1. **State the problem:** We are given a curve $y = \frac{9}{\sqrt{5x+4}}$ and a line $y = 6 - 3x$. They intersect at point $P$ where the $y$-coordinate is 3. We need to find the area of the shaded region between the curve and the line above the $x$-axis. 2. **Find the $x$-coordinate of point $P$:** Since $P$ lies on both the curve and the line and has $y=3$, set the expressions equal to 3: $$3 = \frac{9}{\sqrt{5x+4}}$$ Multiply both sides by $\sqrt{5x+4}$: $$3 \sqrt{5x+4} = 9$$ Divide both sides by 3: $$\sqrt{5x+4} = 3$$ Square both sides: $$5x + 4 = 9$$ Solve for $x$: $$5x = 5 \implies x = 1$$ So, point $P$ is at $(1,3)$. 3. **Find the $x$-intercepts of the curve and the line:** - For the curve, set $y=0$: $$\frac{9}{\sqrt{5x+4}} = 0$$ This never equals zero because numerator is 9 and denominator is positive for $x \geq -\frac{4}{5}$. So curve does not cross $x$-axis. - For the line, set $y=0$: $$6 - 3x = 0 \implies 3x = 6 \implies x = 2$$ So the line crosses the $x$-axis at $(2,0)$. 4. **Find the $x$-coordinate where the curve meets the $y$-axis:** At $x=0$: $$y = \frac{9}{\sqrt{5(0)+4}} = \frac{9}{2} = 4.5$$ 5. **Determine the limits of integration:** The shaded region is bounded between $x=0$ (where the curve starts on the $y$-axis) and $x=1$ (point $P$ where the curve and line intersect). 6. **Set up the integral for the area:** Area $= \int_0^1 \left( \text{upper curve} - \text{lower curve} \right) dx$ Between $x=0$ and $x=1$, the curve $y=\frac{9}{\sqrt{5x+4}}$ is above the line $y=6-3x$. So, $$\text{Area} = \int_0^1 \left( \frac{9}{\sqrt{5x+4}} - (6 - 3x) \right) dx$$ 7. **Calculate the integral:** Split the integral: $$\int_0^1 \frac{9}{\sqrt{5x+4}} dx - \int_0^1 (6 - 3x) dx$$ First integral: Let $u = 5x + 4$, then $du = 5 dx$, so $dx = \frac{du}{5}$. When $x=0$, $u=4$; when $x=1$, $u=9$. So, $$\int_0^1 \frac{9}{\sqrt{5x+4}} dx = 9 \int_4^9 u^{-\frac{1}{2}} \frac{du}{5} = \frac{9}{5} \int_4^9 u^{-\frac{1}{2}} du$$ Integrate: $$\int u^{-\frac{1}{2}} du = 2 u^{\frac{1}{2}} + C$$ Evaluate: $$\frac{9}{5} \times 2 \left( \sqrt{9} - \sqrt{4} \right) = \frac{18}{5} (3 - 2) = \frac{18}{5} \times 1 = \frac{18}{5}$$ Second integral: $$\int_0^1 (6 - 3x) dx = \left[6x - \frac{3x^2}{2} \right]_0^1 = (6 \times 1 - \frac{3}{2} \times 1^2) - 0 = 6 - 1.5 = 4.5$$ 8. **Calculate the area:** $$\text{Area} = \frac{18}{5} - 4.5 = 3.6 - 4.5 = -0.9$$ Since area cannot be negative, this means the line is above the curve in this interval. So we take the absolute value or reverse the order: $$\text{Area} = 4.5 - \frac{18}{5} = 4.5 - 3.6 = 0.9$$ **Final answer:** $$\boxed{0.9}$$ This is the area of the shaded region between the curve and the line above the $x$-axis from $x=0$ to $x=1$.