Approximate the integral below using a right hand sum with 4 rectangles. \[ \int_{0}^{4}\left(16-x^{2}\right) d x \approx[?] \]

To approximate the integral \(\int_{0}^{4}(16-x^{2})\, dx\) using a right-hand sum with 4 rectangles, we first determine the width of each rectangle. The interval from 0 to 4 spans 4 units, and with 4 rectangles, each rectangle will have a width of: \[ \Delta x = \frac{4-0}{4} = 1. \] Next, the right-hand endpoints for these rectangles will be at \(x = 1, 2, 3, 4\). We then evaluate the function \(f(x) = 16 - x^2\) at each of these points: - \(f(1) = 16 - 1^2 = 15\) - \(f(2) = 16 - 2^2 = 12\) - \(f(3) = 16 - 3^2 = 7\) - \(f(4) = 16 - 4^2 = 0\) Now we can calculate the approximate integral using the right-hand sum: \[ \text{Right-hand sum} = f(1) \cdot \Delta x + f(2) \cdot \Delta x + f(3) \cdot \Delta x + f(4) \cdot \Delta x. \] This simplifies to: \[ \text{Right-hand sum} = (15 + 12 + 7 + 0) \cdot 1 = 34. \] Thus, the approximate value of the integral using a right-hand sum with 4 rectangles is: \[ \int_{0}^{4}(16-x^{2})\, dx \approx 34. \]