Let's work through the problem step by step. ### **Given:** - Initial oil at \( t = 4 \) hours: **50 liters** - Rate of oil being pumped in: \( R(t) \), table provided - Find: Approximate oil at \( t = 15 \) hours using a **right Riemann sum with three subintervals** #### **Step 1: Set up the intervals** The time values given are: \( 4, 7, 12, 15 \). This divides the interval \([4, 15]\) into **three subintervals**: 1. \([4, 7]\) 2. \([7, 12]\) 3. \([12, 15]\) #### **Step 2: Right Riemann Sum** For each interval, use the **right endpoint** for \( R(t) \): - Interval 1: \([4, 7]\), width = \(7 - 4 = 3\), use \( R(7) = 6.2 \) - Interval 2: \([7, 12]\), width = \(12 - 7 = 5\), use \( R(12) = 5.9 \) - Interval 3: \([12, 15]\), width = \(15 - 12 = 3\), use \( R(15) = 5.6 \) #### **Step 3: Calculate the sum** \[ \text{Right Riemann Sum} = (3)R(7) + (5)R(12) + (3)R(15) \] \[ = 3 \times 6.2 + 5 \times 5.9 + 3 \times 5.6 \] \[ = 18.6 + 29.5 + 16.8 = 64.9 \] #### **Step 4: Add to initial amount** \[ \text{Total oil at } t = 15 = 50 + 64.9 = 114.9 \] --- ### **Final Answer:** \[ \boxed{114.9} \] So the answer is **(C) 114.9**.