# Mathematics

13.1.3. Definition. Let J = [a, b] be a fixed interval in the real line and P = (x0, x1, . . . , xn) be n+ 1 points of J . Then P is a partition of the interval J if:

(1) x0 = a, (2) xn = b, and (3) xk−1 < xk for k = 1, 2, . . . , n.

We denote the length of the kth subinterval by ∆xk; that is, ∆xk = xk − xk−1. A partition P = (x0, x1, . . . , xn) is regular if all the subintervals [xk−1, xk] have the same length. In this case

∆x1 = ∆x2 = · · · = ∆xn and we write ∆x for their common value.

13.1.4. Notation. Let f be a bounded function defined on the interval [a, b] and P = (x0, x1, . . . , xn) be a partition of [a, b]. Then we define

R(P ) := n∑ k=1

f(xk) ∆xk (13.1)

L(P ) :=

n∑ k=1

f(xk−1) ∆xk (13.2)

M(P ) := n∑ k=1

f ( (12(xk−1 + xk)

) ∆xk. (13.3)

These are, respectively, the right, left, and midpoint sums of f associated with the par- tition P . If P is a regular partition of [a, b] consisting of n subintervals, then we may write Rn for R(P ), Ln for L(P ), and Mn for M(P ).

13.1.5. Definition. The average value of a function f over an interval [a, b] is 1

b− a

∫ b a f .

89

90 13. THE RIEMANN INTEGRAL

13.2. Exercises

(1)

5∑ k=1

k2 = .

(2)

10∑ k=3

4 = .

(3) 100∑ k=1

k(k − 3) = .

(4) 200∑ m=1

m3 − 199∑ m=1

m3 = .

(5)

4∑ k=1

(−1)kkk = .

(6) Let ak = 2 k for each k. Then

8∑ k=3

(ak − ak−1) = .

(7) 50∑ k=3

1

k2 − k =

12

a where a = . Hint. Find numbers p and q such that

1

k2 − k =

p

k − 1 − q k

.

(8) Express 1 + 1

3 +

1

9 +

1

27 +

1

81 +

1

243 in summation notation. Answer:

a∑ k=0

bk where

a = and b = .

(9) 4∑

k=1

(k − 1) k (k + 1) = .

(10) 5∑

k=0

3k+4 = b∑

j=a

3j where a = and b = .

(11) 60∑ k=7

1

3k−2 =

a∑ j=−2

1

3j+b where a = and b = .

(12) 70∑

j=−3

1

5j−7 =

61∑ i=a

1

5i+b where a = and b = .

(13)

18∑ j=−4

1

2j+3 =

7∑ k=a

2k−b where a = and b = .

(14) 60∑ k=4

1

k2 − 1 =

a

3660 where a = . Hint. Write

1

k2 − 1 as the difference of two

simpler fractions.

13.2. EXERCISES 91

(15) 34∑ k=2

1

k2 + 2k =

a

2520 where a = . Hint. Write

1

k2 + 2k as the difference of two

simpler fractions.

(16) Let f(x) = x2 on the interval [0, 4] and let P = (0, 1, 2, 4). Find the right,,left, and midpoint sums of f associated with the partition P .

Answer: R(P ) = ; L(P ) = ; and M(P ) = a

2 where a = .

(17) Let f(x) = x3−x on the interval [−2, 3] and let P = (−2, 0, 1, 3). Find the right, left, and midpoint sums of f associated with the partition P .

Answer: R(P ) = ; L(P ) = ; and M(P ) = a

8 where a = .

(18) Let f(x) = 3 − x on the interval [0, 2] and let Pn be the regular partition of [0, 2] into n subintervals. Then

(a) Rn = a+ b

n where a = and b = .

(b) Ln = c+ d

n where c = and d = .

(c) ∫ 2 0 f = .

(19) Let f(x) = 2x− 3 on the interval [0, 4] and let Pn be the regular partition of [0, 4] into n subintervals. Then

(a) Rn = a+ b

n where a = and b = .

(b) Ln = c+ d

n where c = and d = .

(c) ∫ 4 0 f = .

(20) Let f(x) = x − 2 on the interval [1, 7] and let Pn be the regular partition of [1, 7] into n subintervals. Then

(a) Rn = a+ b

n where a = and b = .

(b) Ln = c+ d

n where c = and d = .

(c) ∫ 7 1 f = .

(21) If

∫ e 1

lnx dx = 1 and

∫ e2 1

lnx dx = 1 + e2, then

∫ e2 e

lnx dx = .

(22) Suppose that

∫ 17 −10

f = 3,

∫ 8 −7 f = 7,

∫ 1 −3 f = −1,

∫ 8 −3 f = 4,

∫ 2 −1 f = 5,

∫ 17 −1

f = 6, and∫ 2 1 f = 1. Then

∫ −7 −10

f = .

(23) For what value of x is

∫ √x 4

f(t) dt sure to be 0? Answer: .

(24) Suppose

∫ 3 −2 f(x) dx = 8. Then

∫ −2 3

f(Ξ) dΞ = .

92 13. THE RIEMANN INTEGRAL

(25) Find the value of the integral

∫ 3 −3

√ 9− x2 dx by regarding it as the area under the graph

of an appropriately chosen function and using an area formula from plane geometry. Answer: .

(26) Find the value of the integral

∫ 2 −2

(4− |x|) dx by regarding it as the area under the graph

of an appropriately chosen function and using area formulas from plane geometry. Answer: .

(27) Let a > 0. Then

∫ a 0

( √ a2 − x2 − a + x) dx = 1

b ap(c − 2) where b = , p = ,

and c = . Hint. Interpret the integral as an area.

(28) If the average value of a continuous function f over the interval [0, 2] is 3 and the average

value of f over [2, 7] is 4, then the average value of f over [0, 7] is a

7 where a = .

(29) Let f(x) = |2− |x− 3| |. Then ∫ 8 0 f(x) dx = .

(30) Let f(x) =

{ 2 + √

2x− x2, for 0 ≤ x ≤ 2 4− x, for x > 2

. Then∫ 2 0 f(x) dx = a+

π

b where a = and b = ;∫ 4

0 f(x) dx = c+

π

d where c = and d = ; and∫ 6

1 f(x) dx = p+

π

q where p = and q = .

(31) Suppose

∫ 3 0 f(x) dx = 4,

∫ 5 2 f(x) dx = 5, and

∫ 3 2 f(x) dx = −1. Then

∫ 2 0 f(x) dx= ,∫ 1

0 f(x + 2) dx = ,

∫ 2 0

( f(x) + 2

) dx = ,

∫ 5 2 f(x − 2) dx = ,∫ 5

0 f(x) dx = ,

∫ 7 5

5f(x− 2) dx = , and ∫ 3 −2 f(x+ 2) dx = .

(32)

∫ 4 −1

(|x|+ |x− 2|) dx = .

(33)

∫ 3 0

(|x− 1|+ |x− 2|) dx = .

13.3. PROBLEMS 93

13.3. Problems

(1) Prove proposition 13.1.1.

(2) Prove proposition 13.1.2.

(3) Show that

n∑ k=1

2−k = 1−2−n for each n. Hint. Let sn = n∑ k=1

2−k and consider the quantity

sn − 12sn.

(4) Let f(x) = x3 + x for 0 ≤ x ≤ 2. Approximate ∫ 2 0 f(x) dx using the midpoint sum. That

is, compute, and simplify, the Riemann sum Mn for arbitrary n. Take the limit as n→∞ of Mn to find the value of

∫ 2 0 f(x) dx. Determine the smallest number of subintervals that

must be used so that the error in the approximation Mn is less than 10 −5.

(5) Without evaluating the integral show that

7

4 ≤ ∫ 2 1/4

( 4

3 x3 − 4×2 + 3x+ 1

) dx ≤ 3.

(6) Let f(x) = x2 sin 1

x if 0 < x ≤ 1 and f(0) = 0. Show that

∣∣∣∣∫ 1 0 f

∣∣∣∣ ≤ 13. (7) Suppose that a < b. Prove that

∫ b a

( f(x)− c

)2 dx is smallest when c is the average value

of f over the interval [a, b].

(8) Show that if f is a continuous function on [a, b], then∣∣∣∣∫ b a f(x) dx

∣∣∣∣ ≤ ∫ b a |f(x)| dx.

Hint. Suppose that d is a positive number and we wish to prove that |c| < d. All we need to do is establish two things: that c < d and that −c < d.

(9) Show that 1 ≤ ∫ 1 0 ex

2 dx ≤ e+ 1

2 . Hint. Examine the concavity properties of the curve

y = ex 2 .

(10) Let 0 ≤ x ≤ 1. Apply the mean value theorem to the function f(x) = ex over the interval [0, x] to show that the curve y = ex lies between the lines y = 1 + x and y = 1 + 3x whenever x is between 0 and 1. Use this result to find useful upper and lower bounds for

the value of ∫ 1 0 e

x dx (that is, numbers m and M such that m ≤ ∫ 1 0 e

x dx ≤M).

(11) Show that

∫ b a

(∫ d c f(x) g(y) dy

) dx =

(∫ b a f

)(∫ d c g

) .

(12) Without evaluating the integral show that

π

3 ≤ ∫ π 0

sinx dx ≤ 5π 6 .

(13) Consider the function f(x) = x2 + 1 defined on the closed interval [0, 2]. For each natural number n let Pn = (x0, x1, . . . , xn) be a regular partition of the interval [0, 2] into n subintervals. Denote the length of the kth subinterval by ∆xk. (Thus for a regular partition ∆x1 = ∆x2 = · · · = ∆xn.)

Definition. Let Pn be a regular partition of [0, 2] as above. For each k between 1 and n let ak be the point in the k

th subinterval [xk−1, xk] where f has its smallest value and bk

94 13. THE RIEMANN INTEGRAL

be the point in [xk−1, xk] where f has its largest value. Then let

L(n) = n∑ k=1

f(ak)∆xk and U(n) = n∑ k=1

f(bk)∆xk.

The number L(n) is the lower sum associated with the partition P and U(n) is the upper sum associated with P .

(a) Let n = 1. (That is, we do not subdivide [0, 2].) Find P1, ∆x1, a1, b1, L(1), and

U(1). How good is L(1) as an approximation to ∫ 2 0 f?

(b) Let n = 2. Find P2. For k = 1, 2 find ∆xk, ak, and bk. Find L(2) and U(2). How

good is L(2) as an approximation to ∫ 2 0 f?

(c) Let n = 3. Find P3. For k = 1, 2, 3 find ∆xk, ak, and bk. Find L(3) and U(3). How

good is L(3) as an approximation to ∫ 2 0 f?

(d) Let n = 4. Find P4. For k = 1, 2, 3, 4 find ∆xk, ak, and bk. Find L(4) and U(4). How

good is L(4) as an approximation to ∫ 2 0 f?

(e) Let n = 8. Find P8. For k = 1, 2, . . . , 8 find ∆xk, ak, and bk. Find L(8) and U(8).

How good is L(8) as an approximation to ∫ 2 0 f?

(f) Let n = 20. Find P20. For k = 1, 2, . . . , 20 find ∆xk, ak, and bk. Find L(20) and

U(20). How good is L(20) as an approximation to ∫ 2 0 f?

(g) Now let n be an arbitrary natural number. (Note: “arbitrary” means “unspecified”.) For k = 1, 2, . . . , n find ∆xk, ak, and bk. Find L(n) and U(n). Explain carefully why

L(n) ≤ ∫ 2 0 f ≤ U(n). How good is L(n) as an approximation to

∫ 2 0 f?

(h) Suppose we wish to approximate ∫ 2 0 f by L(n) for some n and have an error no greater

than 10−5. What is the smallest value of n that our previous calculations guarantee will do the job?

(i) Use the preceding to calculate ∫ 2 0 f with an error of less than 10

−5.

13.4. ANSWERS TO ODD-NUMBERED EXERCISES 95

13.4. Answers to Odd-Numbered Exercises

(1) 55

(3) 323, 200

(5) 232

(7) 25

(9) 90

(11) 51, 7

(13) −15, 6 (15) 979

(17) 48, −12, 93 (19) (a) 4, 16

(b) 4, −16 (c) 4

(21) e2

(23) 16

(25) 9π

2

(27) 4, 2, π

(29) 9

(31) 5, −1, 9, 4, 10, 30, 10 (33) 5

CHAPTER 14

THE FUNDAMENTAL THEOREM OF CALCULUS

14.1. Background

Topics: Fundamental theorem of Calculus, differentiation of indefinite integrals, evaluation of definite integrals using antiderivatives.

The next two results are versions of the most elementary form of the fundamental theorem of calculus. (For a much more sophisticated version see theorem 46.1.1.)

14.1.1. Theorem (Fundamental Theorem Of Calculus – Version I). Let a belong to an open interval J in the real line and f : J → R be a continuous function. Define F (x) =

∫ x a f for all x ∈ J . Then

for each x ∈ J the function F is differentiable at x and DF (x) = f(x).

14.1.2. Theorem (Fundamental Theorem of Calculus – Version II). Let a and b be points in an open interval J ⊆ R with a < b. If f : J → R is continuous and g is an antiderivative of f on J , then ∫ b

a f = g(b)− g(a) .

The next proposition is useful in problem 5. It says that the only circumstance in which a differentiable function F can fail to be continuously differentiable at a point a is when either the right- or left-hand limit of F ′(x) fails to exist at a.

14.1.3. Proposition. Let F be a differentiable real valued function in some open interval contain- ing the point a. If l := limx→a− F

′(x) and r := limx→a+ F ′(x) both exist, then

F ′(a) = r = l = lim x→a

F ′(x).

Proof. Suppose that F is differentiable on the interval (a − δ, a + δ). For x ∈ (a, a + δ) the mean value theorem guarantees the existence of a point c ∈ (a, x) such that

F (x)− F (a) x− a

= F ′(c).

Taking the limit as x approaches a from the right we get F ′(a) = r. A nearly identical argument yields F ′(a) = l. This shows that F is continuously differentiable at a. �

97

98 14. THE FUNDAMENTAL THEOREM OF CALCULUS

14.2. Exercises

(1) Evaluate lim n→∞

n∑ k=1

(n+ 2k)4

n5 by expressing it as an integral and then using the fundamental

theorem of calculus to evaluate the integral. The integral is 1

c

∫ b a xp dx where a = ,

b = , c = , and p = . The value of the integral is q

r where q =

and r = .

(2) Evaluate lim n→∞

n∑ k=1

1

n+ 3k by expressing it as an integral and then using the fundamental

theorem of calculus to evaluate the integral. The integral is 1

c

∫ b a xp dx where a = ,

b = , c = , and p = . The value of the integral is u

v lnu where u =

and v = .

(3) Evaluate lim n→∞

n∑ k=1

n

n2 + k2 by expressing it as an integral and then using the fundamental

theorem of calculus to evaluate the integral. The integral is

∫ b a f(x) dx where a = ,

b = , and f(x) = . The value of the integral is .

(4) Evaluate lim n→∞

n∑ k=1

n

(2n+ 7k)2 by expressing it as an integral and then using the funda-

mental theorem of calculus to evaluate the integral. The integral is 1

c

∫ b a xp dx where

a = , b = , c = , and p = . The value of the integral is 1

r where

r = .

(5) Evaluate lim n→∞

n∑ k=1

(2n+ 5k)2

n3 by expressing it as an integral and then using the funda-

mental theorem of calculus to evaluate the integral. The integral is 1

c

∫ b a xp dx where

a = , b = , c = , and p = . The value of the integral is r

3 where

r = .

(6) Evaluate lim n→∞

n∑ k=1

(2n+ 4k)2

n3 by expressing it as an integral and then using the funda-

mental theorem of calculus to evaluate the integral. The integral is 1

c

∫ b a xp dx where

a = , b = , c = , and p = . The value of the integral is r

3 where

r = .

(7) Let J = ∫ 5 0

√ 3x dx and let P be the regular partition of [0, 5] into n subintervals. Find

the left, right and midpoint approximations to J determined by P .

Answer: Ln = 5

n

q∑ k=p

√ 15k

n where p = and q = .

Rn = 5

n

s∑ k=r

√ 15k

n where r = and s = .

Mn = 5

n

n∑ k=1

√ tk − u vn

where t = , u = , and v = .

14.2. EXERCISES 99

(8) Evaluate lim n→∞

n∑ k=1

ln(5k + n)− lnn n

by expressing it as an integral and then using the

fundamental theorem of calculus to evaluate the integral. The integral is 1

c

∫ b a f(x) dx

where a = , b = , c = , and f(x) = . The value of the

integral is r

s f(r)− 1 where r = and s = .

(9) Let g(x) =

∫ 1 2 x

3

t3 + 4t+ 4

1 + t2 dt. Then Dg(2) =

a

4 where a = .

(10) Let g(x) = (5 + 7 cos2(2πx) − sin(4πx))−1 and f(x) = ∫ 2 x3 g(t) dt. Then Df(12) = −

1

a where a = .

(11) Let g(x) = (1 + (x4 + 7)1/3)−1/2 and f(x) =

∫ x3 x

g(t) dt. Then Df(1) = 2√ a

where

a = .

(12) Let f(x) =

∫ sinπx x2

dt

1 + t4 . Then Df(2) = a− 4

b where a = and b = .

(13) Let g(x) =

∫ x 0

u− 1 u− 2

du. Then

(a) the domain of g is ( , );

(b) g is increasing on ( , ); and

(c) g is concave down on ( , ).

(14) Solve for x:

∫ x 0

(2u− 1)2 du = 14 3

. Answer: x = .

(15) Solve for x:

∫ x+2 x

u du = 0. Answer: x = .

(16) Find a number x > 0 such that

∫ x 1

(u−1) du = 4. Answer: x = 1+a √ a where a = .

(17) Find

∫ 6 3 f ′(x) dx given that the graph of f includes the points (0, 4), (3, 5), (6,−2), and

(8,−9). Answer: .

(18) Let g(x) =

∫ x 0 xf(t) dt where f is a continuous function. Then

Dg(x) = .

(19) Let f(x) =

∫ x 0

1− t2

3 + t4 dt. Then f is increasing on the interval ( , ) and is

concave up on the intervals ( , ) and ( , ) .

(20) If y =

∫ s 0

√ 2 + u3 du, then

dy

ds = .

(21) If y =

∫ t2 2

cos √ x dx and t ≥ 0, then dy

dt = .

(22) If

∫ x −2 f(t) dt = x2 sin(πx) for every x, then f(1/3) =

π

a +

1

b where a = and b = .

100 14. THE FUNDAMENTAL THEOREM OF CALCULUS

(23) Let f(x) =

∫ x3 x lnx

dt

3 + ln t for x ≥ 1. Then Df(e) = 1

a (b2 − 1) where a = and

b = .

(24) Let f(x) =

∫ ln(x2+1) ln(x+1)4

dt

4 + et for x > 0. Then Df(1) =

1

a where a = .

(25) Let f(x) =

∫ 1 2 x2ex−1

−2

t2

(4 + sinπt)2 dt. Then Df(1) =

3

a where a = .

(26) Let f(x) =

∫ xex2 0

dt

5 + (ln t)2 . Then Df(1) =

a

b where a = and b = .

(27) Let f(x) =

∫ ex3 0

dt

6 + (ln t)2 . Then Df(2) =

6ea

b where a = and b = .

(28) Let f(x) =

∫ ln(x2+3) lnx

dt

3 + et for x ≥ 1. Then Df(2) = −3

a where a = .

(29) Let f(x) = (x2+2x+2)−1 for all x ∈ R. Then the interval on which the curve y = ∫ x 0 f(t) dt

is concave up is ( , ).

(30) lim h→0

1

h

∫ 2+h 2

√ 1 + x2 dx = .

(31) lim λ→0+

∫ 2λ λ

e−xx−1 dx = . Hint: e−x

x = e−x − 1

x +

1

x .

(32) lim x→0

1

x

∫ 1+5x 1

(4− cos 2πt)3 dt = .

(33) lim r→0

1

r

∫ e4r 1

√ 3 +

1

x dx = .

(34) lim u→0

1

u

∫ ln(e2+3u) 2

√ 1 + 2t+ 5t2 dt = aeb where a = and b = .

(35)

∫ 9 1

1

x3/2 = a

3 where a = .

(36)

∫ 12e4x dx = aebx + c where a = , b = , and c is an arbitrary constant.

(37)

∫ 40 cos 5x dx = a sin bx+ c where a = , b = , and c is an arbitrary constant.

(38)

∫ 1 0

4√ 4− x2

dx = a

3 where a = .