Finding the Area Between Curves $y = -x^2 + 4x$ and $y = x^2 - 2x$

This problem requires us to calculate the area enclosed between two specific curves, which are given by the equations $y = -x^2 + 4x$ and $y = x^2 - 2x$. To find this area, we will follow these key steps:

• Find the points where the two curves intersect.
• Determine which curve is the 'upper' curve and which is the 'lower' curve within the intersection interval.
• Set up and evaluate a definite integral representing the area.

Calculating Intersection Points

First, we need to find the x-values where the two curves intersect. We do this by setting the two equations equal to each other:

$$ -x^2 + 4x = x^2 - 2x $$

Now, we rearrange the equation to solve for $x$. Move all terms to one side:

$$ 0 = x^2 - 2x + x^2 - 4x $$ $$ 0 = 2x^2 - 6x $$

Factor out the common term, $2x$:

$$ 0 = 2x(x - 3) $$

This equation gives us two solutions:

• $2x = 0 \implies x = 0$
• $x - 3 = 0 \implies x = 3$

So, the curves intersect at $x = 0$ and $x = 3$. These values will be our limits of integration.

Determining the Upper and Lower Curves

To set up the integral correctly, we need to know which function has a larger value (the 'upper' curve) between $x=0$ and $x=3$. We can test a value within this interval, for example, $x=1$:

• For $y = -x^2 + 4x$: $y(1) = -(1)^2 + 4(1) = -1 + 4 = 3$
• For $y = x^2 - 2x$: $y(1) = (1)^2 - 2(1) = 1 - 2 = -1$

Since $3 > -1$ at $x=1$, the curve $y = -x^2 + 4x$ is above the curve $y = x^2 - 2x$ in the interval $(0, 3)$.

Setting Up the Area Integral

The area $A$ between two curves $f(x)$ and $g(x)$ from $x=a$ to $x=b$, where $f(x) \ge g(x)$ on $[a, b]$, is given by the integral:

$$ A = \int_{a}^{b} (f(x) - g(x)) \, dx $$

In our case, $f(x) = -x^2 + 4x$, $g(x) = x^2 - 2x$, $a=0$, and $b=3$. The integral becomes:

$$ A = \int_{0}^{3} ((-x^2 + 4x) - (x^2 - 2x)) \, dx $$

Simplify the integrand:

$$ A = \int_{0}^{3} (-x^2 + 4x - x^2 + 2x) \, dx $$ $$ A = \int_{0}^{3} (-2x^2 + 6x) \, dx $$

Evaluating the Definite Integral

Now, we evaluate the definite integral. First, find the antiderivative of the integrand $(-2x^2 + 6x)$:

$$ \int (-2x^2 + 6x) \, dx = -2 \frac{x^{2+1}}{2+1} + 6 \frac{x^{1+1}}{1+1} = -2 \frac{x^3}{3} + 6 \frac{x^2}{2} = -\frac{2}{3}x^3 + 3x^2 $$

Next, apply the Fundamental Theorem of Calculus using the limits of integration $0$ and $3$:

$$ A = \left[ -\frac{2}{3}x^3 + 3x^2 \right]_{0}^{3} $$

Evaluate the antiderivative at the upper limit ($x=3$) and subtract its value at the lower limit ($x=0$):

$$ A = \left( -\frac{2}{3}(3)^3 + 3(3)^2 \right) - \left( -\frac{2}{3}(0)^3 + 3(0)^2 \right) $$ $$ A = \left( -\frac{2}{3}(27) + 3(9) \right) - (0 + 0) $$ $$ A = (-2 \times 9 + 27) - 0 $$ $$ A = -18 + 27 $$ $$ A = 9 $$

Final Result

The area enclosed between the curves $y = -x^2 + 4x$ and $y = x^2 - 2x$ is 9 square units.