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To determine which graph corresponds to the area enclosed by the curves \( y = 3x^2 \) and \( y = x^2 + 5 \), let's first analyze their intersection points and the general shape of the curves.

### Step 1: Set the equations equal to each other to find intersection points.

\[ 3x^2 = x^2 + 5 \]

Simplifying:

\[ 3x^2 - x^2 = 5 \]

\[ 2x^2 = 5 \]

\[ x^2 = \frac{5}{2} \]

\[ x = \pm \sqrt{\frac{5}{2}} \approx \pm 1.58 \]

So the curves intersect approximately at \( x = -1.58 \) and \( x = 1.58 \).

### Step 2: Sketching the curves
- The curve \( y = 3x^2 \) is a narrower parabola that opens upwards.
- The curve \( y = x^2 + 5 \) is a wider parabola, also opening upwards, but it is shifted 5 units up.

### Step 3: Graph identification
Given this information, the region enclosed by the two curves will be between the intersection points \( x \approx -1.58 \) and \( x \approx 1.58 \). The upper curve is \( y = x^2 + 5 \), and the lower curve is \( y = 3x^2 \), which matches the second graph (the one where the shaded region is between two parabolas at approximately \( x = \pm 1.6 \)).

Thus, **the second graph** corresponds to the area enclosed by the curves.

### Step 4: Find the Area
To find the area between the curves, we integrate the difference between the top function \( y = x^2 + 5 \) and the bottom function \( y = 3x^2 \) over the interval \([- \sqrt{\frac{5}{2}}, \sqrt{\frac{5}{2}}]\):

The area is given by:

\[
A = \int_{-\sqrt{\frac{5}{2}}}^{\sqrt{\frac{5}{2}}} \left[ (x^2 + 5) - 3x^2 \right] dx = \int_{-\sqrt{\frac{5}{2}}}^{\sqrt{\frac{5}{2}}} (5 - 2x^2) dx
\]

Since the function is even, we can double the integral from 0 to \( \sqrt{\frac{5}{2}} \):

\[
A = 2 \int_{0}^{\sqrt{\frac{5}{2}}} (5 - 2x^2) dx
\]

I'll compute this area next.The area of the region enclosed by the curves \( y = 3x^2 \) and \( y = x^2 + 5 \) is approximately **10.54 square units**.

Would you like further clarification or details on any part of the solution?

Here are some related questions you might find helpful:
1. How do you find the intersection points of two curves algebraically?
2. How can you determine which function is on top when calculating areas between curves?
3. What does it mean for a function to be "even" in the context of integration?
4. Can areas between curves be calculated if the curves intersect more than twice?
5. How does shifting a parabola vertically affect the enclosed area?

**Tip:** When calculating the area between curves, always check which curve is on top in the region of interest, as this dictates the order of subtraction for the functions.

Determine which graph corresponds to the area enclosed by the curves y = 3x^2 and y = x^2 + 5. Find the area of the region between y = 3x^2 and y = x^2 + 5.

Learn how to determine the area enclosed between the curves y = 3x^2 and y = x^2 + 5. This solution covers finding intersection points, integrating the difference between the functions, and utilizing properties of even functions for simplification. Suitable for college-level calculus students.

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