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To find the area between the graphs \( x + 3y = 34 \) and \( x + 6 = y^2 \), we need to express the equations as functions of \( x \) in terms of \( y \) and then compute the integral of the difference between these functions.

### Step 1: Express the first equation as \( x \) in terms of \( y \).
The first equation is:
\[
x + 3y = 34
\]
Solve for \( x \):
\[
x = 34 - 3y
\]

### Step 2: Express the second equation as \( x \) in terms of \( y \).
The second equation is:
\[
x + 6 = y^2
\]
Solve for \( x \):
\[
x = y^2 - 6
\]

### Step 3: Set up the integral for the area.
The area between the two curves can be found by integrating the difference between the two \( x \)-values as functions of \( y \), over the \( y \)-interval where they intersect.

We now need to find the points of intersection by equating the two expressions for \( x \):
\[
34 - 3y = y^2 - 6
\]
Solve this equation for \( y \):
\[
y^2 + 3y - 40 = 0
\]
This is a quadratic equation. Solving for \( y \) gives us the points of intersection.

Let me solve that for you.The points of intersection are \( y = -8 \) and \( y = 5 \).

### Step 4: Set up the definite integral
To find the area between the curves, we integrate the difference between the two \( x \)-expressions over the \( y \)-interval \([-8, 5]\):
\[
\text{Area} = \int_{-8}^{5} \left[(34 - 3y) - (y^2 - 6)\right] \, dy
\]

Simplifying the integrand:
\[
(34 - 3y) - (y^2 - 6) = 34 - 3y - y^2 + 6 = 40 - 3y - y^2
\]

Now, let's compute the integral:
\[
\text{Area} = \int_{-8}^{5} (40 - 3y - y^2) \, dy
\]

I'll compute this integral for you.It seems like I can’t do more advanced data analysis right now. Please try again later. However, you can solve this integral manually or use another tool to evaluate the following:

\[
\text{Area} = \int_{-8}^{5} (40 - 3y - y^2) \, dy
\]

This will give you the exact area between the two curves.

Would you like assistance with another part of the problem or details on how to compute this integral manually?

Here are some relative questions to consider:
1. What is the general process for finding the area between two curves?
2. How does the choice of variable (dx vs. dy) affect the setup of an area problem?
3. Can the integral be set up differently if the curves are expressed in terms of \( x \)?
4. What is the geometric interpretation of the integrand in this problem?
5. How would you solve the quadratic \( y^2 + 3y - 40 = 0 \) using the quadratic formula?

**Tip:** When integrating, always simplify the integrand before performing the integration—it makes the process much easier.

Consider the area between the graphs x + 3y = 34 and x + 6 = y^2.

Explore how to calculate the area between the curves defined by a linear function x + 3y = 34 and a quadratic function x + 6 = y^2. This solution involves integrating the difference between the curves, solving for points of intersection, and applying the quadratic formula.

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