Question

How do you evaluate `int (x^2 - 2x)dx` for [0, 1]?

Answer 1

`-2/3`

Explanation:

`int_0^1(x^2-2x)dx=[x^3/3-(2x^2)/2]_0^1`

`=1/3-1=-2/3`

(Note that this will represent the area under the graph of `y=x^2-2x` between the points `x=0 and x=1`)