1. **State the problem:** (a) Given that $$\int_a^{2a} (10 - 6x) \, dx = 1,$$ find the two possible values of $$a$$. (b) Sketch the graph of $$y = 10 - 6x$$ for $$0 \leq x \leq 2$$ with labeled axes and intercepts. (c) Explain why the larger value of $$a$$ found in (a) does not produce an actual area of 1 under the graph between $$a$$ and $$2a$$, and state whether the area is greater or smaller than 1. 2. **Solve the integral in (a):** Calculate the definite integral: $$\int_a^{2a} (10 - 6x) \, dx = \left[10x - 3x^2\right]_a^{2a} = \left(10(2a) - 3(2a)^2\right) - \left(10a - 3a^2\right)$$ Simplify: $$= (20a - 12a^2) - (10a - 3a^2) = 20a - 12a^2 - 10a + 3a^2 = 10a - 9a^2$$ Set equal to 1: $$10a - 9a^2 = 1$$ Rewrite as quadratic: $$9a^2 - 10a + 1 = 0$$ 3. **Solve quadratic equation:** Use quadratic formula: $$a = \frac{10 \pm \sqrt{(-10)^2 - 4 \times 9 \times 1}}{2 \times 9} = \frac{10 \pm \sqrt{100 - 36}}{18} = \frac{10 \pm \sqrt{64}}{18} = \frac{10 \pm 8}{18}$$ Two solutions: $$a_1 = \frac{10 + 8}{18} = \frac{18}{18} = 1$$ $$a_2 = \frac{10 - 8}{18} = \frac{2}{18} = \frac{1}{9}$$ 4. **Sketch the graph in (b):** - The function is $$y = 10 - 6x$$. - At $$x=0$$, $$y=10$$ (y-intercept). - At $$x=2$$, $$y = 10 - 6 \times 2 = 10 - 12 = -2$$. - The graph is a straight line decreasing from (0,10) to (2,-2). - Label axes and intercepts accordingly. 5. **Explain the area discrepancy in (c):** - The integral calculates the net area, which can be positive or negative depending on the function's sign. - For $$a=1$$, the interval is from 1 to 2, where $$y = 10 - 6x$$ is positive at 1 ($$y=4$$) but negative at 2 ($$y=-2$$), so the integral sums positive and negative areas. - The integral equals 1, but the actual geometric area (absolute area) under the curve between 1 and 2 is larger than 1 because the negative part subtracts from the total. - For $$a=\frac{1}{9}$$, the interval is from $$\frac{1}{9}$$ to $$\frac{2}{9}$$, where $$y$$ is positive, so the integral equals the actual area. **Conclusion:** The larger value $$a=1$$ produces an integral of 1 but the actual area under the curve between 1 and 2 is greater than 1 because part of the area is below the x-axis and subtracts from the integral. **Final answers:** - Two possible values of $$a$$: $$1$$ and $$\frac{1}{9}$$. - The larger value $$a=1$$ does not represent the actual area equal to 1; the actual area is greater than 1.