# Triangles and Definite Integrals

1. Jul 4, 2012

### 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:

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

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?

2. Jul 4, 2012

### daveyp225

The triangle's height is not 2.

3. Jul 4, 2012

### apt403

Brain fart. I'm an idiot.