Approximate the area under the curve by evaluating the function at the left-hand endpoints of the subintervals. f(x) = 36 - x from x = 1 to x = 3; 4 subintervals tep 1 Notice that the graph of f(x) is a downward-opening parabola with vertex at (0, 36) and x-intercepts at (-6, 0) and (6, 0). So the graph of f(x) does does lie on or above the x-axis over the interval [1, 3]. To approximate the area under the curve, we use the areas of rectangles whose bases are on the x-axis and whose heights are the vertical distances from points on their bases to the curve. We divide the interval [1, 3] into n = 4 equal subintervals and use them as the bases of n rectangles whose heights are determined by the curve. (See the figure below.) 35 30 25 20 15 10 2 Notice that the figure shows that using left-endpoints will give an approximation that overestimates overestimates the true area. The width of each of these rectangles is the result of dividing the length of the interval [1, 3] by n = 4. MacBook ProThe width of each of these rectangles is the result of dividing the length of the interval [1, 3] by n = 4. Determine the width of each rectangle. width = the length of the interval 3 -1 4 = 15 1/2 Step 2 We divide the interval [1, 3] into 4 subintervals, each of width . The first subinterval has left endpoint 1. Find the remaining left endpoints of the subintervals. The second subinterval has left endpoint 1 + 1 . 32 X 2 2 The third subinterval has left endpoint 1 + 2 . _ =_ 32 X 2 2 .5 The fourth (last) subinterval has left endpoint 1 + 3 . _ X Submit Skip (you cannot come back) Need Help? Read It Submit Answer MacBook Pro 41) esc Q $ % & @ # 8 2 3 4 6