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

Question

Evelyn Agard · Accepted Answer

#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#

Luke Alvarez · Answer

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

1. Find the antiderivative of the function $(x^2 - x)$.
2. Evaluate the antiderivative at the upper and lower bounds of integration.
3. 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.