\( 1 \leftarrow \begin{array}{l}\text { The velocity of an object is given by the following function defined on a specified interval. Approximate the displacement of the object } \\ \text { on this interval by subdividing the interval into the indicated number of subintervals. Use the left endpoint of each subinterval to } \\ \text { compute the height of the rectangles. } \\ \qquad v=1 /(4 \mathrm{t}+4)(\mathrm{m} / \mathrm{s}) \text { for } 0 \leq t \leq 8 ; \mathrm{n}=4 \\ \text { The approximate displacement of the object is } \square \mathrm{m} \text {. } \\ \text { (Round to two decimal places as needed.) }\end{array} \) ?

To approximate the displacement of the object using the left endpoint method, we will follow these steps: 1. **Identify the function and interval**: The velocity function is given by \( v(t) = \frac{1}{4t + 4} \) for the interval \( [0, 8] \). 2. **Determine the number of subintervals**: We are given \( n = 4 \). 3. **Calculate the width of each subinterval**: The width \( \Delta t \) of each subinterval can be calculated as: \[ \Delta t = \frac{b - a}{n} = \frac{8 - 0}{4} = 2 \] 4. **Identify the left endpoints**: The left endpoints of the subintervals are: - For the first subinterval: \( t_0 = 0 \) - For the second subinterval: \( t_1 = 2 \) - For the third subinterval: \( t_2 = 4 \) - For the fourth subinterval: \( t_3 = 6 \) 5. **Evaluate the velocity at each left endpoint**: - \( v(t_0) = v(0) = \frac{1}{4(0) + 4} = \frac{1}{4} = 0.25 \) - \( v(t_1) = v(2) = \frac{1}{4(2) + 4} = \frac{1}{8 + 4} = \frac{1}{12} \approx 0.0833 \) - \( v(t_2) = v(4) = \frac{1}{4(4) + 4} = \frac{1}{16 + 4} = \frac{1}{20} = 0.05 \) - \( v(t_3) = v(6) = \frac{1}{4(6) + 4} = \frac{1}{24 + 4} = \frac{1}{28} \approx 0.0357 \) 6. **Calculate the approximate displacement**: The approximate displacement \( D \) can be calculated using the formula: \[ D \approx \sum_{i=0}^{n-1} v(t_i) \Delta t \] Substituting the values we calculated: \[ D \approx (v(t_0) + v(t_1) + v(t_2) + v(t_3)) \Delta t \] \[ D \approx \left(0.25 + 0.0833 + 0.05 + 0.0357\right) \cdot 2 \] Now, let's calculate the sum of the velocities and then multiply by \( \Delta t \): \[ D \approx (0.25 + 0.0833 + 0.05 + 0.0357) \cdot 2 \] \[ D \approx (0.4190) \cdot 2 \approx 0.8380 \] Finally, rounding to two decimal places, the approximate displacement of the object is: \[ \boxed{0.84} \text{ m} \]