# Math 112 Solutions for 5.1, p. 359

1.
(a) . (b) . (c) . (d) . . (e) . (f) 3 (g) The trapezoid has bases 2 and 4 and altitude 1, so has area 3.
2.
(a) . (b) will be an overstimate, , an underestimate, because function is decreasing. (c) 3.58422; 2.25089. (d) 3.39532; 2.72865. (e) Yes; should decrease as increases, and should increase.
3.
(a) Because the function is increasing, is an underestimate and is an overestimate. (b) 0.715249, 1.238848. (c) 0.863382, 1.125182. (d) As increases, underestimates should increase and overestimates should decrease. That is what we see here. (e) about 1.
4.
.
5.
(a) There is no inscribed polygon, so the inner area is zero. (b) The smallest circumscribed polygon is a square with area 1, so the outer area is 1.
7.
Proof: Suppose that is non-empty. Then has a least element . Now , so . Hence , since is the least element not in , so , a contradiction. Therefore .

8.
(a) Let . since . Suppose . Then . Hence , and . Hence , and the given statement is true for all positive integers .

(b) Let . because . Suppose . Then , and . Hence , and the statement is true for all positive integers .

(e) Let     is divisible by . Then is divisible by 5, so . Suppose . Then is divisible by 5. Hence is divisible by 5 because each term is. Hence , so and is divisible by 5 for every positive integer .

10.
If , then , because .

11.
(a) If is uniformly continuous, then given there exists such that if and then , by the definition in Problem 10.

(b) If , then let and be the points in the th subinterval at which has its maximum and minimum, respectively. Then , by part (a), so that

(c) If , then as , . That means and have the same limit, so has area.