Find the area of the indicated region. (a) The region bounded by \( y=x^{2}-x-2 \) and \( y=2 x^{2}-4 x-6 \) (b) The region bounded by \( y=\sqrt{x}, y=1 \) and \( y=-x \)

To find the area of the indicated regions, we'll start with each part: (a) The area between the curves \( y = x^2 - x - 2 \) and \( y = 2x^2 - 4x - 6 \) requires us to first find the points of intersection. Setting \( x^2 - x - 2 = 2x^2 - 4x - 6 \), rearranging gives \( x^2 - 3x + 4 = 0 \). Applying the quadratic formula, we find the roots \( x = 2 \) and \( x = -1 \). The area can then be computed as: \[ \text{Area} = \int_{-1}^{2} \left( (2x^2 - 4x - 6) - (x^2 - x - 2) \right) \, dx \] Calculating the integral: \[ = \int_{-1}^{2} (x^2 - 3x - 4) \, dx = \left[ \frac{x^3}{3} - \frac{3x^2}{2} - 4x \right]_{-1}^{2} \] Evaluating this gives the area of the region bounded by the two curves. (b) To find the area of the region bounded by \( y = \sqrt{x} \), \( y = 1 \), and \( y = -x \), we first find the points of intersection. \( \sqrt{x} = 1 \) gives \( x = 1 \). For \( y = -x \) and \( y = \sqrt{x} \), squaring gives \( x = -x^2 \) or \( x^2 + x = 0 \), leading to \( x(1 + x) = 0 \), giving \( x = 0 \) and \( x = -1 \). The area is determined from \( x = 0 \) to \( x = 1 \): \[ \text{Area} = \int_{0}^{1} \left( \sqrt{x} - (-x) \right) \, dx \] This simplifies to: \[ = \int_{0}^{1} \left( \sqrt{x} + x \right) \, dx = \left[ \frac{2}{3}x^{3/2} + \frac{x^2}{2} \right]_{0}^{1} = \frac{2}{3} + \frac{1}{2} = \frac{4}{6} + \frac{3}{6} = \frac{7}{6} \] Thus, these calculations provide the areas for the bounded regions.