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 $A(3,6)$. Also, (i) show that the coordinates of point $B$ on the curve where the tangent is parallel to $y=2x$ and $x>0$ are $(\sqrt{3}, -4\sqrt{3})$. (ii) Find the exact area of triangle $AOB$. --- 2. **Find the area of the shaded region:** The area between two curves $y=f(x)$ and $y=g(x)$ from $x=a$ to $x=b$ is given by: $$\text{Area} = \int_a^b |f(x) - g(x)| \, dx$$ Here, $f(x) = x^3 - 7x$ and $g(x) = 2x$. 3. **Find points of intersection:** Given $O(0,0)$ and $A(3,6)$ are intersections. 4. **Determine which curve is on top between $0$ and $3$:** Calculate $f(x) - g(x) = (x^3 - 7x) - 2x = x^3 - 9x$. Check sign at $x=1$: $1 - 9 = -8 < 0$, so $g(x) = 2x$ is above $f(x)$ between $0$ and $3$. 5. **Set up the integral for 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. **Calculate the integral:** $$\int_0^3 (-x^3 + 9x) \, dx = \left[-\frac{x^4}{4} + \frac{9x^2}{2}\right]_0^3 = \left[-\frac{81}{4} + \frac{81}{2}\right] - 0 = -20.25 + 40.5 = 20.25$$ So, the area of the shaded region is $20.25$ square units or $\frac{81}{4}$. --- 7. **(i) Find coordinates of point B where tangent is parallel to $y=2x$:** The slope of $y=2x$ is $2$. The derivative of $y = x^3 - 7x$ is: $$y' = 3x^2 - 7$$ Set $y' = 2$ to find $x$: $$3x^2 - 7 = 2$$ $$3x^2 = 9$$ $$x^2 = 3$$ $$x = \pm \sqrt{3}$$ Since $x$ is positive, $x = \sqrt{3}$. Find $y$ at $x=\sqrt{3}$: $$y = (\sqrt{3})^3 - 7(\sqrt{3}) = 3\sqrt{3} - 7\sqrt{3} = -4\sqrt{3}$$ Thus, $B = (\sqrt{3}, -4\sqrt{3})$. --- 8. **(ii) Find area of triangle $AOB$:** Points: $O = (0,0)$ $A = (3,6)$ $B = (\sqrt{3}, -4\sqrt{3})$ Area of triangle with vertices $(x_1,y_1)$, $(x_2,y_2)$, $(x_3,y_3)$ is: $$\text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right|$$ Substitute: $$= \frac{1}{2} |0(6 + 4\sqrt{3}) + 3(-4\sqrt{3} - 0) + \sqrt{3}(0 - 6)|$$ $$= \frac{1}{2} |0 - 12\sqrt{3} - 6\sqrt{3}| = \frac{1}{2} |-18\sqrt{3}| = 9\sqrt{3}$$ --- **Final answers:** - Area of shaded region = $\frac{81}{4}$ square units. - Coordinates of $B$ are $(\sqrt{3}, -4\sqrt{3})$. - Area of triangle $AOB$ = $9\sqrt{3}$ square units.