Triangles and Definite Integrals

apt403

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?

daveyp225

The triangle's height is not 2.

apt403

Brain fart. I'm an idiot.