Area under a curve.

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Hey everyone :D

I know basic integration and differentiation and I was siting on my bed today thinking how on earth any of it could correlate to finding the area under a curve.
So! I got out my trusty pen and paper and got to work!

After much thought I wrote: " If [tex]dx[/tex] means 'a small amount of x' then, if we graph x then dx in the graph must still be a small amount of x, only... at which point?
So I just assumed that dx one part of x as a whole. Remembering some of my physics class I calculated the area under the graph of "x" from points 0-2 on the x axis which equals 2.
So in my assumtion dx would be from 1-2 or 0-1 on the x axis.
and therefor.
the area under the line must equal: [tex]\int dx[/tex]
but, from 0-2 how do you calculate it? [tex]\int^{2}_{0}dx=[/tex] which does equal two.

So, If the applies for linear equations it must apply for quadratic equations as well, right?

So in the equation [tex]f(x)=-7x^{2}+6[/tex]
The area under the curve for this from 1 to 2 on the x axis must equal:
[tex]\int^{2}_{0}(-7x^{2}+6)dx[/tex]
correct?
 

Answers and Replies

  • #2
rock.freak667
Homework Helper
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After much thought I wrote: " If [tex]dx[/tex] means 'a small amount of x' then, if we graph x then dx in the graph must still be a small amount of x, only... at which point?
So I just assumed that dx one part of x as a whole. Remembering some of my physics class I calculated the area under the graph of "x" from points 0-2 on the x axis which equals 2.
So in my assumtion dx would be from 1-2 or 0-1 on the x axis.
and therefor.
the area under the line must equal: [tex]\int dx[/tex]
but, from 0-2 how do you calculate it? [tex]\int^{2}_{0}dx=[/tex] which does equal two.
If you are finding the area under the line y=x from 0 to 2, the integral should be

[tex]\int_0 ^{2} x dx = 2[/tex]

When you draw the line y=x and draw the verticals x=0 and x=2, the distance Δx, represents a small strip if width 'Δx' and length 'y' . So the area of that strip is 'y Δx'. To find the total area, one sums up all the small strips, from x=0 to x=2. But for the area to be exact, the limit must be taken as Δx tends to 0.

[tex]A= \lim_{x \rightarrow 0} \sum_{x=0} ^{x=2}y Δx = \int_0 ^{2} y dx[/tex]

and in this case y=x. So the integral becomes

[tex]A= \int_0 ^{2} x dx[/tex]


Only in special cases can you find the area using geometric methods, for example, y=x and x=0,x=2, forms a triangle.
 
  • #3
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When you draw the line y=x and draw the verticals x=0 and x=2, the distance Δx, represents a small strip if width 'Δx' and length 'y' . So the area of that strip is 'y Δx'. To find the total area, one sums up all the small strips, from x=0 to x=2. But for the area to be exact, the limit must be taken as Δx tends to 0.
[tex]A=lim_{x\rightarrow0}\sum^{x=2}_{x=0}y916;x=\int^{2}_{0}ydx [/tex]
Yes *cough* what you said.

and I was assuming [tex]\int^{2}_{0}dx[/tex] was the same as [tex]\int^{2}_{0}xdx[/tex]
Sorry for that mistake :)
 
  • #4
Landau
Science Advisor
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I know basic integration and differentiation and I was siting on my bed today thinking how on earth any of it could correlate to finding the area under a curve.
Because of the very definition of the (Riemann) integral.
 

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