How do you find the definite integral of #int (x^2-x)dx# from #[0,2]#?

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Answer 1 · 2015-11-21T16:18:36

#int_0^2(x^2-x)dx = 2/3#

We have

#int_0^2(x^2 - x)dx#

Since it's a sum we can break the integrals into two parts

#int_0^2(x^2-x)dx = int_0^2x^2dx + int_0^2-xdx#

We can integrate these two easily

#int_0^2(x^2-x)dx = x^3/3|_0^2 - x^2/2|_0^2#

Or,

#int_0^2(x^2-x)dx = 2^3/3 - 0^3/3 - 2^2/2 + 0^2/2 = 8/3 - 2 = (8-6)/3 = 2/3#