0
It is still an area. In simple integral calculus, you integrate the function between limits on the x-axis. This will give the area under the curve.
e.g.
Taking y = X^2 between the limits of (x = 0, & x = 2) .
Integrated it is it is [x^3/3] between 2 & o)
Hence
[2^3/3] - [0^3/3] =
8/3 - 0 =
8/3 = 2 2/3 = 2.6666..... units^2
Hope that helps!!!!
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