1. **State the problem:** Find the area of the shaded region enclosed by the curve $y = x^3 - 7x$ and the line $y = 2x$ between points $O(0,0)$ and $A(3,6)$. 2. **Find the points of intersection:** Given points $O(0,0)$ and $A(3,6)$ are intersections of the curve and the line. 3. **Set up the integral for the area:** The area between two curves $y=f(x)$ and $y=g(x)$ from $x=a$ to $x=b$ is $$\text{Area} = \int_a^b |f(x) - g(x)| \, dx$$ Here, $f(x) = x^3 - 7x$ and $g(x) = 2x$. 4. **Determine which function is on top between $0$ and $3$:** Calculate $f(1) = 1 - 7 = -6$ and $g(1) = 2$ so $g(x) > f(x)$ in this interval. 5. **Write the integral for the shaded area:** $$\text{Area} = \int_0^3 (2x - (x^3 - 7x)) \, dx = \int_0^3 (2x - x^3 + 7x) \, dx = \int_0^3 (-x^3 + 9x) \, dx$$ 6. **Integrate:** $$\int_0^3 (-x^3 + 9x) \, dx = \left[-\frac{x^4}{4} + \frac{9x^2}{2}\right]_0^3$$ 7. **Evaluate the definite integral:** $$= \left(-\frac{3^4}{4} + \frac{9 \times 3^2}{2}\right) - \left(-\frac{0}{4} + 0\right) = \left(-\frac{81}{4} + \frac{81}{2}\right) = -20.25 + 40.5 = 20.25$$ 8. **Final answer:** The area of the shaded region is $20.25$ square units. --- **(i) Show that the coordinates of B are $(\sqrt{3}, -4\sqrt{3})$** 1. The tangent to the curve $y = x^3 - 7x$ at point $B$ is parallel to the line $y=2x$, so the slope at $B$ is 2. 2. Find the derivative: $$y' = 3x^2 - 7$$ 3. Set derivative equal to 2: $$3x^2 - 7 = 2 \implies 3x^2 = 9 \implies x^2 = 3 \implies x = \sqrt{3}$$ 4. Find $y$ at $x=\sqrt{3}$: $$y = (\sqrt{3})^3 - 7(\sqrt{3}) = 3\sqrt{3} - 7\sqrt{3} = -4\sqrt{3}$$ 5. So, $B = (\sqrt{3}, -4\sqrt{3})$. --- **(ii) Find the exact area of triangle $AOB$** 1. Coordinates: $A = (3,6)$, $O = (0,0)$, $B = (\sqrt{3}, -4\sqrt{3})$ 2. Area of triangle with vertices $(x_1,y_1)$, $(x_2,y_2)$, $(x_3,y_3)$ is $$\text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right|$$ 3. Substitute: $$= \frac{1}{2} |0(6 + 4\sqrt{3}) + 3(-4\sqrt{3} - 0) + \sqrt{3}(0 - 6)|$$ $$= \frac{1}{2} |0 - 12\sqrt{3} - 6\sqrt{3}| = \frac{1}{2} |-18\sqrt{3}| = 9\sqrt{3}$$ 4. Final answer: The exact area of triangle $AOB$ is $9\sqrt{3}$ square units.