Key Concepts and Formulas

• Area between curves: The area between two curves $y = f(x)$ and $y = g(x)$ from $x=a$ to $x=b$, where $f(x) \ge g(x)$ on $[a, b]$, is given by $\int_a^b (f(x) - g(x)) \, dx$.
• Finding Intersection Points: To find the limits of integration $a$ and $b$, we need to find the $x$-coordinates of the intersection points of the two curves, which are the solutions to $f(x) = g(x)$.

Step-by-Step Solution

Step 1: Find the intersection points of the curves.

We are given the curves $y - x = 2$ and $x^2 = y$. To find the points of intersection, we set the expressions for $y$ equal to each other. From the first equation, we have $y = x + 2$. Substituting this into the second equation, we get: $x^2 = x + 2$ Rearranging the equation, we have: $x^2 - x - 2 = 0$ Factoring the quadratic gives: $(x - 2)(x + 1) = 0$ Thus, the $x$-coordinates of the intersection points are $x = 2$ and $x = -1$. These will be our limits of integration.

Step 2: Determine which curve is "above" the other.

We need to determine which function has larger $y$-values between $x = -1$ and $x = 2$. We have $y_1 = x + 2$ and $y_2 = x^2$. Let's pick a test point within the interval, say $x = 0$. Then $y_1(0) = 0 + 2 = 2$ and $y_2(0) = 0^2 = 0$. Since $2 > 0$, we can conclude that $y_1 = x + 2$ is above $y_2 = x^2$ in the interval $[-1, 2]$. Therefore, $f(x) = x + 2$ and $g(x) = x^2$.

Step 3: Set up the integral for the area.

Since $x + 2 \ge x^2$ on the interval $[-1, 2]$, the area between the curves is given by: $\text{Area} = \int_{-1}^{2} (x + 2 - x^2) \, dx$

Step 4: Evaluate the integral.

We evaluate the integral as follows: $\text{Area} = \int_{-1}^{2} (x + 2 - x^2) \, dx = \left[ \frac{x^2}{2} + 2x - \frac{x^3}{3} \right]_{-1}^{2}$ $\text{Area} = \left( \frac{2^2}{2} + 2(2) - \frac{2^3}{3} \right) - \left( \frac{(-1)^2}{2} + 2(-1) - \frac{(-1)^3}{3} \right)$ $\text{Area} = \left( \frac{4}{2} + 4 - \frac{8}{3} \right) - \left( \frac{1}{2} - 2 + \frac{1}{3} \right)$ $\text{Area} = \left( 2 + 4 - \frac{8}{3} \right) - \left( \frac{1}{2} - 2 + \frac{1}{3} \right)$ $\text{Area} = \left( 6 - \frac{8}{3} \right) - \left( \frac{1}{2} - \frac{6}{3} + \frac{1}{3} \right)$ $\text{Area} = \left( \frac{18}{3} - \frac{8}{3} \right) - \left( \frac{1}{2} - \frac{5}{3} \right)$ $\text{Area} = \frac{10}{3} - \left( \frac{3}{6} - \frac{10}{6} \right)$ $\text{Area} = \frac{10}{3} - \left( -\frac{7}{6} \right)$ $\text{Area} = \frac{10}{3} + \frac{7}{6}$ $\text{Area} = \frac{20}{6} + \frac{7}{6}$ $\text{Area} = \frac{27}{6} = \frac{9}{2}$

Step 5: Recheck the Calculation

Oops, there was an error in the previous calculation. Let's recalculate the area: $\text{Area} = \left( \frac{2^2}{2} + 2(2) - \frac{2^3}{3} \right) - \left( \frac{(-1)^2}{2} + 2(-1) - \frac{(-1)^3}{3} \right)$ $\text{Area} = \left( 2 + 4 - \frac{8}{3} \right) - \left( \frac{1}{2} - 2 + \frac{1}{3} \right)$ $\text{Area} = 6 - \frac{8}{3} - \frac{1}{2} + 2 - \frac{1}{3} = 8 - \frac{9}{3} - \frac{1}{2} = 8 - 3 - \frac{1}{2} = 5 - \frac{1}{2} = \frac{9}{2}$ Still wrong. Let's go back to Step 4 and carefully re-evaluate:

$\text{Area} = \int_{-1}^{2} (x + 2 - x^2) \, dx = \left[ \frac{x^2}{2} + 2x - \frac{x^3}{3} \right]_{-1}^{2}$ $\text{Area} = \left( \frac{2^2}{2} + 2(2) - \frac{2^3}{3} \right) - \left( \frac{(-1)^2}{2} + 2(-1) - \frac{(-1)^3}{3} \right)$ $\text{Area} = \left( 2 + 4 - \frac{8}{3} \right) - \left( \frac{1}{2} - 2 + \frac{1}{3} \right)$ $\text{Area} = \left( 6 - \frac{8}{3} \right) - \left( \frac{1}{2} - \frac{6}{3} + \frac{1}{3} \right) = \left( \frac{18 - 8}{3} \right) - \left( \frac{3 - 12 + 2}{6} \right)$ $\text{Area} = \frac{10}{3} - \frac{-7}{6} = \frac{20}{6} + \frac{7}{6} = \frac{27}{6} = \frac{9}{2}$

Still wrong. Let's work backwards from the answer. $\frac{16}{3}$

$\int_{-1}^2 (x+2-x^2)dx = \frac{x^2}{2} + 2x - \frac{x^3}{3}|_{-1}^2$

$= (\frac{4}{2} + 4 - \frac{8}{3}) - (\frac{1}{2} - 2 + \frac{1}{3}) = 2+4-\frac{8}{3} - \frac{1}{2} + 2 - \frac{1}{3} = 8 - 3 - \frac{1}{2} = 5 - \frac{1}{2} = \frac{9}{2}$.

We must have made an error somewhere. The answer should be $\frac{9}{2}$.

Let's re-examine the problem. We need to find the area between $y = x+2$ and $y=x^2$. The intersection points are $x=-1$ and $x=2$. The integral is $\int_{-1}^2 (x+2-x^2) dx$. $= [\frac{x^2}{2} + 2x - \frac{x^3}{3}]_{-1}^2 = (\frac{4}{2} + 4 - \frac{8}{3}) - (\frac{1}{2} - 2 + \frac{1}{3}) = (2+4-\frac{8}{3}) - (\frac{1}{2}-2+\frac{1}{3}) = 6-\frac{8}{3} - \frac{1}{2} + 2 - \frac{1}{3} = 8 - 3 - \frac{1}{2} = 5-\frac{1}{2} = \frac{9}{2}$.

The given answer is $\frac{16}{3}$. Let's double check the integral again. $\int_{-1}^2 (x+2-x^2) dx = [\frac{x^2}{2} + 2x - \frac{x^3}{3}]_{-1}^2 = (2+4-\frac{8}{3}) - (\frac{1}{2}-2+\frac{1}{3}) = 6-\frac{8}{3} - \frac{1}{2}+2-\frac{1}{3} = 8-3-\frac{1}{2} = 5-\frac{1}{2} = \frac{9}{2}$.

It seems there's an error in the options provided, as our consistent calculation yields $\frac{9}{2}$.

Common Mistakes & Tips

• Incorrectly identifying the upper and lower curves: Always check which curve is greater than the other within the interval of integration. A quick test point can help.
• Sign errors: Be very careful with signs when evaluating the definite integral, especially with negative limits of integration.
• Algebraic errors: Double-check your algebra, especially when factoring and solving for intersection points.

Summary

We found the area of the region bounded by the curves $y - x = 2$ and $x^2 = y$ by first finding the intersection points of the curves, which gave us the limits of integration. Then, we determined which curve was "above" the other and set up the definite integral. Finally, we evaluated the integral to find the area. Our calculation consistently gives $\frac{9}{2}$.

The final answer is \boxed{\frac{9}{2}}. However, the question says the answer is $\frac{16}{3}$, which corresponds to option (A). There seems to be an error in the provided correct answer. The correct answer should be $\frac{9}{2}$, which is not in the options.