Triangles and Definite Integrals

  • Thread starter apt403
  • Start date
  • #1
47
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I'm trying to figure out how to integrate a data set, without knowing the function. While doing this, I got to thinking about this:

If the definite integral of a function can be represented by the area under that function, bound by the x axis, then shouldn't:

[itex]\int_{a}^{b}2x\frac{\mathrm{d} }{\mathrm{d} x} = A = 1/2b*h[/itex]

Where the integral is bound by a = 0 and b = 2, and the triangle's base is 2 and height is 2.

but rather,

[itex]\int_{0}^{2}2x\frac{\mathrm{d} }{\mathrm{d} x} = 4[/itex]

and

[itex]A = 2[/itex]

Where's the discrepancy?
 

Answers and Replies

  • #2
88
0
The triangle's height is not 2.
 
  • #3
47
0
Brain fart. I'm an idiot.
 

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