1. **State the problem:** We are given three functions: $$y = f(x) = -x^2 + 6x,$$ $$y = g(x) = x^2 - 6x + 10,$$ $$y = x.$$ We want to find the values of $a, b, c$ and coefficients $p, q, r, s$ such that the shaded area $S$ is expressed as $$S = \int_a^b (px^2 + qx + r) \, dx + \int_b^c (-x^2 + sx) \, dx.$$ 2. **Find the intersection points $a, b, c$: ** - Find intersections of $f(x)$ and $g(x)$: $$-x^2 + 6x = x^2 - 6x + 10$$ $$\Rightarrow -x^2 + 6x - x^2 + 6x - 10 = 0$$ $$\Rightarrow -2x^2 + 12x - 10 = 0$$ Divide both sides by $-2$: $$\cancel{-2}x^2 + \cancel{-12}x + 5 = 0$$ $$x^2 - 6x + 5 = 0$$ Factor: $$(x - 5)(x - 1) = 0$$ So, $$x = 1, 5$$ - Find intersections of $g(x)$ and $y = x$: $$x^2 - 6x + 10 = x$$ $$x^2 - 7x + 10 = 0$$ Factor: $$(x - 5)(x - 2) = 0$$ So, $$x = 2, 5$$ - Find intersections of $f(x)$ and $y = x$: $$-x^2 + 6x = x$$ $$-x^2 + 5x = 0$$ $$x(-x + 5) = 0$$ So, $$x = 0, 5$$ 3. **Determine the order of $a, b, c$: ** From the graph and intersections, the shaded region is between $x=1$ and $x=5$, with a change at $x=2$ where the bounding curves switch. Thus, $$a = 1, b = 2, c = 5$$ 4. **Identify the functions forming the area in each interval:** - For $x \in [1, 2]$, the upper curve is $g(x) = x^2 - 6x + 10$ and the lower curve is $y = x$. - For $x \in [2, 5]$, the upper curve is $f(x) = -x^2 + 6x$ and the lower curve is $y = x$. 5. **Express the integrand in the form $px^2 + qx + r$: ** - For $x \in [1, 2]$, the integrand is upper minus lower: $$g(x) - x = (x^2 - 6x + 10) - x = x^2 - 7x + 10$$ So, $$p = 1, q = -7, r = 10$$ - For $x \in [2, 5]$, the integrand is upper minus lower: $$f(x) - x = (-x^2 + 6x) - x = -x^2 + 5x$$ So, $$s = 5$$ 6. **Final answers:** $$a = 1, b = 2, c = 5$$ $$p = 1, q = -7, r = 10, s = 5$$