Sketch the region bounded by the graphs of the equations. Find the area of the region. \[ \begin{array}{l} f(x)=-\frac{2}{x^{3}} \\ y=0 \\ x=-4 \\ x=-1\end{array} \]

To sketch the region bounded by the given equations, start by plotting the graph of \( f(x) = -\frac{2}{x^3} \). This function is negative for all \( x \) values, and it approaches 0 as \( x \) moves away from 0 in either direction. Specifically, when \( x = -4 \) and \( x = -1 \), you can calculate the function values: - For \( x = -4 \): \[ f(-4) = -\frac{2}{(-4)^3} = -\frac{2}{-64} = \frac{1}{32} \] - For \( x = -1 \): \[ f(-1) = -\frac{2}{(-1)^3} = -\frac{2}{-1} = 2 \] The area you are interested in is the region bounded by \( f(x) \), the horizontal line \( y=0 \), and the vertical lines \( x=-4 \) and \( x=-1 \). To find the area, compute the definite integral of \( f(x) \) from \( x=-4 \) to \( x=-1 \): \[ \text{Area} = \int_{-4}^{-1} -f(x) \, dx = \int_{-4}^{-1} \frac{2}{x^3} \, dx \] Now, compute the integral: \[ \int \frac{2}{x^3} \, dx = -\frac{2}{2x^2} = -\frac{1}{x^2} \] So, \[ \text{Area} = \left[-\frac{1}{x^2}\right]_{-4}^{-1} \] Evaluating at the bounds: \[ \text{Area} = -\frac{1}{(-1)^2} - \left(-\frac{1}{(-4)^2}\right) = -1 - \left(-\frac{1}{16}\right) = -1 + \frac{1}{16} = -\frac{16}{16} + \frac{1}{16} = -\frac{15}{16} \] Since the area must be positive, we take the absolute value: \[ \text{Area} = \frac{15}{16} \] Thus, the area of the bounded region is \( \frac{15}{16} \) square units.