1. Problem statement: Find the area bounded by the given curves for problems 1 through 5. 2. Problem 1: Find the area bounded by $y = x^2$ and $y = x$. Formula: $A=\int_a^b (\text{top}-\text{bottom})\,dx$. Intersections: Solve $x^2=x$ gives $x( x-1)=0$ so $x=0$ and $x=1$. Top function: For $0x^2$ so top is $y=x$ and bottom is $y=x^2$. Computation: $A=\int_0^1 (x - x^2)\,dx=\left[\frac{1}{2}x^2-\frac{1}{3}x^3\right]_0^1=\frac{1}{2}-\frac{1}{3}=\frac{1}{6}$. 3. Problem 2: Find the area bounded by $y=x+7$ and $y=9-x^2$. Intersections: Solve $x+7=9-x^2$ giving $x^2+x-2=0$ and roots $x=-2,\;x=1$. Top function: Test $x=0$ gives $y_{1}=7$ and $y_{2}=9$ so $y=9-x^2$ is top. Computation: $A=\int_{-2}^{1}\big((9-x^2)-(x+7)\big)\,dx=\int_{-2}^{1}(2-x-x^2)\,dx$. Evaluate: $\left[2x-\tfrac{1}{2}x^2-\tfrac{1}{3}x^3\right]_{-2}^{1}=\left(2-\tfrac{1}{2}-\tfrac{1}{3}\right)-\left(-4-2+\tfrac{8}{3}\right)=\frac{9}{2}$. 4. Problem 3: Find the area bounded by $x=1-y^2$ and $x=y^2-1$. Intersections: Solve $1-y^2=y^2-1$ giving $2y^2=2$ so $y=\pm 1$. Orientation: For a given $y$ the right curve is $x=1-y^2$ and the left is $x=y^2-1$. Computation: $A=\int_{-1}^{1}\big((1-y^2)-(y^2-1)\big)\,dy=\int_{-1}^{1}(2-2y^2)\,dy$. Evaluate: $\left[2y-\tfrac{2}{3}y^3\right]_{-1}^{1}=\left(2-\tfrac{2}{3}\right)-\left(-2+\tfrac{2}{3}\right)=\frac{8}{3}$. 5. Problem 4: Find the area bounded by $y=x^3$ and $y=4x^2$. Intersections: Solve $x^3=4x^2$ gives $x^2(x-4)=0$ so $x=0$ and $x=4$. Top function: For $0x^3$ so top is $y=4x^2$. Computation: $A=\int_0^4 (4x^2-x^3)\,dx=\left[\tfrac{4}{3}x^3-\tfrac{1}{4}x^4\right]_0^4=\frac{64}{3}$. 6. Problem 5: Find the area bounded by the lines $x+2y=2$, $y-x=1$, and $2x+y=7$. Convert lines: $L_1:\;y=\tfrac{2-x}{2}$, $L_2:\;y=x+1$, $L_3:\;y=7-2x$. Intersections: $L_1\cap L_2:\;x+2(x+1)=2\Rightarrow x=0,\;y=1$ so $(0,1)$. $L_2\cap L_3:\;x+1=7-2x\Rightarrow x=2,\;y=3$ so $(2,3)$. $L_3\cap L_1:\;x+2(7-2x)=2\Rightarrow x=4,\;y=-1$ so $(4,-1)$. Area (shoelace): For vertices $(0,1),(2,3),(4,-1)$ compute $\tfrac{1}{2}|\sum x_i y_{i+1}-\sum y_i x_{i+1}|=6$. Final answers table: | Problem | Area | |---|---:| | 1 | $\frac{1}{6}$ | | 2 | $\frac{9}{2}$ | | 3 | $\frac{8}{3}$ | | 4 | $\frac{64}{3}$ | | 5 | $6$ |