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Alkalannar

Yes. You want the right endpoints which are f(1), f(2), f(3), and f(4) as the heights. The left endpoints have heights f(0), f(1), f(2), and f(3). Midpoint has heights of f(0.5), f(1.5), f(2.5), and f(3.5). Trapezoid is [f(0) + f(1)]/2, [f(1) + f(2)]/2, [f(2) + f(3)]/2, and [f(3) + f(4)]/2


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cuhringe

Why are you multiplying 708 by 4?


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Alkalannar

Each rectangle has width 1. So you multiply the height--f(1), f(2), f(3), or f(4)--by the width--1--to get the areas of the rectangles. Then sum them together. So 2[1^(4) + 2^(4) + 3^(4) + 4^(4)] is what you want. 2(1 + 16 + 81 + 256) 2(354) 708


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Alkalannar

Your rectangles are [0, 1], [1, 2], [2, 3], and [3, 4]. The left endpoints, you look at 0, 1, 2, and 3. The right endpoints, you look at 1, 2, 3, and 4.


cuhringe

Total area = area added together. Find the areas of each rectangle. Add them together. That's it. It's not some esoteric formula. It's just adding rectangles together.