EXAM

Midterm Math 157

November 14, 2013

ANSWERS

Problem 1. Define. [2 points each] (a) Let s : [a, b] → ℝ be a function. What does it mean to say that s is a step function?

Answer: A function s : [a, b] → ℝ is a step function if and only if there exists a partition P = {x_{0}, x_{1}, …, x_{n}} of the interval [a, b] for which s is constant on the open subintervals (x_{k−1}, x_{k}).

(b) Let s : [a, b] → ℝ be a step function. Define ∫_{a}^{b} s.

Answer: Let P = {x_{0}, x_{1}, …, x_{n}} be a partition of [a, b] for which s is constant on the open subintervals (x_{k−1}, x_{k}). Let s_{k} be the constant value of s on (x_{k−1}, x_{k}). Then

∫_{a}^{b} s := ∑_{k=1}^{n} s_{k}(x_{k} − x_{k−1}).

(c) Let f : [a, b] → ℝ be any bounded function. Define the statement f is integrable, and the expression ∫_{a}^{b} f.

Answer: If there exists one and only one number I satisfying

∫_{a}^{b} s < I < ∫_{a}^{b} t

for all step functions s and t with s < f < t, then we say f is integrable and we denote the number I by ∫_{a}^{b} f.

Problem 2. [2 points each] Compute. Do three out of four, or do all four for a bonus.

(a) ∫_{1}^{5} [\frac{1}{2}x + 1] dx

Answer: The function s(x) = [\frac{1}{2}x + 1] is a step function taking the value 1 on (1, 2), the value 2 on (2, 4) and the value 3 on (4, 5).

So,

∫_{1}^{5} [\frac{1}{2}x + 1] dx = 1(1) + 2(2) + 3(1) = 8.

(b) ∫_{0}^{5} f where f(x) = \begin{cases}1 & if 3 ≤ |x^{2} − 4| ≤ 5 \\ 0 & otherwise. \end{cases} .

Answer: Note that

|x^{2} − 4| ≤ 5 ⇔ −3 ≤ x ≤ 3.

and

3 ≤ |x^{2} − 4| ⇔ −1 ≤ x ≤ 1 or x ≤ −\sqrt{7} or x ≥ \sqrt{7}.

Therefore, 3 ≤ |x^{2} − 4| ≤ 5 ⇔ −3 ≤ x ≤ −\sqrt{7} or − 1 ≤ x ≤ 1 or \sqrt{7} ≤ x ≤ 3.

So,

∫_{0}^{5} f = 1(1 − 0) + 1(3 − \sqrt{7}) = 4 − \sqrt{7}.

Problem 2. Continued.

(c) ∫_{0}^{4} |2x − 6|dx

Answer:

Since the function f : [0, 4] → ℝ given by f(x) = |2x − 6| is piecewise monotonic, it is integrable and since it’s nonegative, the integral is the area of the ordinate set of f. Here’s a picture of the ordinate set, which is the union of two triangles of area \frac{1}{2}(6)(3) = 9 and \frac{1}{2}(2)(1) = 1.

So,

∫_{0}^{4} |2x − 6|dx = 10.

(d) ∫_{1}^{3}(3x^{2} − 5x)dx We compute:

∫_{1}^{3}(3x^{2} − 5x)dx = ∫_{0}^{3}(3x^{2} − 5x)dx − ∫_{0}^{1}(3x^{2} − 5x)dx

= 3 ∫_{0}^{3} x^{2}dx − 5 ∫_{0}^{3} xdx − 3 ∫_{0}^{1} x^{2}dx + 5 ∫_{0}^{1} xdx

= 3\frac{3^{3}}{3} − 5\frac{3^{2}}{2} − 3\frac{1^{3}}{3} + 5\frac{1^{2}}{2}

= 6.

Problem 3. True or false : [Right answer +1, wrong answer −1, no answer +\frac{1}{2}]

(a) For all functions f, g, h : ℝ → ℝ, f ∘ (g + h) = f ∘ g + f ∘ h.

Answer:

False. For example, if f(x) = 1, g(x) = 2, h(x) = 2, then f ∘ (g + h)(x) = 1 and f ∘ g + f ∘ h = 2.

(b) For every function f : [a, b] → ℝ, there exist step functions s, t : [a, b] → ℝ with s ≤ f ≤ t.

Answer:

False. The function f : [−1, 1] → ℝ given by f(x) = \frac{1}{x} if x ≠ 0 and f(0) = 5 is unbounded and there are no step functions s with s ≤ f and no step function t with f ≤ t.

(c) If f is integrable and even, then ∫_{−b}^{b} f(x)dx = 2 ∫_{0}^{b} f(x)dx.

Answer:

True.

(d) If s : [a, b] → ℝ is a step function, then s([a, b]) is a finite set.

Answer:

True. By definition, there’s a partition {x_{0}, …, x_{n}} of [a, b] for which s is constant on the n open subintervals (x_{k−1}, x_{k}). Accounting for the possibly distinct values of s at the n + 1 endpoints shows that s takes on at most 2n + 1 values.

(e) If s : [a, b] → ℝ is a step function, then for any y ∈ ℝ, the set s^{−1}({y}) is either empty or an interval.

Answer:

False. For example, if s : [0, 3] → ℝ is given by s(x) = 1 if 0 ≤ x < 1, x(x) = 2 if 1 ≤ x ≤ 2 and s(x) = 1 if 2 < x ≤ 3. Then we have s^{−1}({1}) = [0, 1) ∪ (2, 3] which is not an interval.

Problem 3. Continued.

(f) Let π_{1} : ℝ × ℝ → ℝ and π_{2} : ℝ × ℝ → ℝ be the canonical projections. Let A ⊆ ℝ × ℝ be the ordinate set of a nonegative function f. Then

π_{1}(A) = the domain of f and π_{2}(A) = the range of f.

Answer:

False. Say f : [0, 1] → ℝ is given by f(x) = 2. Then the ordinate set A of f is the rectangle A = {(x, y) : 0 ≤ x ≤ 1 and 0 ≤ y ≤ 2}. We see π_{1}(A) = [0, 1] = the domain of f but π_{2}(A) = [0, 2] ≠ the range of f = {2}.

(g) For every ϵ > 0, there exists a step function s : [0, 1] → ℝ with

s(x) ≤ x^{2} for all x ∈ [0, 1] and ∫_{0}^{1} s > \frac{1}{3} − ϵ.

Answer:

True. We know that the function given by f(x) = x^{2} is integrable on [0, 1] and ∫_{0}^{1} x^{2} = \frac{1}{3}. So, the lower integral \underline{I}(f) = \frac{1}{3}. Since \frac{1}{3} − ϵ is less than \frac{1}{3}, which is the least upper bound of

S = {∫_{0}^{1} s : s is a step function with s ≤ f}

there must be an element of S greater than \frac{1}{3} − ϵ. That is, a step function s with s ≤ f and ∫_{0}^{1} s > \frac{1}{3} − ϵ.

(h) If f, g : [a, b] → ℝ are integrable, then so is fg and ∫_{a}^{b} fg = (∫_{a}^{b} f) (∫_{a}^{b} g)

Answer:

False. If f = g = the constant function 1 and [a, b] = [0, 3], then

∫_{a}^{b} fg = ∫_{0}^{3} 1 = 3 and (∫_{a}^{b} f) (∫_{a}^{b} g) = (∫_{0}^{3} 1) (∫_{0}^{3} 1) = (3)(3) = 9.

Midterm : November 14, 2013 6

Problem 4. Consider the function f : [0, 1] → ℝ whose graph consists of straight lines

connecting

(1, 0) and (\frac{1}{2}, 1)
(\frac{1}{2}, 1) and (\frac{1}{3}, 0)
(\frac{1}{3}, 0) and (\frac{1}{4}, 1)
(\frac{1}{4}, 1) and (\frac{1}{5}, 0)
⋮

The value of f at zero is immaterial, say f(0) = 0 if you like. One could give a formula for f on each interval (\frac{1}{n}, \frac{1}{n + 1}) but a picture is worth a thousand words:

(a) [2 points] Is f is piecewise monotonic on [0, 1]? Answer: No.

(b) [2 points] Is f integrable on [0, 1]? Answer: Yes.

Problem 4. Continued.

(c) [2 bonus points] Determine \underline{I}(f) and \overline{I}(f). Answer: f is integrable, so \underline{I}(f) = \overline{I}(f) = ∫_{0}^{1} f = \frac{1}{2}. To see that ∫_{0}^{1} f = \frac{1}{2}, first note that the ordinate set of f on any interval of the form [\frac{1}{n}, 1] consists of the union of a finite number of triangles, hence ∫ 1_{\frac{1}{n}} f is the sum of the areas of these finitely many rectangles. Adding them up (from right to left) we have

∫ 1_{\frac{1}{n}} f = \frac{1}{2} (1 − \frac{1}{3}) + \frac{1}{2} (\frac{1}{3} − \frac{1}{4}) + \frac{1}{2} (\frac{1}{4} − \frac{1}{5}) + ⋯ \frac{1}{2} (\frac{1}{n − 1} − \frac{1}{n})

= \frac{1}{2} (1 − \frac{1}{n}) .

Intuitively, as n gets larger and larger, these ordinates sets exhaust more and more of the ordinate set of f and we guess that ∫_{0}^{1} f = \frac{1}{2}. To make this idea rigorous, we use step functions. Since \frac{1}{2} (1 − \frac{1}{n}) − \frac{1}{n} < ∫ 1_{\frac{1}{n}} f, we can find a step function s with s < f and ∫ 1_{\frac{1}{n}} s ≥ \frac{1}{2} (1 − \frac{1}{n}) − \frac{1}{n}. By extending s to take the value 0 on [0, \frac{1}{n}), we’ve found a step function s̃ less than f with

∫_{0}^{1} s̃ ≥ \frac{1}{2} (1 − \frac{1}{n}) − \frac{1}{n} = \frac{1}{2} − \frac{3}{2n}.

Similarly, we can find a step function t with t > f on [\frac{1}{n}, 1] with ∫_{\frac{1}{n}} t < \frac{1}{2} (1 − \frac{1}{n}) + \frac{1}{n}. By extending t to take the value 1 on [0, \frac{1}{n}), we’ve found a step function t̃, greater than f with

∫_{0}^{1} t̃≤ \frac{1}{2} (1 − \frac{1}{n}) + \frac{1}{n} + \frac{1}{n} = \frac{1}{2} + \frac{3}{2n}.

Since \frac{1}{2} is the only number that lies between \frac{1}{2} − \frac{3}{2n} and \frac{1}{2} + \frac{3}{2n} for every n, we conclude that f is integrable on [0, 1] and ∫_{0}^{1} f = \frac{1}{2}.

Problem 5. Short answer. [2 points each] Do three out of four. Or do all four for a bonus. (a) Find a rational number and an irrational number in the interval (0, \frac{1}{2}).

Answer: \frac{1}{4} is a rational number, and \frac{\sqrt{3}}{4} is an irrational number, in (0, \frac{1}{2}). The number \frac{\sqrt{3}}{4} is irrational since \sqrt{3} is irrational and \frac{1}{4} is rational and the product of an irrational number and a nonzero rational number is irrational. The number \frac{\sqrt{3}}{4} lies in (0, \frac{1}{2}) since

0 < \sqrt{3} < \sqrt{4} ⇒ 0 < \frac{\sqrt{3}}{4} ≤ \frac{\sqrt{4}}{2} = \frac{1}{2}.

(b) If possible, find a left inverse and a right inverse of the function f : ℕ → ℕ given by f(n) = n + 2. If either is not possible, be sure to say so.

Answer: Since f is injective, it has a left inverse. Since f is not surjective (the number 1 is not in the image of f) it does not have a right inverse. The function g : ℕ → ℕ given by

g(n) = \begin{cases}n − 2 & if n ≥ 3 \\ 8 & if n = 1, 2 \end{cases}

is a left inverse of f.

Problem 5. Continued.

(c) Compute \underline{I}(f) and \overline{I}(f) for f : [0, 1] → ℝ given by f(x) = \begin{cases}1 + x & if x ∈ ℚ, \\ 2 − x & if x ∉ ℚ. \end{cases}

Answer:

Here the lines {(x, y) : y = 1 + x} and {(x, y) : y = 2 − x} are sketched.

Any step function less than f has an ordinate set contained in the region pictured below, and whose area is less than, but can be as close as desired, to the area of this region. Therefore \underline{I}(f) is the area of this region, which is \frac{5}{4}.

Any step function greater than f has an ordinate set containing the region pictured below and whose area is greater than, but can be as close as desired, to the area of this region. Therefore \overline{I}(f) is the area of this region, which is \frac{7}{4}.

Problem 5. Continued.

(d) Pictured below is the graph of an increasing function g.

Let {x_{0}, x_{1}, …, x_{n}} be a partition of the interval [0, 8] into n subintervals of equal length and let

A = ∑_{k=1}^{n} g(x_{k−1})(x_{k} − x_{k−1}) and B = ∑_{k=1}^{n} g(x_{k})(x_{k} − x_{k−1}).

Note that

A ≤ ∫_{0}^{8} g ≤ B.

How large must n be in order for B − A ≤ \frac{1}{10}. Answer: The difference

B − A = (g(x_{n}) − g(x_{0})) (\frac{8}{n}) = (12 − 2) (\frac{8}{n}) = \frac{80}{n}.

Setting

\frac{80}{n} ≤ \frac{1}{10} ⇒ n ≥ 800.