The given region is bounded by the parabola \(y^2 = 2x\) and the line \(y = 4x - 1\).
Step 1: Find intersection points.
Substituting \(x = \frac{y+1}{4}\) (from \(y = 4x - 1\)) into \(y^2 = 2x\):
\[ y^2 = 2 \cdot \frac{y+1}{4} \implies y^2 = \frac{y+1}{2}. \] Simplify to: \[ 2y^2 - y - 1 = 0 \implies (2y + 1)(y - 1) = 0. \] Thus, \(y = -\frac{1}{2}\) and \(y = 1\).
Step 2: Set up integral for the shaded area.
The shaded area is calculated as: \[ \text{Area} = \int_{-\frac{1}{2}}^1 (x_{\text{right}} - x_{\text{left}}) \, dy, \] where \(x_{\text{right}} = \frac{y+1}{4}\) (line) and \(x_{\text{left}} = \frac{y^2}{2}\) (parabola).
Step 3: Solve the integral.
\[ \text{Area} = \int_{-\frac{1}{2}}^1 \left( \frac{y+1}{4} - \frac{y^2}{2} \right) dy = \int_{-\frac{1}{2}}^1 \frac{y+1}{4} \, dy - \int_{-\frac{1}{2}}^1 \frac{y^2}{2} \, dy. \] Simplify: \[ \text{Area} = \left[ \frac{y^2}{8} + \frac{y}{4} \right]_{-\frac{1}{2}}^1 - \left[ \frac{y^3}{6} \right]_{-\frac{1}{2}}^1. \] Compute each term:
Simplify to find the area: \[ \text{Area} = \frac{9}{32}. \]
A body of mass 1000 kg is moving horizontally with a velocity of 6 m/s. If 200 kg extra mass is added, the final velocity (in m/s) is: