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Triangles and Definite Integrals 
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#1
Jul412, 01:47 AM

P: 46

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? 


#2
Jul412, 02:04 AM

P: 88

The triangle's height is not 2.



#3
Jul412, 02:06 AM

P: 46

Brain fart. I'm an idiot.



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