1. **Problem 1a:** Find the area bounded by the curves $f(x) = 2x - x^2$ and $g(x) = x - 2$. 2. **Find the points of intersection:** Solve $2x - x^2 = x - 2$. $$2x - x^2 = x - 2$$ $$-x^2 + 2x - x + 2 = 0$$ $$-x^2 + x + 2 = 0$$ Multiply both sides by $-1$: $$\cancel{-}x^2 + x + 2 = 0 \Rightarrow x^2 - x - 2 = 0$$ 3. **Factor the quadratic:** $$x^2 - x - 2 = (x - 2)(x + 1) = 0$$ 4. **Solutions:** $$x = 2 \quad \text{or} \quad x = -1$$ 5. **Set up the integral for the area:** The area $A$ between two curves $f(x)$ and $g(x)$ from $a$ to $b$ is $$A = \int_a^b |f(x) - g(x)| \, dx$$ Since $f(x) = 2x - x^2$ and $g(x) = x - 2$, check which is on top between $x=-1$ and $x=2$. At $x=0$: $$f(0) = 0, \quad g(0) = -2$$ So $f(x) \ge g(x)$ in this interval. 6. **Calculate the area:** $$A = \int_{-1}^2 \bigl( (2x - x^2) - (x - 2) \bigr) dx = \int_{-1}^2 (-x^2 + x + 2) dx$$ 7. **Integrate:** $$\int (-x^2 + x + 2) dx = -\frac{x^3}{3} + \frac{x^2}{2} + 2x + C$$ 8. **Evaluate definite integral:** $$A = \left[-\frac{x^3}{3} + \frac{x^2}{2} + 2x \right]_{-1}^2$$ Calculate at $x=2$: $$-\frac{8}{3} + 2 + 4 = -\frac{8}{3} + 6 = \frac{10}{3}$$ Calculate at $x=-1$: $$-\frac{-1}{3} + \frac{1}{2} - 2 = \frac{1}{3} + \frac{1}{2} - 2 = \frac{2}{6} + \frac{3}{6} - 2 = \frac{5}{6} - 2 = -\frac{7}{6}$$ 9. **Subtract:** $$A = \frac{10}{3} - \left(-\frac{7}{6}\right) = \frac{10}{3} + \frac{7}{6} = \frac{20}{6} + \frac{7}{6} = \frac{27}{6} = \frac{9}{2} = 4.5$$ --- "slug": "area bounded 1a", "subject": "calculus", "svg": "", "desmos": {"latex": "y=2x - x^2", "features": {"intercepts": true, "extrema": true}}, "q_count": 2 }