To set up the integral to find the area of the region bounded by the functions $f(x) = x^2 - 9x + 20$ and $g(x) = x + 75$, and below by the $x$-axis, we need to follow these steps:

1. Express $f(x)$ and $g(x)$ in terms of $y$.

   Since we want to integrate with respect to $y$, we need to solve the functions for $x$:

   • For $f(x) = x^2 - 9x + 20$:

     Set $y = x^2 - 9x + 20$:

     $y = x^2 - 9x + 20 \Rightarrow x^2 - 9x + (20 - y) = 0$

     This is a quadratic equation in $x$. We can use the quadratic formula to solve for $x$:

     $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

     Here, $a = 1$, $b = -9$, and $c = 20 - y$:

     $x = \frac{-(-9) \pm \sqrt{(-9)^2 - 4(1)(20 - y)}}{2(1)}$ $x = \frac{9 \pm \sqrt{81 - 80 + 4y}}{2}$ $x = \frac{9 \pm \sqrt{1 + 4y}}{2}$

     Therefore, $x$ can be expressed as two functions of $y$:

     $x_1(y) = \frac{9 + \sqrt{1 + 4y}}{2} \quad \text{and} \quad x_2(y) = \frac{9 - \sqrt{1 + 4y}}{2}$

   • For $g(x) = x + 75$:

     Set $y = x + 75$:

     $x = y - 75$

2. Find the points of intersection in terms of $y$.

   To find the region bounded by $f(x)$ and $g(x)$, we set the equations equal to each other:

   $x^2 - 9x + 20 = x + 75$ $x^2 - 10x - 55 = 0$

   Solving this quadratic equation:

   $x = \frac{-(-10) \pm \sqrt{(-10)^2 - 4(1)(-55)}}{2(1)}$ $x = \frac{10 \pm \sqrt{100 + 220}}{2}$ $x = \frac{10 \pm \sqrt{320}}{2}$ $x = \frac{10 \pm 8\sqrt{5}}{2}$ $x = 5 \pm 4\sqrt{5}$

   These give the $x$-values of the intersections. Plugging these back into $g(x) = x + 75$ gives the $y$-values at the points of intersection:

   $y_1 = (5 + 4\sqrt{5}) + 75$ $y_2 = (5 - 4\sqrt{5}) + 75$

   So, $y_1 = 80 + 4\sqrt{5}$ and $y_2 = 80 - 4\sqrt{5}$.

3. Set up the integral to find the area with respect to $y$.

   The area is bounded above by $f(x)$ and $g(x)$ and below by the $x$-axis. We use $x_1(y)$ and $x_2(y)$ to represent the rightmost and leftmost $x$-values, respectively, for the quadratic function, and $x = y - 75$ for the linear function.

   The integral for the area is:

   $\text{Area} = \int_{y_2}^{y_1} \left( x_1(y) - x_2(y) \right) \, dy$

   Substitute $x_1(y) = \frac{9 + \sqrt{1 + 4y}}{2}$ and $x_2(y) = y - 75$:

   $\text{Area} = \int_{80 - 4\sqrt{5}}^{80 + 4\sqrt{5}} \left( \frac{9 + \sqrt{1 + 4y}}{2} - (y - 75) \right) \, dy$

This integral represents the area of the region bounded above by the functions $f(x)$ and $g(x)$ and bounded below by the $x$-axis when integrating with respect to $y$.

Would you like further details or have any questions?

Here are 5 related questions:

1. How would you set up the integral if the region were bounded by different functions?
2. How can you solve for the points of intersection of more complex functions?
3. What is the process for finding the area bounded by curves using vertical slices?
4. How does changing the bounds of integration affect the area calculated?
5. What are some real-world applications of finding areas between curves?

Tip: When dealing with areas between curves, always check which function is above or below over the given interval to set up the correct integral.