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 $\int_a^b [f(x) - g(x)] \, dx$.
• Definite Integral: The definite integral $\int_a^b f(x) \, dx$ represents the signed area under the curve $y = f(x)$ from $x = a$ to $x = b$.
• Finding Intersection Points: To find where two curves intersect, set their equations equal to each other and solve for $x$.

Step-by-Step Solution

Step 1: Understanding the Region

We are asked to find the area of the region $A$ defined by the inequalities $0 \le x \le 3$, $0 \le y \le 4$, and $y \le x^2 + 3x$. This region is bounded by the x-axis, the y-axis, the line $x=3$, the line $y=4$, and the parabola $y = x^2 + 3x$. We need to find the area of the region enclosed by these boundaries.

Step 2: Analyzing the Boundaries

The inequalities define a rectangular region with vertices at (0,0), (3,0), (3,4), and (0,4). The parabola $y = x^2 + 3x$ intersects the x-axis at $x=0$ and $x=-3$. Since we are only concerned with $0 \le x \le 3$, the relevant intersection point is at the origin (0,0). The parabola opens upwards. We need to determine where the parabola intersects the line $y=4$.

Step 3: Finding the Intersection of the Parabola and the Line y=4

To find the intersection points, we set $x^2 + 3x = 4$. This gives us the quadratic equation $x^2 + 3x - 4 = 0$. Factoring, we get $(x+4)(x-1) = 0$, so $x = -4$ or $x = 1$. Since $0 \le x \le 3$, the relevant intersection point is at $x=1$. The corresponding y-value is $y=4$.

Step 4: Setting up the Integrals

Since the parabola $y = x^2 + 3x$ is below the line $y = 4$ for $0 \le x \le 1$, and the line $y=4$ is below the parabola for $1 \le x \le 3$, we need to split the integral into two parts. From $x=0$ to $x=1$, the upper bound is $y = x^2 + 3x$ and the lower bound is $y=0$. From $x=1$ to $x=3$, the upper bound is $y=4$ and the lower bound is $y=0$. Therefore, the area is given by:

$A = \int_0^1 (x^2 + 3x) \, dx + \int_1^3 4 \, dx$

Step 5: Evaluating the Integrals

First Integral: $\int_0^1 (x^2 + 3x) \, dx = \left[ \frac{x^3}{3} + \frac{3x^2}{2} \right]_0^1 = \left( \frac{1}{3} + \frac{3}{2} \right) - (0) = \frac{2}{6} + \frac{9}{6} = \frac{11}{6}$

Second Integral: $\int_1^3 4 \, dx = [4x]_1^3 = 4(3) - 4(1) = 12 - 4 = 8$

Step 6: Calculating the Total Area

Adding the two integrals, we get: $A = \frac{11}{6} + 8 = \frac{11}{6} + \frac{48}{6} = \frac{59}{6}$

Common Mistakes & Tips

• Sketching: Always sketch the region to visualize the boundaries and intersection points.
• Correct Bounds: Ensure the limits of integration are correct based on the intersection points.
• Identify Upper and Lower Functions: Be careful to identify the upper and lower functions correctly in each interval.

Summary

The area of the region A is found by splitting the integral into two parts, based on the intersection of the parabola and the line $y=4$. Evaluating each integral and summing them gives the total area. The area is $\frac{59}{6}$ sq. units.

Final Answer

The final answer is \boxed{\frac{59}{6}}, which corresponds to option (A).