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

We have

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

We can integrate these two easily

Or,

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To find the definite integral of (\int_{0}^{2} (x^2 - x) , dx), follow these steps:

- Find the antiderivative of the function ((x^2 - x)).
- Evaluate the antiderivative at the upper and lower bounds of integration.
- Subtract the value of the antiderivative at the lower bound from the value at the upper bound to find the definite integral.

The antiderivative of ((x^2 - x)) is (\frac{1}{3}x^3 - \frac{1}{2}x^2).

Evaluate the antiderivative at (x = 2) and (x = 0):

- At (x = 2): (\left(\frac{1}{3}(2)^3 - \frac{1}{2}(2)^2\right))
- At (x = 0): (\left(\frac{1}{3}(0)^3 - \frac{1}{2}(0)^2\right))

Subtract the value at the lower bound from the value at the upper bound: (\left(\frac{1}{3}(2)^3 - \frac{1}{2}(2)^2\right) - \left(\frac{1}{3}(0)^3 - \frac{1}{2}(0)^2\right))

Simplify the expression to find the definite integral.

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