Let’s solve this problem step by step. The goal is to determine the area of the shaded region by setting up and evaluating an integral.

The curve is given as $y = 5x - x^2$, and the line is $y = x$. The region is bounded below by the curve and above by the line.

Steps to Solve

1. Find the points of intersection between the curve $y = 5x - x^2$ and the line $y = x$: $5x - x^2 = x$ Rearrange into standard quadratic form: $-x^2 + 4x = 0 \quad \Rightarrow \quad x(-x + 4) = 0$ So, $x = 0$ and $x = 4$.

2. Set up the integral for the area: The area of the shaded region is the integral of the difference between the top function (line $y = x$) and the bottom function (curve $y = 5x - x^2$), over the interval $[0, 4]$: $\text{Area} = \int_{0}^{4} \left[ (x) - (5x - x^2) \right] dx$

3. Simplify the integrand: $\text{Integrand} = x - (5x - x^2) = x - 5x + x^2 = x^2 - 4x$

4. Evaluate the integral: $\text{Area} = \int_{0}^{4} (x^2 - 4x) dx$ Split the integral: $\text{Area} = \int_{0}^{4} x^2 dx - \int_{0}^{4} 4x dx$ Compute each term:

   • For $\int_{0}^{4} x^2 dx$: $\int x^2 dx = \frac{x^3}{3} \quad \Rightarrow \quad \left[ \frac{x^3}{3} \right]_{0}^{4} = \frac{4^3}{3} - \frac{0^3}{3} = \frac{64}{3}$
   • For $\int_{0}^{4} 4x dx$: $\int 4x dx = 2x^2 \quad \Rightarrow \quad \left[ 2x^2 \right]_{0}^{4} = 2(4^2) - 2(0^2) = 32$

5. Combine the results: $\text{Area} = \frac{64}{3} - 32 = \frac{64}{3} - \frac{96}{3} = -\frac{32}{3}$

The shaded area is $\frac{32}{3}$ units squared.

Let me know if you'd like details on any of the above!

Here are some related questions for practice:

1. How do you calculate the area between two curves in general?
2. What happens if the intersection points change?
3. How would you set up the integral if $y = -x + 6$ replaced $y = x$?
4. Can you verify the solution using a numerical approximation method?
5. How would this problem change if the region were rotated around an axis?

Tip: Always sketch the curves and intersections before setting up integrals for bounded regions.