4.3
	1. It turns out that for many functions, the limit of the Riemann sums is independent of the choice of evaluation points. Discuss why this is a somewhat surprising result. To make the result more believable, consider a continuous function f (x). As the number of partition points gets larger, the distance between the endpoints gets smaller. For the continuous function f (x) , explain why the difference between the function values at any two points in a given subinterval will have to get smaller.
	2. Rectangles are not the only basic geometric shapes for which we have an area formula. Discuss how you might approximate the area under a parabola using circles or triangles. Which geometric shape do you think is the easiest to use?
	In exercises 3-10, list the evaluation points corresponding to the midpoint of each subinterval, sketch the function and approximating rectangles and evaluate the Riemann sum.
	3. f (x) = x2+1, [0, 1], n = 4
	4. f (x) = x2+1, [0, 2], n = 4
	5. f (x) = x3-1, [1, 2], n = 4
	6. f (x) = x3-1, [1, 3], n = 4
	7. f (x) = sin x, [0, ], n = 4
	8. f (x) = sin x, [0, ], n = 8
	9. f (x) = 4-x2, [-1, 1], n = 4
	10. f (x) = 4-x2, [-3, -1], n = 4
	In exercises 11-26, approximate the area under the curve on the given interval using n rectangles and the indicated evaluation rule.
	11. y = x2 on [0, 1], n = 8 , midpoint evaluation
	12. y = x2 on [0, 1], n = 8 , right-endpoint evaluation
	13. y = x2 on [-1, 1], n = 8 , left-endpoint evaluation
	14. y = x2 on [-1, 1], n = 8 , midpoint evaluation
	15. , midpoint evaluation
	16. , right-endpoint evaluation
	17. y = e-2x on [-1, 1], n = 16 , left-endpoint evaluation
	18. y = e-2x on [-1, 1], n = 16 , midpoint evaluation
	19. y = cos x on [0, /2], n = 50 , midpoint evaluation
	20. y = cos x on [0, /2], n = 100 , right-endpoint evaluation
	21. y = 3x-2 on [1, 4], n = 4 , midpoint evaluation
	22. y = 3x-2 on [1, 4], n = 40 , midpoint evaluation
	23. y = x3-1 on [1, 3], n = 100 , midpoint evaluation
	24. y = x3-1 on [1, 3], n = 100 , right-endpoint evaluation
	25. y = x3-1 on [-1, 1], n = 100 , left-endpoint evaluation
	26. y = x3-1 on [-1, 1], n = 100 , right-endpoint evaluation
	In exercises 27-30, construct a table of Riemann sums as in Example 3.5 to show that sums with right-endpoint, midpoint and left-endpoint evaluation all converge to the same value as n .
	27. f (x) = 4-x2, [-2, 2]	28. f (x) = sin x, [0, /2]
	29. f (x) = x3-1, [1, 3]	30. f (x) = x3-1, [-1, 1]
	In exercises 31-34, use Riemann sums and a limit to compute the exact area under the curve.
	31. y = x2 + 2 on [0, 1]	32. y = x2 + 3x on [0, 1]
	33. y = 2x2 + 1 on [1, 3]	34. y = 4x + 2 on [1, 3]
	In exercises 35-40, use the given function values to estimate the area under the curve using left-endpoint and right-endpoint evaluation.
	35.
	x 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 f(x) 2.0 2.4 2.6 2.7 2.6 2.4 2.0 1.4 0.6
	36.
	x 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 f(x) 3.0 2.2 1.6 0.7 0.6 0.4 -0.2 0.4 0.6
	37.
	x 0.0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 f(x) 1.0 1.4 2.1 2.7 2.6 2.8 3.0 3.4 3.6
	38.
	x 0.0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 f(x) 2.0 2.2 1.6 1.4 1.6 2.0 2.2 2.4 2.0
	39.
	x 1.0 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 f(x) 1.8 1.4 1.1 0.7 1.2 1.4 1.8 2.4 2.6
	40.
	x 1.0 1.2 1.4 1.6 1.8 2.0 2.2 2.4 2.6 f(x) 0.0 0.4 0.8 1.2 1.4 1.2 1.4 1.4 1.0
	In exercises 41-45, graphically determine whether a Rieman sum with (a) left-endpoint, (b) midpoint and (c) right-endpoint evaluation points will be greater than or less than the area under the curve y = f (x) on [a, b].
	41. f (x) is increasing and concave up on [a, b].
	42. f (x) is increasing and concave down on [a, b].
	43. f (x) is decreasing and concave up on [a, b].
	44. f (x) is decreasing and concave down on [a, b].
	45. For the function f (x) = x2 on the interval [0, 1] , by trial and error find evaluation points for n = 2 such that the Riemann sum equals the exact area of 2/3.
	46. For the function f (x) = on the interval [0, 1] , by trial and error find evaluation points for n = 2 such that the Riemann sum equals the exact area of 2.
	47. Show that for right-endpoint evaluation on the interval [a, b] with each subinterval of length x = (b-a)/n , the evaluation points are ci = a+i x , for i = 1, 2, , n.
	48. Show that for left-endpoint evaluation on the interval [a, b] with each subinterval of length x = (b-a)/n , the evaluation points are ci = a+(i-1) x , for i = 1, 2, , n.
	49. As in exercises 47 and 48, find a formula for the evaluation points for midpoint evaluation.
	50. As in exercises 47 and 48, find a formula for evaluation points which are one-third of the way from the left-endpoint to the right-endpoint.
	51. Riemann sums can also be defined on irregular partitions for which subintervals are not of equal size. An example of an irregular partition of the interval [0, 1] is x0 = 0 , x1 = 0.2, x2 = 0.6, x3 = 0.9, x4 = 1. Explain why the corresponding Riemann sum would be f (c1)(0.2)+f (c2)(0.4)+f (c3)(0.3)+f (c4)(0.1),
	for evaluation points c1 , c2 , c3 and c4. Identify the interval from which each ci must be chosen and give examples of evaluation points. To see why irregular partitions might be useful,
	consider the function on the interval [0, 2]. One way to approximate the area under the graph of this function is to compute Riemann sums using midpoint evaluation for n = 10 , n = 50 , n = 100 and so on. Show graphically and numerically that with midpoint evaluation, the Riemann sum with n = 2 gives the correct area on the subinterval [0, 1]. Then explain why it would be wasteful to compute Riemann sums on this subinterval for larger and larger values of n. A more efficient strategy would be to compute the areas on [0, 1] and [ 1, 2 ] separately and add them together. The area on [0, 1] can be computed exactly using a small value of n, while the area on [1, 2] must be approximated using larger and larger values of n. Use this technique to estimate the area for f (x) on the interval [0, 2]. Try to determine the area to within an error of 0.01 (discuss why you believe your answer is this accurate).
	52. Graph the function f (x) = e-x2. You may recognize this curve as the so-called “bell curve, ” which is of fundamental importance in statistics. We define the area function g(t ) to be the area between this graph and the x - axis between x = 0 and x = t  (for now, assume that t > 0 ). Sketch the area that defines g(1) and g(2) and argue that g(2) > g(1). Explain why the function g(x) is increasing and hence g' (x) > 0 for x > 0. Further, argue that g'(2) < g'(1). Explain why g' (x) is a decreasing function. Thus, g' (x) has the same general properties (positive, decreasing) that f (x) does. In fact, we will discover in Section 4.5 that g' (x) = f (x). To collect some evidence for this result, use Riemann sums to estimate g(2) , g(1.1) , g(1.01) and g(1). Use these values to estimate g' (1) and compare to f (1).