1. **State the problem:** Find the area between the curves $y=x$, $y=4x$, and $y=-x+2$. 2. **Find the points of intersection:** - Between $y=x$ and $y=4x$: Set $x=4x \Rightarrow 3x=0 \Rightarrow x=0$, so point is $(0,0)$. - Between $y=x$ and $y=-x+2$: Set $x=-x+2 \Rightarrow 2x=2 \Rightarrow x=1$, so point is $(1,1)$. - Between $y=4x$ and $y=-x+2$: Set $4x=-x+2 \Rightarrow 5x=2 \Rightarrow x=\frac{2}{5}$, so point is $(\frac{2}{5}, \frac{8}{5})$. 3. **Determine the region:** The three lines form a triangle with vertices at $(0,0)$, $(1,1)$, and $(\frac{2}{5}, \frac{8}{5})$. 4. **Set up the integrals:** The area can be found by integrating the difference between the upper and lower curves over the intervals: - From $x=0$ to $x=\frac{2}{5}$, top curve is $y=-x+2$, bottom curve is $y=4x$. - From $x=\frac{2}{5}$ to $x=1$, top curve is $y=-x+2$, bottom curve is $y=x$. 5. **Calculate the area:** $$\text{Area} = \int_0^{\frac{2}{5}} ((-x+2) - 4x) \, dx + \int_{\frac{2}{5}}^1 ((-x+2) - x) \, dx$$ Simplify integrands: - First integral: $-x+2 - 4x = -5x + 2$ - Second integral: $-x+2 - x = -2x + 2$ Calculate each integral: $$\int_0^{\frac{2}{5}} (-5x + 2) \, dx = \left[-\frac{5}{2}x^2 + 2x\right]_0^{\frac{2}{5}} = \left(-\frac{5}{2} \times \frac{4}{25} + 2 \times \frac{2}{5}\right) - 0 = \left(-\frac{10}{25} + \frac{4}{5}\right) = \left(-0.4 + 0.8\right) = 0.4$$ $$\int_{\frac{2}{5}}^1 (-2x + 2) \, dx = \left[-x^2 + 2x\right]_{\frac{2}{5}}^1 = \left(-1 + 2\right) - \left(-\frac{4}{25} + \frac{4}{5}\right) = (1) - \left(-0.16 + 0.8\right) = 1 - 0.64 = 0.36$$ 6. **Add the areas:** $$\text{Total area} = 0.4 + 0.36 = 0.76$$ **Final answer:** The area between the curves is $0.76$ square units.