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

Question

1709 views around the world

You can reuse this answer
Creative Commons License

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,

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

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