Math 432 - Real Analysis II Solutions to Test 2

Instructions: On a separate sheet of paper, answer the below questions as completely and clearly as possible. Cite theorems when appropriate.

Question 1. Suppose that f and g are continuous functions on [a, b] and that

Z b Z b f(x) dx = g(x) dx. a a Prove that there exists a c ∈ [a, b] such that f(c) = g(c). As a hint, you may want to consider using the Intermediate Value Theorem for Interals.

Solution 1. Since f and g are continuous, then so is f − g. Applying the Intermediate Value Theorem for Integrals, we have that there exists some c ∈ [a, b] such that

1 Z b Z b Z b f(c) − g(c) = f(x) − g(x) dx = f(x) dx − g(x) dx = 0. b − a a a a Thus, f(c) = g(c).

Question 2. Consider the function f deﬁned by   0 if x < 0 f(x) = x if 0 ≤ x ≤ 1  4 if x > 1

(a) Sketch a graph of f(x). For which values of x is f(x) continuous? discontinuous? (b) Let Z x F (x) = f(t) dt. 0 Compute F (x) for all values of x (including when x < 0). Write your final answer in piecewise form. (c) Sketch the graph of F (x). (d) Identify x values for which F (x) is discontinuous. (e) Identify x values for which F is non-differentiable. For x values where F is differentiable, what is F 0(x)?

Solution 2.

(a) f is continuous for all x 6= 1. (b) Computing, we have that  0 if x < 0  1 2 F (x) = 2 x if 0 ≤ x < 1  1 2 + 4(x − 1) if x ≥ 1 (c) (d) Note that F is continuous for all values of x. Of course, this is not surprising since it is guaranteed by the FTC.

1 (e) F is diﬀerentiable at all x except for x = 1. For all x 6= 1, we have that F 0(x) = f(x).

Question 3. Consider the function   1 if − 1 ≤ x < 0 f(x) = 0 if x = 0  1 if 0 < x ≤ 1

deﬁned on [−1, 1].

(a) Sketch a graph of f(x). (b) Use (and cite) theorems from class to prove that f(x) is integrable on [−1, 1]. Compute this integral. (c) Now, use Cauchy’s ε − P criterion to show that f is integrable. To do this, for any ε > 0, ﬁnd a partition P of [−1, 1] such that U(f, P ) − L(f, P ) < ε. Note: Your partition (which you will give as some {−1 = t0 < t1 < t2 ··· < tn = 1}) should be in terms of ε. (d) Compute Z x F (x) = f(t) dt −1 for −1 ≤ x ≤ 1.

(e) Notice that it is not true that F 0(x) = f(x) for all x ∈ [−1, 1]. Why is this not a violation of the Fundamental Theorem of Calculus II?

Solution 3.

(a) (b) Notice that f(x) is equal to the constant function 1 except at a single point. Thus, using the theorem that says that a function diﬀering from an integrable function on only a ﬁnite number of points is integrable. Thus, f is integrable. Furthermore, the integral of f is equal to the integral of the other function. So, we have that Z 1 Z 1 f(x) dx = 1 dx = 2. −1 −1 (c) Given ε > 0, consider the partition P = {−1, −ε/3, ε/3, 1}. On the subintervals, we have that M(f, [tk−1, tk]) = 1. However, we have that

m(f, [−1, −ε/3]) = 1

m(f, [−ε/3, ε/3]) = 0 m(f, [ε/3, 1]) = 1. . Computing the Darboux sums, we have that

U(f, P ) = 2

and that L(f, P ) = 1 · (−ε/3 − (−1)) + 0 · (ε/3 − (−ε/3)) + 1 · (1 − ε/3) = 2 − 2ε/3. Thus, U(f, P ) − L(f, P ) = 2 − (2 − 2ε/3) = 2ε/3 < ε.

2 (d) Since f(x) is the same as the constant function 1 for all x 6= 1, we have that Z x Z x F (x) = f(t) = 1 dt = x + 1. −1 −1

(e) Notice that F is always diﬀerentiable and that, in particular F 0(0) = 1 6= f(0). This is not a contra- diction, because the FTC II says that if f is continuous at c, then F 0(c) = f(c). However, this is not satisﬁed at c = 0, so the theorem does not apply.

Question 4. Consider the function x if x ∈ f(x) = Q x2 if x 6∈ Q deﬁned on [0, 1]

(a) Let P = {0 = t0 < t1 < ··· < tn = 1} be any partition. For this f, compute M(f, [tk−1, tk]) and m(f, [tk−1, tk]). Be sure to justify your answers. (b) Using (a), compute U(f) and L(f). (c) Use your result from (b) to prove that f is not integrable.

Solution 4.

2 (a) We have that M(f, [tk−1, tk]) = tk and that m(f, [tk−1, tk]) = tk−1. To see, this, we make an argument similar to one in a homework assignment. First, notice that it is also true that tk = M(x, [tk−1, tk]) and 2 2 that tk−1 = m(x , [tk−1, tk]). (b) Computing Darboux sums, we get that

n n X X U(f, P ) = M(f, [tk−1, tk])(tk − tk−1) = M(x, [tk − tk−1]). k=1 k=1 Similarly, we have that

n n X X 2 L(f, P ) = m(f, [tk−1, tk])(tk − tk−1) = M(x , [tk − tk−1])(tk − tk−1). k=1 k=1

Thus, when we compute the upper and lower Darboux integrals, we have that U(f) = U(x) = 1/2 and that L(f) = L(x2) = 1/3. (c) Since L(f) 6= U(f), we have that f is not integrable.