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

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Answer 1 · 2015-10-15T14:57:25

$-2/3$

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

$=1/3-1=-2/3$

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

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