1. Find the area of the finite region between the parabola y = 2 and the...

 

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1. Find the area of the finite region between the parabola y = 2 and the line y = 4 by setting up and evaluating a suitable definite integral. The evaluation should be done both by hand and by using SageMath. [2] Second, a harder way, using a simplified version of Riemann integration: 2. Compute the definite integral you set up in solving 1 using the Right-Hand Rule formula, as described in the accompanying handout Right-Hand Rule Riemann Sums. The evaluation of the formula, once you've set it up, should be done both by hand and by using SageMath. [4] NOTE. You may the find the following summation formulas useful in doing 2: i=1 72 i=1+2+3+ + n = n(n + 1) 2 2 = 1 +2+3+...+ n = n(n+1) (2n+1) 6 i=1 Third, and hardest, is Archimedes' way, though we'll let him do the really heavy lifting. How did he do it? The key was the following result: THEOREM. Suppose a straight line intersects a parabola at points P and Q. Let R be the point on the part of the parabola between P and Q that is furthest from the line, and let S and T be the points on the parabola between P and R, and between Q and R. respectively, that are farthest from the line PR and the line QR, respectively. Then each of the triangles APSR and AQTR has one-eighth of the area of the triangle APQR. 1. Find the area of the finite region between the parabola y = 2 and the line y = 4 by setting up and evaluating a suitable definite integral. The evaluation should be done both by hand and by using SageMath. [2] Second, a harder way, using a simplified version of Riemann integration: 2. Compute the definite integral you set up in solving 1 using the Right-Hand Rule formula, as described in the accompanying handout Right-Hand Rule Riemann Sums. The evaluation of the formula, once you've set it up, should be done both by hand and by using SageMath. [4] NOTE. You may the find the following summation formulas useful in doing 2: i=1 72 i=1+2+3+ + n = n(n + 1) 2 2 = 1 +2+3+...+ n = n(n+1) (2n+1) 6 i=1 Third, and hardest, is Archimedes' way, though we'll let him do the really heavy lifting. How did he do it? The key was the following result: THEOREM. Suppose a straight line intersects a parabola at points P and Q. Let R be the point on the part of the parabola between P and Q that is furthest from the line, and let S and T be the points on the parabola between P and R, and between Q and R. respectively, that are farthest from the line PR and the line QR, respectively. Then each of the triangles APSR and AQTR has one-eighth of the area of the triangle APQR.