find the area enclosed between the parabola y=3x-x² and y=x²-x

To find the area enclosed between the curves \( y = 3x - x^2 \) and \( y = x^2 - x \), follow these steps:

### 1. Find the points of intersection:
Set the two equations equal to find where the curves intersect:
\[
3x - x^2 = x^2 - x
\]
Rearrange the equation:
\[
3x - x^2 - x^2 + x = 0
\]
\[
-2x^2 + 4x = 0
\]
Factor the quadratic equation:
\[
-2x(x - 2) = 0
\]
The solutions are:
\[
x = 0 \quad \text{or} \quad x = 2
\]
So, the curves intersect at \( x = 0 \) and \( x = 2 \).

### 2. Set up the integral for the area:
The area between the curves is given by the integral of the difference between the two functions, from \( x = 0 \) to \( x = 2 \):
\[
A = \int_0^2 \left( (3x - x^2) - (x^2 - x) \right) \, dx
\]
Simplify the integrand:
\[
(3x - x^2) - (x^2 - x) = 3x - x^2 - x^2 + x = 4x - 2x^2
\]
Thus, the integral becomes:
\[
A = \int_0^2 (4x - 2x^2) \, dx
\]

### 3. Compute the integral:
Now, calculate the integral:
\[
A = \left[ 2x^2 - \frac{2x^3}{3} \right]_0^2
\]
Evaluate at the bounds:
\[
A = \left( 2(2)^2 - \frac{2(2)^3}{3} \right) - \left( 2(0)^2 - \frac{2(0)^3}{3} \right)
\]
Simplify:
\[
A = \left( 2(4) - \frac{2(8)}{3} \right) = 8 - \frac{16}{3}
\]
\[
A = \frac{24}{3} - \frac{16}{3} = \frac{8}{3}
\]

### Final Answer:
The area enclosed between the curves is \( \frac{8}{3} \).

Would you like more details on any step?

Here are 5 related questions:
1. How do you calculate the area between more complex curves?
2. What happens if the curves don't intersect?
3. How do you determine the boundaries of integration for other curves?
4. What if the curves are given in parametric form?
5. How can you find areas between curves using polar coordinates?

**Tip:** When setting up an integral for areas between curves, always ensure that the "top" function is subtracted from the "bottom" function over the correct interval.

Explore how to find the area enclosed between the curves y=3x-x² and y=x²-x. This solution involves solving for intersection points and calculating the definite integral of the difference between two functions. Suitable for students in Grades 11-12 studying integral calculus.

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