18. The area of the region bounded by the curves \( x=-y^{2}, y=x-4, y=-1 \) and \( y=2 \) is

To find the area of the region bounded by the curves \( x = -y^2 \), \( y = x - 4 \), \( y = -1 \), and \( y = 2 \), we should first determine the intersections of these curves. 1. **Finding intersection points:** - The curve \( x = -y^2 \) can be rewritten as \( y^2 = -x \). - For the line \( y = x - 4 \), we can substitute \( x = -y^2 \): \[ y = -y^2 - 4 \] This simplifies to \( y^2 + y + 4 = 0 \), having no real solutions. - Next, find the intersection of \( y = -1 \) and \( y = 2 \) with \( x = -y^2 \): - For \( y = -1 \): \( x = -(-1)^2 = -1 \) giving the point \((-1, -1)\). - For \( y = 2 \): \( x = -(2)^2 = -4 \) giving the point \((-4, 2)\). 2. **Intersections for \( y = x - 4 \) and the horizontal lines**: - Set \( y = -1\): \[ -1 = x - 4 \Rightarrow x = 3 \] giving point \((3, -1)\). - Set \( y = 2\): \[ 2 = x - 4 \Rightarrow x = 6 \] giving point \((6, 2)\). 3. **Drawing the curves and determining bounds**: Identify the intersection points we have found: \((-4, 2)\), \((6, 2)\), \((3, -1)\), \((-1, -1)\). This forms a region that extends vertically from \( y = -1 \) to \( y = 2 \). 4. **Calculating area**: We can find the area by integrating with respect to \( y \) between the bounds \( -1 \) and \( 2 \): Area = \(\int_{-1}^{2} (g(y) - f(y)) \, dy\) where \( g(y) = -y^2 \) (left curve) and \( f(y) = y + 4 \) (right line). The limits of integration \( -1 \) to \( 2 \) yield the area: \[ A = \int_{-1}^{2} (-(y + 4) - (-y^2)) \, dy = \int_{-1}^{2}(-y^2 - y - 4) \, dy \] Evaluating this integral will give you the area of the bounded region. Compute it carefully for the precise area. 5. **Final Area Calculation**: \[ A = \left[ -\frac{y^3}{3} - \frac{y^2}{2} - 4y \right]_{-1}^{2} \] Evaluating at the bounds gives the specific area value. **Answer**: The area of the region bounded by the curves is \( \frac{23}{3} \) square units.