Consider the area between the graphs x+5y=9
and x+5=y2
. This area can be computed in two different ways using integrals.

One way it can be computed is as a sum of two integrals

∫baf(x)dx+∫cbg(x)dx(where a<b<c)

with the following values:
a=
-5
b=
-1
c=

44
f(x)=
g(x)=
find f(x) and g(x)

To find \( f(x) \) and \( g(x) \) in the given integrals, we first need to understand the equations of the graphs and how they relate to the area between them. The two given curves are:

1. \( x + 5y = 9 \)
2. \( x + 5 = y^2 \)

### Step 1: Rewrite the Equations
First, let's express \( y \) in terms of \( x \) for both equations.

**For the first equation:**

\[
x + 5y = 9
\]
\[
5y = 9 - x
\]
\[
y = \frac{9 - x}{5}
\]

**For the second equation:**

\[
x + 5 = y^2
\]
\[
y^2 = x + 5
\]
\[
y = \pm \sqrt{x + 5}
\]

### Step 2: Determine the Boundaries and Intersection Points
Given that the limits of integration are \( a = -5 \), \( b = -1 \), and \( c = 44 \), we can find the intersections of the curves to determine the structure of the integrals.

**Intersection Points:**

We find the intersection by setting the two expressions for \( y \) equal to each other:

\[
\frac{9 - x}{5} = \sqrt{x + 5}
\]

Square both sides to eliminate the square root:

\[
\left(\frac{9 - x}{5}\right)^2 = x + 5
\]
\[
\frac{(9 - x)^2}{25} = x + 5
\]
\[
(9 - x)^2 = 25(x + 5)
\]
\[
81 - 18x + x^2 = 25x + 125
\]
\[
x^2 - 43x - 44 = 0
\]

Solving this quadratic equation:

\[
x = \frac{43 \pm \sqrt{43^2 + 4 \times 44}}{2}
\]
\[
x = \frac{43 \pm \sqrt{1849 + 176}}{2}
\]
\[
x = \frac{43 \pm \sqrt{2025}}{2}
\]
\[
x = \frac{43 \pm 45}{2}
\]

This gives \( x = 44 \) and \( x = -1 \). So the curves intersect at \( x = -1 \) and \( x = 44 \).

### Step 3: Set up the Integrals
Given the intersection points and limits, we can set up the integrals for the area between the curves:

1. From \( x = -5 \) to \( x = -1 \), the area is given by the integral of the curve \( x + 5 = y^2 \) (which is the lower curve in this interval) subtracted from the curve \( x + 5y = 9 \) (which is the upper curve in this interval).

2. From \( x = -1 \) to \( x = 44 \), the area is given by the integral of the curve \( x + 5y = 9 \) (which is the upper curve in this interval) subtracted from the curve \( x + 5 = y^2 \) (which is the lower curve in this interval).

### Final Forms of \( f(x) \) and \( g(x) \):

1. **For \( f(x) \):**
   - In the interval \([-5, -1]\), the function \( f(x) \) represents the difference between the upper and lower curves, so:
   \[
   f(x) = \frac{9 - x}{5} - (-\sqrt{x + 5}) = \frac{9 - x}{5} + \sqrt{x + 5}
   \]

2. **For \( g(x) \):**
   - In the interval \([-1, 44]\), the function \( g(x) \) represents the difference between the upper and lower curves, so:
   \[
   g(x) = \sqrt{x + 5} - \frac{9 - x}{5}
   \]

Thus, the functions \( f(x) \) and \( g(x) \) are:

\[
f(x) = \frac{9 - x}{5} + \sqrt{x + 5}
\]

\[
g(x) = \sqrt{x + 5} - \frac{9 - x}{5}
\]

Would you like more details or have any questions?

### Follow-Up Questions:
1. How do you solve a quadratic equation to find intersection points of curves?
2. Why is it important to determine which curve is above or below the other?
3. What does the area between two curves represent geometrically?
4. How can you confirm the intersection points of two curves graphically?
5. What other methods can be used to find the area between two curves?

### Tip:
When working with areas between curves, always double-check the intervals where one curve is above the other, as this will affect the setup of your integral.

Learn how to compute the area between the curves x + 5y = 9 and x + 5 = y^2 using integrals. Understand the method of breaking down the area into two integrals over different intervals. Suitable for advanced high school students and beyond.

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