1. Problem 1: Find the area bounded by the coordinate axes and the line $x + y = 2$. 2. The line $x + y = 2$ can be rewritten as $y = 2 - x$. 3. The region is bounded by $x=0$, $y=0$, and $y=2-x$. 4. The area $A$ can be expressed as the double integral: $$A = \int_0^2 \int_0^{2-x} dy \, dx$$ 5. Evaluate the inner integral: $$\int_0^{2-x} dy = (2 - x) - 0 = 2 - x$$ 6. Now evaluate the outer integral: $$\int_0^2 (2 - x) dx = \left[2x - \frac{x^2}{2}\right]_0^2 = (4 - 2) - 0 = 2$$ --- 1. Problem 2: Find the area bounded by the lines $x=0$, $y=2x$, and $y=4$. 2. The region is bounded below by $y=2x$, above by $y=4$, and on the left by $x=0$. 3. Solve $y=2x$ for $x$: $x = \frac{y}{2}$. 4. The limits for $y$ are from $0$ to $4$. 5. The area $A$ can be expressed as: $$A = \int_0^4 \int_0^{\frac{y}{2}} dx \, dy$$ 6. Evaluate the inner integral: $$\int_0^{\frac{y}{2}} dx = \frac{y}{2} - 0 = \frac{y}{2}$$ 7. Evaluate the outer integral: $$\int_0^4 \frac{y}{2} dy = \frac{1}{2} \int_0^4 y dy = \frac{1}{2} \left[ \frac{y^2}{2} \right]_0^4 = \frac{1}{2} \times \frac{16}{2} = 4$$ --- 1. Problem 3: Find the area bounded by the parabola $x = -y^2$ and the line $y = x + 2$. 2. Rewrite the line as $x = y - 2$. 3. The region is bounded between $x = -y^2$ and $x = y - 2$. 4. Find the intersection points by setting $-y^2 = y - 2$: $$-y^2 - y + 2 = 0 \implies y^2 + y - 2 = 0$$ 5. Solve the quadratic: $$y = \frac{-1 \pm \sqrt{1 + 8}}{2} = \frac{-1 \pm 3}{2}$$ 6. Roots are $y=1$ and $y=-2$. 7. The area $A$ is: $$A = \int_{-2}^1 \left[(y - 2) - (-y^2)\right] dy = \int_{-2}^1 (y - 2 + y^2) dy = \int_{-2}^1 (y^2 + y - 2) dy$$ 8. Evaluate the integral: $$\left[ \frac{y^3}{3} + \frac{y^2}{2} - 2y \right]_{-2}^1 = \left( \frac{1}{3} + \frac{1}{2} - 2 \right) - \left( \frac{-8}{3} + 2 - (-4) \right)$$ 9. Simplify: $$\left( \frac{1}{3} + \frac{1}{2} - 2 \right) - \left( -\frac{8}{3} + 2 + 4 \right) = \left( \frac{5}{6} - 2 \right) - \left( \frac{14}{3} \right) = \left( -\frac{7}{6} \right) - \frac{14}{3} = -\frac{7}{6} - \frac{28}{6} = -\frac{35}{6}$$ 10. Since area is positive, take absolute value: $$A = \frac{35}{6}$$ Final answers: 1. Area = 2 2. Area = 4 3. Area = $\frac{35}{6}$