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 where two curves intersect, set their equations equal to each other and solve for $x$.
• Definite Integration: The definite integral $\int_a^b f(x) \, dx$ represents the signed area under the curve $y = f(x)$ from $x = a$ to $x = b$.

Step-by-Step Solution

Step 1: Identify the curves and the region.

We are given the region defined by $x \ge 0$, $2x^2 \le y \le 4 - 2x$. This means we need to find the area between the curves $y = 2x^2$ and $y = 4 - 2x$ in the first quadrant ($x \ge 0$). $y = 2x^2$ is a parabola opening upwards, and $y = 4 - 2x$ is a straight line with a negative slope.

Step 2: Find the intersection points of the curves.

To find where the curves $y = 2x^2$ and $y = 4 - 2x$ intersect, we set them equal to each other: $2x^2 = 4 - 2x$ $2x^2 + 2x - 4 = 0$ $x^2 + x - 2 = 0$ $(x + 2)(x - 1) = 0$ This gives us $x = -2$ and $x = 1$. Since we are given $x \ge 0$, we only consider the intersection point at $x = 1$. The curves intersect at $x=1$. When $x=1$, $y = 2(1)^2 = 2$, so the intersection point is $(1, 2)$.

Step 3: Determine the limits of integration.

Since $x \ge 0$, our lower limit of integration is $x = 0$. The intersection point we found in Step 2 gives us the upper limit of integration, $x = 1$.

Step 4: Determine which curve is above the other in the interval $[0, 1]$.

We need to determine which function is greater than the other in the interval $[0, 1]$. Let's test a value within the interval, say $x = 0.5$:

• $y = 2x^2 = 2(0.5)^2 = 2(0.25) = 0.5$
• $y = 4 - 2x = 4 - 2(0.5) = 4 - 1 = 3$

Since $3 > 0.5$, the line $y = 4 - 2x$ is above the parabola $y = 2x^2$ in the interval $[0, 1]$.

Step 5: Set up and evaluate the integral.

The area between the curves is given by: $A = \int_0^1 [(4 - 2x) - (2x^2)] \, dx$ $A = \int_0^1 (4 - 2x - 2x^2) \, dx$ $A = \left[ 4x - x^2 - \frac{2}{3}x^3 \right]_0^1$ $A = \left( 4(1) - (1)^2 - \frac{2}{3}(1)^3 \right) - \left( 4(0) - (0)^2 - \frac{2}{3}(0)^3 \right)$ $A = 4 - 1 - \frac{2}{3} - 0$ $A = 3 - \frac{2}{3}$ $A = \frac{9}{3} - \frac{2}{3}$ $A = \frac{7}{3}$

Step 6: Verify the intersection with $x=0$

We need to check the region where $x=0$. When $x=0$, $y=2x^2 = 0$ and $y=4-2x = 4$. Therefore, the region is bounded between $y=0$ and $y=4$ when $x=0$.

Step 7: Recalculate the area

There was an error in Step 5. The area is given by: $A = \int_0^1 [(4 - 2x) - (2x^2)] \, dx$ $A = \int_0^1 (4 - 2x - 2x^2) \, dx$ $A = \left[ 4x - x^2 - \frac{2}{3}x^3 \right]_0^1$ $A = \left( 4(1) - (1)^2 - \frac{2}{3}(1)^3 \right) - \left( 4(0) - (0)^2 - \frac{2}{3}(0)^3 \right)$ $A = 4 - 1 - \frac{2}{3} - 0$ $A = 3 - \frac{2}{3}$ $A = \frac{9}{3} - \frac{2}{3}$ $A = \frac{7}{3}$

I was wrong. The correct answer is 7/3.

Common Mistakes & Tips

• Incorrect Limits of Integration: Carefully determine the points of intersection and the region of interest to set the correct limits for the integral.
• Incorrect Order of Subtraction: Always subtract the lower curve from the upper curve. If you get a negative area, you've likely reversed the order.
• Sign Errors: Be meticulous with signs during integration and evaluation.

Summary

To find the area of the region bounded by $y = 2x^2$ and $y = 4 - 2x$ for $x \ge 0$, we first found the intersection points of the curves. Then, we determined which curve was above the other in the interval defined by the intersection point and the constraint $x \ge 0$. Finally, we set up and evaluated the definite integral of the difference between the upper and lower curves to find the area. The area is $\frac{7}{3}$ square units.

Final Answer

The final answer is \boxed{7/3}, which corresponds to option (D).