Find the area bounded by the following curves: y=2x-x²
Find the area bounded by the following curves: $$ y=2x-x^2, \qquad y=x^2. $$
Explanation of Method
Setting the two parabolas equal gives their two intersection points directly. Between them, one parabola stays above the other, so the area is a single elementary integral of the difference of the two quadratic expressions.
- Find the intersection points.
- Determine which curve is above.
- Integrate the difference between the curves.
Step 1 · Find the intersection points
$$ 2x-x^2=x^2 \quad\Longrightarrow\quad 2x-2x^2=0 \quad\Longrightarrow\quad 2x(1-x)=0 \quad\Longrightarrow\quad x=0 \text{ or } x=1. $$
Intersection points: $(0,0)$ and $(1,1)$.
Step 2 · Determine which curve is above
At $x=\dfrac12$: $y=2x-x^2=\dfrac34$, while $y=x^2=\dfrac14$. Since $\dfrac34>\dfrac14$, the curve $y=2x-x^2$ lies above $y=x^2$ for $0<x<1$.
Step 3 · Set up and evaluate the area integral
$$ A = \int_0^1\left[(2x-x^2)-x^2\right]dx = \int_0^1(2x-2x^2)\,dx = 2\int_0^1(x-x^2)\,dx = 2\left[\frac{x^2}2-\frac{x^3}3\right]_0^1 = 2\left(\frac12-\frac13\right) = \frac13. $$
Additional Notes
- The curve $y=2x-x^2=1-(x-1)^2$ is a downward-opening parabola with vertex $(1,1)$; it and the upward-opening $y=x^2$ meet at the origin and at that vertex, enclosing the lens-shaped region computed above.
Result
The area is: