1. Problem 1: Find the area bounded by the curves $y = x^2$ and $y = x$. 2. Formula used: Area between curves is $$\int_a^b (\text{top}-\text{bottom})\,dx$$. 3. Intersections: Solve $x^2=x$ giving $x(x-1)=0$ so $x=0$ and $x=1$. 4. Order on interval: For $0\le x\le1$ we have $x\ge x^2$ so top is $y=x$ and bottom is $y=x^2$. 5. Compute: $$A_1=\int_0^1 (x-x^2)\,dx=\Big[\tfrac{1}{2}x^2-\tfrac{1}{3}x^3\Big]_0^1=\tfrac{1}{2}-\tfrac{1}{3}=\tfrac{1}{6}.$$ 1. Problem 2: Find the area bounded by the curves $y=x+7$ and $y=9-x^2$. 2. Formula used: Area between curves is $$\int_a^b (\text{top}-\text{bottom})\,dx$$. 3. Intersections: Solve $x+7=9-x^2$ giving $x^2+x-2=0$ so $(x+2)(x-1)=0$ and $x=-2,1$. 4. Order on interval: For $x\in[-2,1]$ we have $9-x^2\ge x+7$ so top is $y=9-x^2$ and bottom is $y=x+7$. 5. Compute integrand and integral: $$A_2=\int_{-2}^1 \big((9-x^2)-(x+7)\big)\,dx=\int_{-2}^1 (-x^2-x+2)\,dx$$ and $$=\Big[-\tfrac{1}{3}x^3-\tfrac{1}{2}x^2+2x\Big]_{-2}^1=\tfrac{9}{2}.$$ 1. Problem 3: Find the area bounded by the curves $x=1-y^2$ and $x=y^2-1$. 2. Formula used (integration with respect to $y$): Area is $$\int_c^d (\text{right}-\text{left})\,dy$$. 3. Intersections: Solve $1-y^2=y^2-1$ giving $2-2y^2=0$ so $y=\pm1$. 4. Order on interval: For $-1\le y\le1$ we have $1-y^2\ge y^2-1$ so right is $x=1-y^2$ and left is $x=y^2-1$. 5. Compute: $$A_3=\int_{-1}^1 \big((1-y^2)-(y^2-1)\big)\,dy=\int_{-1}^1 (2-2y^2)\,dy=2\int_{-1}^1 (1-y^2)\,dy$$ and $$=2\Big[y-\tfrac{1}{3}y^3\Big]_{-1}^1=\tfrac{8}{3}.$$ 1. Problem 4: Find the area bounded by the curves $y=x^3$ and $y=4x^2$. 2. Formula used: Area between curves is $$\int_a^b (\text{top}-\text{bottom})\,dx$$. 3. Intersections: Solve $x^3=4x^2$ giving $x^2(x-4)=0$ so $x=0$ and $x=4$. 4. Order on interval: For $0\le x\le4$ we have $4x^2\ge x^3$ so top is $y=4x^2$ and bottom is $y=x^3$. 5. Compute: $$A_4=\int_0^4 (4x^2-x^3)\,dx=\Big[\tfrac{4}{3}x^3-\tfrac{1}{4}x^4\Big]_0^4=\tfrac{64}{3}.$$ 1. Problem 5: Find the area enclosed by the lines $x+2y=2$, $y-x=1$, and $2x+y=7$. 2. Formula used: For a polygon use the shoelace formula $$\text{Area}=\tfrac{1}{2}\big|\sum_{i=1}^n x_i y_{i+1}-y_i x_{i+1}\big|$$ with cyclic indices. 3. Intersections (vertices): Solve pairs to get $A=(0,1)$ from $x+2y=2$ and $y-x=1$, $B=(2,3)$ from $y-x=1$ and $2x+y=7$, and $C=(4,-1)$ from $x+2y=2$ and $2x+y=7$. 4. Apply shoelace to $A(0,1),B(2,3),C(4,-1)$: Compute $\sum x_i y_{i+1}=0\cdot3+2\cdot(-1)+4\cdot1=2$ and $\sum y_i x_{i+1}=1\cdot2+3\cdot4+(-1)\cdot0=14$. 5. Compute area: $$A_5=\tfrac{1}{2}|2-14|=6.$$ Final answers (areas): 1. $\tfrac{1}{6}$. 2. $\tfrac{9}{2}$. 3. $\tfrac{8}{3}$. 4. $\tfrac{64}{3}$. 5. $6$.