point, if someone could ans and the I will ask the other half of the trouble I am having. Or if someone feel like it explain from the beginning how the intergration of the curve works. I am new to integration so I would appreciate all the help anyone can offer. 


#2
Jan1813, 02:54 PM

P: 1,583




#3
Jan1813, 02:57 PM

P: 106

Oh yeah that would help, give me sec they will be uploaded



#4
Jan1813, 03:06 PM

P: 106

Understanding the area under a curv and trapizum rule
Here they are, sorry about that



#5
Jan1813, 03:32 PM

Mentor
P: 21,397

If you approximate the area by a rectangle there are a couple of possibilities: 1. Use the left endpoint as the height of the rectangle, getting an area of f(x_{i})* Δx. 2. Use the right endpoint as the height of the rectangle, getting an area of f(x_{i + 1})* Δx. If you approximate the area by a trapezoid instead of a rectangle, the top side of the trapezoid is a straight line between (x_{i}, f(x_{i})) and (x_{i+1}, f(x_{i+1})). The trazezoid approximation uses the idea that the area of a trapezoid is the base times the average of the two unequal sides. IOW, the approximate area is (1/2)[ f(x_{i+1}) + f(x_{i+1})] * Δx. That's really all there is to it. 


#6
Jan1813, 03:44 PM

P: 1,583




#7
Jan2013, 06:14 AM

P: 106

Right, first of all thanks for the replies. The have been a big help. I see how the rule work, but really its only and approximation, not a definite integral, because not matter how many rectangles or trap you will always have gaps inbetween the curve. I think this is the real part that causing the confusion. I watch a couple of vids and read more articles but, nothing really ans my question.
Once again big for any advice that anyone could give. 


#8
Jan2013, 08:32 AM

P: 718




#9
Jan2013, 09:15 AM

P: 106

So for example if, I workout the area underneath a curve of f(x) between two point a and b and take the integral. I am really getting a very close approximation to the actuall area under the curve. So when you say If your willing to do the calculation, you mean your are referring to the amount of ordinates, under a given curve?



#10
Jan2013, 09:22 AM

P: 1,583




#11
Jan2013, 09:38 AM

P: 718




#12
Jan2013, 10:53 AM

P: 106

Right okay I have clocked it now. Thanks for clearing that, up for me. Some of my confusion stems from the book I read, that don't really show how, the integral dose what is dose more of here how to workout the math. If you catch my drift.



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