# Triangles and Definite Integrals

by apt403
Tags: definite, integrals, triangles
 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: $\int_{a}^{b}2x\frac{\mathrm{d} }{\mathrm{d} x} = A = 1/2b*h$ Where the integral is bound by a = 0 and b = 2, and the triangle's base is 2 and height is 2. but rather, $\int_{0}^{2}2x\frac{\mathrm{d} }{\mathrm{d} x} = 4$ and $A = 2$ Where's the discrepancy?
 P: 88 The triangle's height is not 2.
 P: 46 Brain fart. I'm an idiot.

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