- #1
apt403
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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:
\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?
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?