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## Main Question or Discussion Point

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?

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?