# P31 - sums of squares cubes etc are in the notes or book...

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Project 31 MAC 2311 YOU MUST SHOW YOUR WORK TO RECEIVE FULL CREDIT!! 1. We continue to look at the parabolic arch formed by the function f ( x ) = 3 x - 1 2 x 2 on [ 0 , 6 ] . We would like to calculate its area using Riemann Sums of rectangles. Sketch the parabola below on [ 0 , 6 ] and shade the area beneath it. Suppose we divide the interval [ 0 , 6 ] into n subintervals: Interval 1, Interval 2, . . . , Interval n . What is the width of each interval? Show or explain why the right endpoint of Interval i is 0 + i ± 6 n ² . If we approximate the area of the arch with n rectangles of equal width and x * i = right endpoint, what is the area of the rectangle on Interval i (in terms of i and n )? What is the limit that represents the area of the arch? Area = lim n X i =1

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Evaluate this limit using the technique learned in class (the formulas for
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Unformatted text preview: sums of squares, cubes, etc. are in the notes or book). Evaluate these deﬁnite integrals using ONLY your answer from above and properties of integrals: Z 6 3 x-1 2 x 2 d x = Z 6 x 2-6 x d x = Z 6 x 2-6 x + 2 d x = 2. If f ( x ) is negative on [ a, b ], then R b a f ( x ) d x is negative. Why? Weren’t deﬁnite integrals deﬁned to be areas? (Review your deﬁnitions.) Sketch the following piecewise-deﬁned function, and evaluate the given def-inite integral in terms of area. Shade the regions whose areas you use to evaluate the deﬁnite integral. f ( x ) = √ 16-x 2 x ≤ 4-2 x < x < 3-2 x ≥ 3 Z 4-4 f ( x ) d x =...
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P31 - sums of squares cubes etc are in the notes or book...

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