1. **State the problem:** We are given two functions $f(x) = x^2 - 7x + 12$ and $g(x) = \frac{3x}{4} + \frac{21}{4}$. We want to write the integral with respect to $y$ to find the area of the region bounded above by $f(x)$ and $g(x)$ and bounded below by the $x$-axis. 2. **Understand the problem:** The region is bounded below by the $x$-axis ($y=0$), and above by the curves $f(x)$ and $g(x)$. Since we integrate with respect to $y$, we need to express $x$ as a function of $y$ for both curves. 3. **Rewrite the functions in terms of $y$:** - For the line $g(x)$: $$y = \frac{3x}{4} + \frac{21}{4} \implies y - \frac{21}{4} = \frac{3x}{4} \implies x = \frac{4}{3}\left(y - \frac{21}{4}\right) = \frac{4y}{3} - 7$$ - For the parabola $f(x)$: $$y = x^2 - 7x + 12$$ Rewrite as: $$x^2 - 7x + (12 - y) = 0$$ 4. **Solve for $x$ in terms of $y$ using the quadratic formula:** $$x = \frac{7 \pm \sqrt{(-7)^2 - 4 \cdot 1 \cdot (12 - y)}}{2} = \frac{7 \pm \sqrt{49 - 48 + 4y}}{2} = \frac{7 \pm \sqrt{1 + 4y}}{2}$$ 5. **Determine the relevant branch of the parabola:** Since the parabola opens upward and the region is between $x=2$ and $x=6$, the left branch corresponds to the minus sign: $$x = \frac{7 - \sqrt{1 + 4y}}{2}$$ 6. **Set the limits for $y$:** The region is bounded below by $y=0$ (the $x$-axis) and above by the intersection of $f(x)$ and $g(x)$. 7. **Find the intersection points of $f(x)$ and $g(x)$ to find the upper limit for $y$:** Set $f(x) = g(x)$: $$x^2 - 7x + 12 = \frac{3x}{4} + \frac{21}{4}$$ Multiply both sides by 4: $$4x^2 - 28x + 48 = 3x + 21$$ Rearrange: $$4x^2 - 31x + 27 = 0$$ 8. **Solve quadratic for $x$:** $$x = \frac{31 \pm \sqrt{31^2 - 4 \cdot 4 \cdot 27}}{2 \cdot 4} = \frac{31 \pm \sqrt{961}}{8} = \frac{31 \pm 31}{8}$$ So $x= \frac{31 - 31}{8} = 0$ or $x= \frac{31 + 31}{8} = \frac{62}{8} = 7.75$. 9. **Find corresponding $y$ values at these $x$ values:** For $x=0$: $$y = f(0) = 0^2 - 7 \cdot 0 + 12 = 12$$ For $x=7.75$: $$y = f(7.75) = (7.75)^2 - 7 \cdot 7.75 + 12 = 60.0625 - 54.25 + 12 = 17.8125$$ 10. **Check the region's $y$ bounds:** The graph description says the region is between approximately $y=4$ and $y=6$, so we consider the intersection points relevant to that range. 11. **Write the integral with respect to $y$:** The horizontal slices run from the parabola (left) to the line (right), so the length of each slice is: $$x_{right} - x_{left} = \left(\frac{4y}{3} - 7\right) - \frac{7 - \sqrt{1 + 4y}}{2}$$ 12. **Integral for the area:** $$\text{Area} = \int_{y=0}^{y=6} \left[ \left(\frac{4y}{3} - 7\right) - \frac{7 - \sqrt{1 + 4y}}{2} \right] dy$$ **Final answer:** $$\boxed{\int_0^6 \left( \frac{4y}{3} - 7 - \frac{7}{2} + \frac{\sqrt{1 + 4y}}{2} \right) dy}$$ or equivalently $$\int_0^6 \left( \frac{4y}{3} - \frac{21}{2} + \frac{\sqrt{1 + 4y}}{2} \right) dy$$