14.3. Problems

(1) Estimate 104∑ k=1

√ k by interpreting it as a Riemann sum for an appropriate integral.

(2) Let f(x) = x3 + x, let n be an arbitrary natural number, and let P = (x0, x1, . . . xn) be a regular partition of the interval [0, 2] into n subintervals. (Note: “arbitrary” means “unspecified”.) For each k between 1 and n let ck be the midpoint of the k

th subinterval [xk−1, xk]. (a) Find the width ∆xk of each subinterval. (b) Find xk for each k = 0, . . . , n. (c) Find ck for each k = 1, . . . , n. (d) Find the corresponding Riemann midpoint sum

∑n k=1f(ck)∆xk. Simplify the expres-

sion and put it in the form a+ b/n+ c/n2 + . . . . (e) Find the limit of the Riemann sums in part (d) as n→∞. (f) Compute

∫ 2 0 f using the fundamental theorem of calculus

(g) What is the smallest number of subintervals we can use so that the Riemann sum found in (d) approximates the true value of the integral found in (f) with an error of less than 10−5?

(3) Let f(x) = −12x + 3 2 for −1 ≤ x ≤ 3. Partition the interval [−1, 3] into n subintervals of

equal length. Write down the corresponding right approximating sum Rn. Show how this expression can be simplified to the form a+ bn for appropriate numbers a and b. Take the

limit of this expression as n gets large to find the value of ∫ 3 −1 f(x) dx. Check your answer

in two different ways: using a geometrical argument and using the fundamental theorem of calculus.

(4) Let f(x) = x2 + 1 for 0 ≤ x ≤ 3. Partition the interval [0, 3] into n subintervals of equal length. Write down the corresponding right approximating sum Rn. Show how this expression can be simplified to the form a + bcn +

d cn2

for appropriate numbers a, b, c,

and d. Take the limit of this expression as n gets large to find the value of ∫ 3 0 f(x) dx.

Check your answer using the fundamental theorem of calculus.

(5) Define functions f , g, and h as follows:

h(x) =

{ 1, for 0 ≤ x ≤ 2 x, for 2 < x ≤ 4.

g(x) =

∫ x2 x

h(t) dt for 0 ≤ x ≤ 2

f(x) =

∫ x 0 g(t) dt for 0 ≤ x ≤ 2.

(a) For each of the functions h, g, and f answer the following questions: (i) Where is the function continuous? differentiable? twice differentiable? (ii) Where is the function positive? negative? increasing? decreasing? concave up?

concave down? (iii) Where are the x-intercepts? maxima? minima? points of inflection?

(b) Make a careful sketch of the graph of each of the functions. (c) What is the moral of this problem? That is, what do these examples suggest about

the process of integration in general?

14.3. PROBLEMS 103

Hints for solution. When working with the first function h it is possible to get the “right answers” to questions (i)—(iii) but at the same time fail to give coherent reasons for the assertions made. This part of the problem is meant to encourage paying attention to the precise definitions of some of the terms. Indeed, the correct answers will vary from text to text. Some texts, for instance, distinguish between functions that are increasing and those that are strictly increasing. Other texts replace these terms by nondecreasing and increasing, respectively. Some texts define concavity only for functions which are twice differentiable; others define it in terms of the first derivative; still others define it geometrically.

This first part of the problem also provides an opportunity to review a few basic facts: differentiability implies continuity; continuity can be characterized (or defined) in terms of limits; and so on.

Unraveling the properties of the second function g is rather harder. Try not to be put off by the odd looking definition of g. The crucial insight here is that by carrying out the indicated integration it is possible to express g, at least piecewise, as a polynomial. From a polynomial expression it is a simple matter to extract the required information. Impatience at this stage is not a reliable friend. It is not a good idea to try to carry out the integration before you have thought through the problem and discovered the necessity of dividing the interval into two pieces. It may be helpful to compute the values of g at x = 1.0, 1.1, 1.2, . . . , 1.9, 2.0. Notice that about midway in these computations something odd happens. What is the precise point p where things change? Eventually one sees that g too is expressible as one polynomial on [0, p] and as another polynomial on (p , 2]. Once g has been expressed piecewise by polynomials it is possible to proceed with questions (i)–(iii). To determine whether g is continuous at p, compute the right- and left-hand limits of g there.

The question of the differentiability of g is subtle and deserves some serious thought. It may be tempting to carry over the format of continuity argument to decide about the differentiability of g at p. Suppose we compute the right- and left-hand limits of the derivative of g at p and find that they are not equal. Can we then conclude that g is not differentiable at p? At first one is inclined to say no, that all we have shown is that the derivative of g is not continuous at p, which does not address the issue of the existence of g′(p). Interestingly enough, it turns out that the only way in which a differentiable function F can fail to be continuously differentiable at a point a is for either the right- or left-hand limit of F ′(x) to fail to exist at a. The crucial result, which is a bit hard to find in beginning calculus texts, is proposition 14.1.3. Thus when we discover that a function F is differentiable at all points other than a, and that the limits limx→a− F

′(x) and limx→a+ F

′(x) both exist but fail to be equal, there is only one possible explanation: F fails to be differentiable at a.

After finding a piecewise polynomial expression for g, another difficulty arises in deter- mining whether g is concave up. It is easy to see that g is concave up on the intervals (0, p) and (p , 2). But this isn’t enough to establish the property for the entire interval (0, 2). In fact, according to the definition of concavity given in many texts g is not concave up. Why? Because, according to Finney and Thomas (see [2], page 237), for example, concav- ity is defined only for differentiable functions. A function is concave up on an interval only if its derivative is increasing on the interval. So if our function g fails to be differentiable at some point it can not be concave up. On the other hand, under any reasonable geometric definition of concavity g certainly is concave up on (0, 2)—although it is a bit hard to show. The solution to this dilemma is straightforward: pick a definition and stick to it.

Analysis of the last function f proceeds pretty much as for g. One new wrinkle is the difficulty in determining where f is positive. The point at which f changes sign is a root


of a fifth degree polynomial. An approximation based either on the intermediate value theorem or Newton’s method goes smoothly.

As with g, conclusions concerning the concavity of f may differ depending on the defi- nitions used. This time both a geometrical definition and one based on the first derivative lead to one conclusion while a definition based on the second derivative leads to another.

Finally, for part (c) does it make any sense to regard integration as a “smoothing” operation? In what way?

(6) Show that if f is continuous, then∫ x 0 f(u)(x− u) du =

∫ x 0

∫ u 0 f(t) dt du.

Hint. What can you say about functions F and G if you know that F ′(x) = G′(x) for all x and that F (x0) = G(x0) at some point x0?

(7) Let f be a continuous function and a < b. Show that

∫ b a f(−x) dx =

∫ −a −b

f(x) dx.

Hint. Show that if F is an antiderivative of f , then the function G : x 7→ −F (−x) is an antiderivative of the function g : x 7→ f(−x).

(8) Let a < b, f be a continuous function defined on the interval [a, b], and g be the function

defined by g(t) =

∫ b a

(f(x) − t)2 dx for t in R. Find the value for t at which g assumes a

minimum. How do you know that this point is the location of a minimum (rather than a maximum)?

(9) Let λ be a positive constant. Define F (x) =

∫ λx x


t dt for all x > 0. Without mentioning

logarithms show that F is a constant function.

(10) Without computing the integrals give a simple geometric argument that shows that the

sum of ∫ 1 0

√ x dx and

∫ 1 0 x

2 dx is 1. Then carry out the integrations.


14.4. Answers to Odd-Numbered Exercises

(1) 1, 3, 2, 4, 121, 5

(3) 0, 1, 1

1 + x2 , π


(5) 2, 7, 5, 2, 67

(7) 0, n− 1, 1, n, 30, 15, 2 (9) 9

(11) 3

(13) (a) −∞, 2 (b) −∞, 1 (c) −∞, 2

(15) −1 (17) −7 (19) −1, 1, −

√ 3, 0,

√ 3. ∞

(21) 2t cos t

(23) 2, e

(25) 200

(27) 8, 35

(29) −∞, −1 (31) ln 2

(33) 8

(35) 4

(37) 8, 5

(39) π


(41) 4

(43) 3, 5, 2

(45) 4

(47) −38 (49) −3 (51)

√ 3

(53) 4, 3

(55) 5


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