Key Concepts and Formulas

• Area Between Curves: The area between two curves $x=f(y)$ and $x=g(y)$ from $y=c$ to $y=d$ where $f(y) \ge g(y)$ is given by $\int_c^d [f(y) - g(y)] \, dy$.
• Finding Intersection Points: To find where two curves intersect, set their equations equal to each other and solve for the variable.
• Fundamental Theorem of Calculus: $\int_a^b F'(x) \, dx = F(b) - F(a)$.

Step-by-Step Solution

Step 1: Analyze the Equations and Choose Integration Variable

We are given the parabola $y^2 = 2x$ and the line $x + y = 4$. We need to find the area enclosed between these two curves. We can express both equations in terms of $x$ as a function of $y$.

• Parabola: $x = \frac{y^2}{2}$
• Line: $x = 4 - y$

Integrating with respect to $y$ is simpler because we don't have to split the parabola into two functions. We will use the formula $\text{Area} = \int_c^d [f(y) - g(y)] \, dy$, where $f(y)$ is the "right" function and $g(y)$ is the "left" function.

Why this step? Converting the equations to $x = f(y)$ form is crucial for applying our chosen integration strategy. Expressing both curves in terms of $y$ allows us to integrate with respect to $y$ easily.

Step 2: Find the Points of Intersection

To find the limits of integration, we need to find the $y$-values where the parabola and line intersect. We set the $x$-values equal to each other:

$\frac{y^2}{2} = 4 - y$

Multiplying both sides by 2, we get:

$y^2 = 8 - 2y$

Rearranging into a quadratic equation:

$y^2 + 2y - 8 = 0$

Factoring the quadratic:

$(y + 4)(y - 2) = 0$

So, the solutions are $y = -4$ and $y = 2$. These are our limits of integration.

Why this step? The points of intersection define the boundaries of our enclosed region. The $y$-coordinates serve as the lower and upper bounds of our definite integral with respect to $y$.

Step 3: Determine Which Curve is "Right" and Which is "Left"

We need to determine which function is greater (i.e., to the right) in the interval $[-4, 2]$. Let's test a value, say $y = 0$:

• Line: $x = 4 - 0 = 4$
• Parabola: $x = \frac{0^2}{2} = 0$

Since $4 > 0$, the line is to the right of the parabola. Thus, $f(y) = 4 - y$ and $g(y) = \frac{y^2}{2}$.

Now, we set up the integral:

$\text{Area} = \int_{-4}^2 \left( (4 - y) - \frac{y^2}{2} \right) \, dy$

Why this step? Correctly identifying $f(y)$ and $g(y)$ (right and left functions) is important. The test point method verifies the relative positions of the curves within the integration interval.

Step 4: Evaluate the Integral

Now, we evaluate the definite integral:

$\text{Area} = \int_{-4}^2 \left( 4 - y - \frac{y^2}{2} \right) \, dy$

First, find the antiderivative:

$\int \left( 4 - y - \frac{y^2}{2} \right) \, dy = 4y - \frac{y^2}{2} - \frac{y^3}{6} + C$

Now, evaluate at the limits of integration:

$\text{Area} = \left[ 4y - \frac{y^2}{2} - \frac{y^3}{6} \right]_{-4}^2 = \left( 4(2) - \frac{2^2}{2} - \frac{2^3}{6} \right) - \left( 4(-4) - \frac{(-4)^2}{2} - \frac{(-4)^3}{6} \right)$

$= \left( 8 - 2 - \frac{8}{6} \right) - \left( -16 - 8 + \frac{64}{6} \right) = \left( 6 - \frac{4}{3} \right) - \left( -24 + \frac{32}{3} \right)$

$= \frac{18}{3} - \frac{4}{3} + \frac{72}{3} - \frac{32}{3} = \frac{18 - 4 + 72 - 32}{3} = \frac{54}{3} = 18$

The area is 18 square units.

Why this step? This step computes the definite integral to arrive at the numerical answer.

Common Mistakes & Tips:

• Always sketch the graph to visualize the region and identify the correct "right" and "left" functions.
• Double-check the limits of integration.
• Be careful with signs when evaluating the antiderivative at the limits.

Summary

We found the area enclosed by the parabola $y^2 = 2x$ and the line $x + y = 4$ by expressing both equations as $x$ in terms of $y$, finding the intersection points to determine the limits of integration, setting up the definite integral, and evaluating it. The area is 18 square units. The correct answer given was 4, but the correct area is 18.

Final Answer: The final answer is \boxed{18}.