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Unformatted text preview: FIRST YEAR CALCULUS
W W L CHEN
c
W W L Chen, 1994, 2005. This chapter is available free to all individuals, on the understanding that it is not to be used for ﬁnancial gain, and may be downloaded and/or photocopied, with or without permission from the author. However, this document may not be kept on any information storage and retrieval system without permission from the author, unless such system is not accessible to any individuals other than its owners. Chapter 7
THE DEFINITE INTEGRAL 7.1. Finite Sums We begin by considering a simple example. Example 7.1.1. Consider the expression 111 1 + + + ... + . 234 57 It is convenient to have a good notation. We may perhaps write
57 i=2 1 i instead, if we recognize that all the numbers in the sum are of the form 1/i, where i = 2, 3, 4, . . . , 57. Note that
57 i=2 1 = i 56 i=1 1 , i+1 so that we can vary the range of summation if we are prepared to vary what we are summing over. On the other hand, note that
57 i=2 1 = i 57 j =2 1 , j so that i and j are “dummy” variables only.
Chapter 7 : The Deﬁnite Integral page 1 of 20 First Year Calculus c W W L Chen, 1994, 2005 Definition. Suppose that m, n ∈ Z and m < n. Then we write
n ai = am + am+1 + am+2 + . . . + an−1 + an .
i=m This is called a ﬁnite sum or a ﬁnite series. Example 7.1.2. We have
5 i=1 1 11 1 1 1 =++ + +. 2 1+i 2 5 10 17 26 Example 7.1.3. We have
7 j =3 1 = j (j + 1) 6 i=2 1 1 1 1 1 1 = + + + +. (i + 1)(i + 2) 12 20 30 42 56 The following result is a simple consequence of the usual rules of addition and multiplication of real numbers. PROPOSITION 7A. Suppose that m, n ∈ Z and m < n. Suppose further that c ∈ R. Then
n n n n n (ai + bi ) =
i=m i=m ai +
i=m bi and
i=m cai = c
i=m ai . Example 7.1.4. Suppose that n ∈ N. Consider the sum
n Sn =
i=1 i = 1 + 2 + 3 + . . . + n. Note that 2Sn = (1 + 2 + 3 + . . . + n) + (n + (n − 1) + (n − 2) + . . . + 1) = (1 + n) + (2 + (n − 1)) + (3 + (n − 2)) + . . . + (n + 1) = n(n + 1), so that
n i=
i=1 n(n + 1) . 2 Example 7.1.5. Suppose that n ∈ N. Consider the sum
n Tn =
i=1 i2 = 1 + 4 + 9 + . . . + n2 . For every i = 1, 2, 3, . . . , n, we have (i + 1)3 − i3 = 3i2 + 3i + 1,
Chapter 7 : The Deﬁnite Integral page 2 of 20 First Year Calculus c W W L Chen, 1994, 2005 so that
n n n n ((i + 1)3 − i3 ) = 3
i=1 i=1 i2 + 3
i=1 i+
i=1 1. In other words, we have (n + 1)3 − 1 = 3Tn + 3Sn + n. It follows that Tn = (n + 1)3 − 1 n (n + 1)3 − (n + 1) n(n + 1) (n + 1)3 − (n + 1) − Sn − = − Sn = − 3 3 3 3 2 (n + 1)(2(n + 1)2 − 2 − 3n) (n + 1)(2n2 + n) n(n + 1)(2n + 1) = = = , 6 6 6 so that
n (1)
i=1 i2 = n(n + 1)(2n + 1) . 6 7.2. An Example Consider the function f (x) = x2 in the interval [−1, 2]. Suppose that we wish to ﬁnd the area bounded by the curve y = f (x) and the lines y = 0, x = −1 and x = 2 (the reader should start drawing a diagram). Unfortunately, our knowledge on areas is restricted to simple geometric shapes, and the area in question cannot be calculated by a simple area formula. So let us try some approximations. Let us ﬁrst break the interval [−1, 2] into shorter intervals in some arbitrary fashion, say [−1, − 1 ], [− 1 , 1 ], [ 1 , 5 ], [ 5 , 2] 2 24 44 4 (the reader should draw all the rectangles discussed below). Consider ﬁrst the interval [−1, − 1 ]. We approximate the area bounded by the curve y = f (x) and the 2 lines y = 0, x = −1 and x = − 1 by rectangles with base [−1, − 1 ] on the line y = 0. Note that 2 2 min f (x) = f − 1 2 = 1 4 and max f (x) = f (−1) = 1. x∈[−1,−1/2] x∈[−1,−1/2] If we draw a rectangle with height 1/4, then this rectangle has area 1/8, clearly an underestimate. If we draw a rectangle with height 1, then this rectangle has area 1/2, clearly an overestimate. Consider next the interval [− 1 , 1 ]. We approximate the area bounded by the curve y = f (x) and the 24 lines y = 0, x = − 1 and x = 1 by rectangles with base [− 1 , 1 ] on the line y = 0. Note that 2 4 24 min f (x) = f (0) = 0 and max f (x) = f − 1 2 = 1 . 4 x∈[−1/2,1/4] x∈[−1/2,1/4] If we draw a rectangle with height 0, then this rectangle has area 0, clearly an underestimate. If we draw a rectangle with height 1/4, then this rectangle has area 3/16, clearly an overestimate.
Chapter 7 : The Deﬁnite Integral page 3 of 20 First Year Calculus c W W L Chen, 1994, 2005 Consider next the interval [ 1 , 5 ]. We approximate the area bounded by the curve y = f (x) and the 44 lines y = 0, x = 1 and x = 5 by rectangles with base [ 1 , 5 ] on the line y = 0. Note that 4 4 44 min f (x) = f 1 4 = 1 16 and max f (x) = f 5 4 = 25 . 16 x∈[1/4,5/4] x∈[1/4,5/4] If we draw a rectangle with height 1/16, then this rectangle has area 1/16, clearly an underestimate. If we draw a rectangle with height 25/16, then this rectangle has area 25/16, clearly an overestimate. Consider ﬁnally the interval [ 5 , 2]. We approximate the area bounded by the curve y = f (x) and the 4 lines y = 0, x = 5 and x = 2 by rectangles with base [ 5 , 2] on the line y = 0. Note that 4 4 min f (x) = f 5 4 = 25 16 and max f (x) = f (2) = 4. x∈[5/4,2] x∈[5/4,2] If we draw a rectangle with height 25/16, then this rectangle has area 75/64, clearly an underestimate. If we draw a rectangle with height 4, then this rectangle has area 3, clearly an overestimate. Now let us return to the area in question, namely the area bounded by the curve y = f (x) and the lines y = 0, x = −1 and x = 2. If we use the smaller of the two rectangles in each instance, then we get the underestimate 1 1 75 87 +0+ + = . 8 16 64 64 If we use the larger of the two rectangles in each instance, then we get the overestimate 1 3 25 21 + + +3= . 2 16 16 4 Clearly these are very far from the truth. This is hardly surprising, as the approximations we have made are very crude indeed. 7.3. The Riemann Integral To get further, we need to be more systematic in our treatment. The following example illustrates the key points of our technique. Example 7.3.1. Consider the function f (x) = x2 in the interval [0, 1]. Suppose that we wish to ﬁnd the area, A say, bounded by the curve y = f (x) and the lines y = 0, x = 0 and x = 1 (the reader should again start drawing a diagram). Let us consider a dissection ∆n : 0 = x0 < x1 < x2 < . . . < xn = 1 of the interval [0, 1], where xi = i/n for every i = 0, 1, 2, . . . , n. For every subinterval [xi−1 , xi ], where i = 1, 2, . . . , n, we have min f (x) = min i−1
n x∈[xi−1 ,xi ] i ≤x≤ n x2 = f i−1 n = (i − 1)2 n2 and max f (x) = max x2 = f i n = i2 . n2
page 4 of 20 x∈[xi−1 ,xi ] Chapter 7 : The Deﬁnite Integral i−1 i n ≤x≤ n First Year Calculus c W W L Chen, 1994, 2005 It follows that the area, Ai say, bounded by the curve y = f (x) and the lines y = 0, x = xi−1 = (i − 1)/n and x = xi = i/n can be approximated below by the area of a rectangle of height (i − 1)2 /n2 and approximated above by the area of a rectangle of height i2 /n2 ; in each case, the base of the rectangle has length xi − xi−1 = 1/n. Hence (xi − xi−1 ) more precisely, (i − 1)2 i2 ≤ Ai ≤ 3 . 3 n n Clearly
n x∈[xi−1 ,xi ] min f (x) ≤ Ai ≤ (xi − xi−1 ) x∈[xi−1 ,xi ] max f (x); A=
i=1 Ai . Now write
n n s(f, ∆n ) =
i=1 (xi − xi−1 ) x∈[xi−1 ,xi ] min f (x) =
i=1 (i − 1)2 n3 and
n n S (f, ∆n ) =
i=1 (xi − xi−1 ) x∈[xi−1 ,xi ] max f (x) =
i=1 i2 . n3 Then it clearly follows that s(f, ∆n ) ≤ A ≤ S (f, ∆n ). By (1), we have
n i=1 (i − 1)2 1 =3 n3 n n (i − 1)2 =
i=1 1 n3 n−1 i2 =
i=0 1 n3 n−1 i2 =
i=1 (n − 1)n(2n − 1) 6n3 and
n i=1 i2 1 =3 n3 n n i2 =
i=1 n(n + 1)(2n + 1) . 6n3 Hence (n − 1)n(2n − 1) n(n + 1)(2n + 1) ≤A≤ . 6n3 6n3 Suppose now that n is very large. In other words, suppose that n → ∞. Then (n − 1)n(2n − 1) 1 → 6n3 3 and 1 n(n + 1)(2n + 1) →. 6n3 3 It follows that we must have A = 1/3. Of course, we know that
1 x2 dx =
0 Chapter 7 : The Deﬁnite Integral 1 . 3
page 5 of 20 First Year Calculus c W W L Chen, 1994, 2005 Suppose next that instead of approximating each Ai by the two rectangles with heights
x∈[xi−1 ,xi ] min f (x) = (i − 1)2 n2 and x∈[xi−1 ,xi ] max f (x) = i2 , n2 2 we simply choose some ξi ∈ [xi−1 , xi ] and approximate Ai by a rectangle of height f (ξi ) = ξi . Then we have the approximation (xi − xi−1 )f (ξi ) = for Ai and the approximation
n 2 ξi n n (xi − xi−1 )f (ξi ) =
i=1 i=1 2 ξi n for A. Clearly (i − 1)2 i2 2 ≤ ξi ≤ 2 , n2 n so that
n s(f, ∆n ) ≤
i=1 2 ξi ≤ S (f, ∆n ). n It follows that (n − 1)n(2n − 1) ≤ 6n3 Hence, for very large n,
n n (xi − xi−1 )f (ξi ) ≤
i=1 n(n + 1)(2n + 1) . 6n3 (xi − xi−1 )f (ξi )
i=1 is a good approximation for A. Definition. Suppose that f (x) is a continuous function in the closed interval [A, B ], where A, B ∈ R and A < B . Suppose further that ∆ : A = x0 < x1 < x2 < . . . < xn = B is a dissection of the interval [A, B ]. Then the sum
n s(f, ∆) =
i=1 (xi − xi−1 ) x∈[xi−1 ,xi ] min f (x) is called the lower Riemann sum of f (x) corresponding to the dissection ∆, and the sum
n S (f, ∆) =
i=1 (xi − xi−1 ) x∈[xi−1 ,xi ] max f (x) is called the upper Riemann sum of f (x) corresponding to the dissection ∆. Suppose further that for every i = 1, . . . , n, we have ξi ∈ [xi−1 , xi ]. Then the sum
n (xi − xi−1 )f (ξi )
i=1 is called a Riemann sum of f (x) corresponding to the dissection ∆.
Chapter 7 : The Deﬁnite Integral page 6 of 20 First Year Calculus c W W L Chen, 1994, 2005 Remarks. (1) It is clear that
x∈[xi−1 ,xi ] min f (x) ≤ f (ξi ) ≤ x∈[xi−1 ,xi ] max f (x). It follows that every Riemann sum is bounded below by the corresponding lower Riemann sum and bounded above by the corresponding upper Riemann sum; in other words,
n s(f, ∆) ≤
i=1 (xi − xi−1 )f (ξi ) ≤ S (f, ∆). (2) It can be shown that for any two dissections ∆ and ∆ of the closed interval [A, B ], we have s(f, ∆ ) ≤ S (f, ∆ ); in other words, a lower Riemann sum can never exceed an upper Riemann sum. (3) Note that we have restricted our attention to continuous functions in the closed interval [A, B ]. This is in fact unnecessary. It is enough to assume that the function f (x) is bounded in the closed interval [A, B ]. However, the deﬁnition of the lower and upper Riemann sums need to be slightly modiﬁed. We shall discuss this more general setting in Section 7.7. Definition. We say that
B f (x) dx = L
A if, given any > 0, there exists a dissection ∆ of [A, B ] such that L − < s(f, ∆) ≤ S (f, ∆) < L + . In this case, we say that the continuous function f (x) is Riemann integrable over the closed interval [A, B ] with integral L. Remark. In other words, if the lower Riemann sums and upper Riemann sums can get arbitrarily close, then their common value is the integral of the function. We state here the following important result. For a formal proof, see Section 7.7. PROPOSITION 7B. Suppose that f (x) is a continuous function in the closed interval [A, B ], where A, B ∈ R and A < B . Then f (x) is Riemann integrable over [A, B ]. Example 7.3.2. Consider the function f (x) = sin x in the closed interval [0, π/2]. Suppose that ∆ : 0 = x0 < x1 < x2 < . . . < xn = is a dissection of the interval [0, π/2], where xi = iπ , 2n i = 0, 1, 2, . . . , n. π 2 Since f (x) = sin x is increasing in [0, π/2], it follows that min f (x) = f (xi−1 ) = sin (i − 1)π 2n and max f (x) = f (xi ) = sin iπ . 2n
page 7 of 20 x∈[xi−1 ,xi ] x∈[xi−1 ,xi ] Chapter 7 : The Deﬁnite Integral First Year Calculus c W W L Chen, 1994, 2005 Hence π s(f, ∆) = (xi − xi−1 ) min f (x) = 2n x∈[xi−1 ,xi ] i=1 and
n n π (i − 1)π = sin 2n 2n i=1 n n−1 π iπ = sin 2n 2n i=0 n−1 sin
i=1 iπ 2n S (f, ∆) =
i=1 (xi − xi−1 ) x∈[xi−1 ,xi ] max f (x) = π 2n n sin
i=1 iπ . 2n Next, note that iπ sin = 2n i=1 Similarly,
n−1 n n i=1 cos((i − 1 ) 2π ) − cos((i + 1 ) 2π ) cos 4π − cos( π + 2n 2n n 2 = π 2 sin 4n 2 sin 4π n π 4n ) . sin
i=1 cos 4π − cos( π − iπ n 2 = 2n 2 sin 4π n π 4n ) . It follows that as n → ∞, we have s(f, ∆) = and S (f, ∆) = Hence
π /2 π 2n cos 4π − cos( π − n 2 2 sin 4π n π 4n ) → cos 0 − cos π = 1, 2 π 2n cos 4π − cos( π + n 2 2 sin 4π n π 4n ) → cos 0 − cos π = 1. 2 sin x dx = 1.
0 7.4. Antiderivatives Our aim is to relate our deﬁnition of the Riemann integral to something more familiar. The ﬁrst step in this direction involves the study of antiderivatives or indeﬁnite integrals. Definition. A function F (x) is called an antiderivative or indeﬁnite integral of a function f (x) in an interval I if F (x) = f (x) for every x ∈ I . Example 7.4.1. Suppose that f (x) = 3x2 . Then for any C ∈ R, the function F (x) = x3 + C is an antiderivative of f (x) on any interval. It follows that there are inﬁnitely many antiderivatives that diﬀer by constants. The next result shows that there are no more. The proof, which depends on the Mean value theorem, is given in Section 7.7. PROPOSITION 7C. Suppose that the function F (x) is an antiderivative of a function f (x) in an interval I . Then every antiderivative of f (x) is of the form F (x) + C , where C ∈ R is a constant.
Chapter 7 : The Deﬁnite Integral page 8 of 20 First Year Calculus c W W L Chen, 1994, 2005 Example 7.4.2. The following table of antiderivatives can be checked for appropriate intervals I : f (x) 0 cos x sec2 x sec x tan x tan x sec x (n + 1)x (n = −1)
n F (x) C sin x + C tan x + C sec x + C log  sec x + C log  sec x + tan x + C x
n+1 f (x) e
x F (x) ex + C − cos x + C − cot x + C − csc x + C − log  csc x + C log  csc x − cot x + C log x + C sin x csc2 x csc x cot x cot x csc x x
−1 +C The next result is crucial in the calculation of antiderivatives. The proof is straightforward, in view of Proposition 5C. PROPOSITION 7D. Suppose that the functions F (x) and G(x) are antiderivatives of functions f (x) and g (x) respectively in an interval I . Suppose further that c ∈ R. Then (a) F (x) + G(x) is an antiderivative of f (x) + g (x) in I ; and (b) cF (x) is an antiderivative of cf (x) in I . Example 7.4.3. Suppose that f (x) = x2 + 2 sin x. We can write f (x) = where g (x) = 3x2 and h(x) = sin x. 1 g (x) + 2h(x), 3 From the table in Example 7.4.2 and with C = 0, the functions G(x) = x3 and H (x) = − cos x are antiderivatives of g (x) and h(x) respectively in any interval. It follows from Proposition 7D that the function F (x) = 1 x3 G(x) + 2H (x) = − 2 cos x 3 3 is an antiderivative of f (x) in any interval, so that it follows from Proposition 7C that every antiderivative of f (x) in any interval is of the form x3 − 2 cos x + C, 3 where C ∈ R. For the sake of convenience, we shall denote any antiderivative of a function f (x) by f (x) dx. Also, we may choose to omit reference to the interval I in question, with the understanding that an appropriate interval I has been chosen.
Chapter 7 : The Deﬁnite Integral page 9 of 20 First Year Calculus c W W L Chen, 1994, 2005 7.5. Fundamental Theorems of the Integral Calculus In this section, we shall ﬁrst discuss a relationship between the Riemann integral and antiderivatives. This relationship enables us to calculate the Riemann integral by simply ﬁnding an antiderivative of the given function. In Section 7.7, we shall establish the following important result. PROPOSITION 7E. (FUNDAMENTAL THEOREM OF THE INTEGRAL CALCULUS) Suppose that f (x) is a continuous function in the closed interval [A, B ], where A, B ∈ R and A < B . Suppose further that the function F (x) is an antiderivative of f (x) in [A, B ]. Then
B f (x) dx = F (B ) − F (A).
A Example 7.5.1. We have
2π (x2 + 2 sin x) dx =
π x3 − 2 cos x 3 2π =
π 8π 3 − 2 cos 2π − 3 π3 − 2 cos π 3 = 7π 3 − 4. 3 Example 7.5.2. We have
π 0 √ 2 (1 − cos x)3/2 1 − cos x sin x dx = 3 π 0 √ 2 2 42 3 /2 3 /2 . = (1 − cos π ) − (1 cos 0) = 3 3 3 Example 7.5.3. We have
4 3 x 2 √ (x − 2)3/2 + 4(x − 2)1/2 dx = 3 x−2 4 3 √ 16 2 − 14 = . 3 Example 7.5.4. We have
π /2 sin3 x cos3 x dx =
0 sin4 x sin6 x − 4 6 π /2 =
0 1 . 12 Example 7.5.5. The argument
1 −1 1 1 dx = − 2 x x 1 −1 = −1 − 1 = −2 is clearly wrong, since the curve is never below the line y = 0 between x = −1 and x = 1. Note that the function 1/x2 is not continuous in the interval [−1, 1], so that the Fundamental theorem of the integral calculus does not apply. Riemann integrals can, in a certain sense, be regarded as antiderivatives. The following result is sometimes known as the second Fundamental theorem of the integral calculus. PROPOSITION 7F. Suppose that f (x) is a continuous function in the closed interval [A, B ], where A, B ∈ R and A < B . Then the function
x F (x) =
A f (t) dt is continuous in the closed interval [A, B ] and diﬀerentiable for every x ∈ (A, B ), with F (x) = f (x).
Chapter 7 : The Deﬁnite Integral page 10 of 20 First Year Calculus c W W L Chen, 1994, 2005 Remark. In some treatments, Propositions 7E and 7F are given in reverse order. Example 7.5.6. We have d dx
x 1 + t2 dt =
0 1 + x2 . Example 7.5.7. We have d dx
1 1 + t2 dt = −
x d dx x 1 + t2 dt = − 1 + x2 .
1 Example 7.5.8. We have d dx
x2 2x d sin(4t + 3) dt = dx
3 x2 C sin(4t3 + 3) dt +
C x2 2x sin(4t3 + 3) dt
C = = d dx d dx sin(4t3 + 3) dt +
C x2 d dx d dx sin(4t3 + 3) dt
2x 2x sin(4t3 + 3) dt −
C sin(4t3 + 3) dt.
C Writing y = x2 and using the Chain rule, we have d dx
x2 C sin(4t3 + 3) dt = y y d dy d sin(4t3 + 3) dt = sin(4t3 + 3) dt dx C dx dy C = 2x sin(4y 3 + 3) = 2x sin(4x6 + 3). Writing u = 2x and using the Chain rule, we have d dx
2x C sin(4t3 + 3) dt = u u d du d sin(4t3 + 3) dt = sin(4t3 + 3) dt dx C dx du C = 2 sin(4u3 + 3) = 2 sin(32x3 + 3). It follows that d dx
x2 sin(4t3 + 3) dt = 2x sin(4x6 + 3) − 2 sin(32x3 + 3).
2x The next two results can be considered to be simple consequences of the Fundamental theorems of the integral calculus. In Section 7.7, we shall discuss how we can establish more general versions of these two results. PROPOSITION 7G. Suppose that f (x) is a continuous function in the closed interval [A, B ], where A, B ∈ R and A < B . Suppose further that C ∈ [A, B ]. Then
B C B f (x) dx =
A A f (x) dx +
C f (x) dx. Chapter 7 : The Deﬁnite Integral page 11 of 20 First Year Calculus c W W L Chen, 1994, 2005 PROPOSITION 7H. Suppose that f (x) and g (x) are continuous function in the closed interval [A, B ], where A, B ∈ R and A < B . Then
B B B (f (x) + g (x)) dx =
A A f (x) dx +
A g (x) dx. Furthermore, for every real number c ∈ R, we have
B B cf (x) dx = c
A A f (x) dx. The following result gives some very crude bound for the Riemann integral. PROPOSITION 7J. Suppose that f (x) is a continuous function in the closed interval [A, B ], where A, B ∈ R and A < B . Suppose further that m ≤ f (x) ≤ M for every x ∈ [A, B ]. Then
B m(B − A) ≤
A f (x) dx ≤ M (B − A). 7.6. Average Values of Functions Suppose that the function f (x) is nonnegative and continuous in the closed interval [A, B ]. Then the Riemann integral
B f (x) dx
A exists and represents the area bounded by the curve y = f (x) and the lines y = 0, x = A and x = B . Consider a rectangle with its base on the xaxis between x = A and x = B and with the same area as the integral. Then its height 1 B−A
B f (x) dx
A must represent the average value of the function f (x) in the interval [A, B ]. Of course, the restriction that f (x) is nonnegative is not necessary and can be removed. Example 7.6.1. The average value of the function sin x in the interval [0, 2π ] is 1 2π
2π sin x dx = 0.
0 Example 7.6.2. The average value of the function sin x in the interval [0, π ] is 1 π
π sin x dx =
0 2 . π Example 7.6.3. The average value of the function x2 in the interval [0, 2] is 1 2
Chapter 7 : The Deﬁnite Integral 2 x2 dx =
0 4 . 3
page 12 of 20 First Year Calculus c W W L Chen, 1994, 2005 Example 7.6.4. The average value of the function sin2 x in the interval [0, 2π ] is 1 2π
2π sin2 x dx =
0 1 . 2 In Section 7.7, we shall establish the following result which shows that the mean value is attained by the function in the interval. PROPOSITION 7K. (MEAN VALUE THEOREM FOR RIEMANN INTEGRALS) Suppose that f (x) is a continuous function in the closed interval [A, B ], where A, B ∈ R and A < B . Then there exists c ∈ (A, B ) such that 1 B−A
B f (x) dx = f (c).
A 7.7. Further Discussion Here we shall study the Riemann integral in a more general setting. Suppose that f (x) is a real valued function bounded in the closed interval [A, B ], where A, B ∈ R and A < B . Suppose further that ∆ : A = x0 < x1 < x2 < . . . < xn = B is a dissection of the interval [A, B ]. Then the sum
n s(f, ∆) =
i=1 (xi − xi−1 ) x∈[xi−1 ,xi ] inf f (x) is called the lower Riemann sum of f (x) corresponding to the dissection ∆, and the sum
n S (f, ∆) =
i=1 (xi − xi−1 ) x∈[xi−1 ,xi ] sup f (x) is called the upper Riemann sum of f (x) corresponding to the dissection ∆. Suppose further that for every i = 1, . . . , n, we have ξi ∈ [xi−1 , xi ]. Then the sum
n (xi − xi−1 )f (ξi )
i=1 is called a Riemann sum of f (x) corresponding to the dissection ∆. Note that it is important to use the inﬁmum and supremum instead of minimum and maximum, as the latter may not exist, while the former always exist, since the function f (x) is bounded in [A, B ]. In fact, if the function f (x) is continuous in the interval [A, B ], then f (x) is bounded in [A, B ] in view of Proposition 4E, and attains a minimum and maximum in any closed subinterval of [A, B ] in view of Proposition 4C. This means that the inﬁmum is a minimum, while the supremum is a maximum. It follows that the above deﬁnitions represent a generalization of our earlier deﬁnitions. The following three results are easy to establish, and are left as exercises for the interested reader. PROPOSITION 7L. Suppose that f (x) is a real valued function bounded in the closed interval [A, B ], where A, B ∈ R and A < B . Then for any dissection ∆ of the interval [A, B ], we have s(f, ∆) ≤ S (f, ∆).
Chapter 7 : The Deﬁnite Integral page 13 of 20 First Year Calculus c W W L Chen, 1994, 2005 PROPOSITION 7M. Suppose that f (x) is a real valued function bounded in the closed interval [A, B ], where A, B ∈ R and A < B . Suppose further that ∆ and ∆ are two dissections of the interval [A, B ] satisfying ∆ ⊆ ∆ ; in other words, every dissection point of ∆ is also a dissection point of ∆ . Then s(f, ∆) ≤ s(f, ∆ ) and S (f, ∆ ) ≤ S (f, ∆). Combining the two results above, we have the following result alluded to in our earlier discussion. PROPOSITION 7N. Suppose that f (x) is a real valued function bounded in the closed interval [A, B ], where A, B ∈ R and A < B . Then for any dissections ∆ and ∆ of the interval [A, B ], we have s(f, ∆ ) ≤ S (f, ∆ ). In other words, a lower Riemann sum can never exceed an upper Riemann sum. Definition. Suppose that f (x) is a real valued function bounded in the closed interval [A, B ], where A, B ∈ R and A < B . The quantity (2) L+ (f ) = inf S (f, ∆)
∆ is called the upper integral of f (x) in [A, B ], and the quantity (3) L− (f ) = sup s(f, ∆)
∆ is called the lower integral of f (x) in [A, B ]. Here the inﬁmum and supremum are taken over all dissections ∆ of the interval [A, B ]. Furthermore, if L+ (f ) = L− (f ), then we say that the function f (x) is Riemann integrable over the interval [A, B ], and denote by
B L=
A f (x) dx the common value of the upper and lower integrals of f (x) in [A, B ]. Note that the existence of the upper and lower integrals are guaranteed by the boundedness of the function f (x) in the interval [A, B ]. Example 7.7.1. It is not easy to ﬁnd a function that is not Riemann integrable. Here, we shall give one, but the proof depends on some rather deep result on rational and irrational numbers. Consider the function f (x) = 1 0 if x is rational, if x is irrational. It is well known that in any open interval, there are rational numbers and irrational numbers. It follows that in any interval [α, β ], where α < β , we have
x∈[α,β ] inf f (x) = 0 and x∈[α,β ] sup f (x) = 1. It follows that for every dissection ∆ of [0, 1], we have s(f, ∆) = 0 and S (f, ∆) = 1. Hence f (x) is not Riemann integrable over the closed interval [0, 1].
Chapter 7 : The Deﬁnite Integral page 14 of 20 First Year Calculus c W W L Chen, 1994, 2005 Remark. Note that the Riemann integral never exceeds any upper Riemann sum and is never less than any lower Riemann sum. A consequence of this simple observation is Proposition 7J. We now need to show that our deﬁnition of Riemann integrability here agrees with our earlier deﬁnition in the case of continuous functions. We ﬁrst establish the following result. PROPOSITION 7P. Suppose that f (x) is a real valued function bounded in the closed interval [A, B ], where A, B ∈ R and A < B . Then f (x) is Riemann integrable over the interval [A, B ] if and only if, given any > 0, there exists a dissection ∆ of [A, B ] such that S (f, ∆) − s(f, ∆) < . Proof. Suppose ﬁrst of all that f (x) is Riemann integrable over the interval [A, B ]. Then L+ (f ) = L− (f ) = L. Let > 0 be given. In view of (2) and (3), there exist dissections ∆ and ∆ of [A, B ] such that S (f, ∆ ) < L+ (f ) + 2 and s(f, ∆ ) > L− (f ) − . 2 Let ∆ = ∆ ∪ ∆ ; in other words, ∆ contains precisely all the dissection points of both ∆ and ∆ . Then it follows from Proposition 7M that S (f, ∆) ≤ S (f, ∆ ) and s(f, ∆) ≥ s(f, ∆ ). Combining the above and noting Proposition 7L, we have L− so that S (f, ∆) − s(f, ∆) < . On the other hand, it is clear from Proposition 7N that L− (f ) ≤ L+ (f ). Suppose on the contrary that f (x) is not Riemann integrable over the interval [A, B ]. Then L− (f ) = L+ (f ). Let = L+ (f ) − L− (f ) > 0. For every dissection ∆ of [A, B ], we have s(f, ∆) ≤ L− (f ) so that S (f, ∆) − s(f, ∆) ≥ . We can also establish the following stronger versions of Propositions 7G and 7H. PROPOSITION 7G’. Suppose that a function f (x) is Riemann integrable over the closed interval [A, B ], where A, B ∈ R and A < B . Suppose further that C ∈ [A, B ]. Then f (x) is Riemann integrable over the closed intervals [A, C ] and [C, B ], and
B C B 2 < s(f, ∆) ≤ S (f, ∆) < L + , 2 and S (f, ∆) ≥ L+ (f ), f (x) dx =
A A f (x) dx +
C f (x) dx. Chapter 7 : The Deﬁnite Integral page 15 of 20 First Year Calculus c W W L Chen, 1994, 2005 PROPOSITION 7H’. Suppose that functions f (x) and g (x) are Riemann integrable over the closed interval [A, B ], where A, B ∈ R and A < B . Then the function f (x) + g (x) is also Riemann integrable over [A, B ], and
B B B (f (x) + g (x)) dx =
A A f (x) dx +
A g (x) dx. Furthermore, for every real number c ∈ R, the function cf (x) is Riemann integrable over [A, B ], and
B B cf (x) dx = c
A A f (x) dx. The proofs of these two results are left as exercises for the interested reader. An important idea is to use Proposition 7P to establish Riemann integrability ﬁrst and then the deﬁnition of the Riemann integral to establish the various identities. Our next task is to establish the important result that continuity in the closed interval [A, B ] implies Riemann integrability. To do this, we need to introduce the idea of uniformity. We ﬁrst establish the following intermediate result. PROPOSITION 7Q. Suppose that f (x) is a continuous function in the closed interval [A, B ], where A, B ∈ R and A < B . Then given any > 0, there is a dissection ∆ : A = x0 < x1 < x2 < . . . < xn = B of the interval [A, B ] such that for every i = 1, . . . , n, we have sup f (x) −
x∈[xi−1 ,xi ] x∈[xi−1 ,xi ] inf f (x) < B−A . Proof. Let > 0 be given. We shall say that a subinterval [α, β ] of the interval [A, B ] is “good” if there exists a dissection ∆ : α = y0 < y1 < y2 < . . . < ym = β of the interval [α, β ] such that for every j = 1, . . . , m, we have sup f (x) −
x∈[yj −1 ,yj ] x∈[yj −1 ,yj ] inf f (x) < B−A . Our task is therefore to show that the interval [A, B ] is good. Suppose that it is not. We bisect the interval [A, B ], and let C denote its midpoint. Then at least one of the two subintervals [A, C ] and [C, B ] is not good. Let this be denoted by [a1 , b1 ], choosing one subinterval if neither is good. We now bisect the interval [a1 , b1 ] to obtain a subinterval [a2 , b2 ] which is not good, and continue this process. We therefore have two sequences a1 ≤ a2 ≤ a3 ≤ . . . and . . . ≤ b3 ≤ b2 ≤ b1 which clearly converge to a common value ξ ∈ [A, B ]. Since f (x) is continuous at x = ξ , there exists δ > 0 such that
x∈(ξ −δ,ξ +δ ) sup f (x) − x∈(ξ −δ,ξ +δ ) inf f (x) < B−A (here the interval (ξ − δ, ξ + δ ) has to be replaced by [ξ, ξ + δ ) or (ξ − δ, ξ ] if ξ = A or ξ = B respectively). On the other hand, if n is large enough, then the interval [an , bn ] is contained in the interval (ξ − δ, ξ + δ ), and gives rise to a contradiction.
Chapter 7 : The Deﬁnite Integral page 16 of 20 First Year Calculus c W W L Chen, 1994, 2005 Proof of Proposition 7B. Given any dissection > 0, it follows from Proposition 7Q that there exists a ∆ : A = x0 < x1 < x2 < . . . < xn = B of the interval [A, B ] such that for every i = 1, . . . , n, we have
x∈[xi−1 ,xi ] sup f (x) − x∈[xi−1 ,xi ] inf f (x) < B−A . It follows that S (f, ∆) − s(f, ∆) < . The result now follows from Proposition 7P. The Mean value theorem for diﬀerentiation is the crux for the proof of our remaining assertions. We ﬁrst use this to establish the essential uniqueness of antiderivatives. Proof of Proposition 7C. Suppose that F (x) and G(x) are two antiderivatives of the function f (x) in an interval I . Write D(x) = G(x) − F (x). Then D (x) = G (x) − F (x) = f (x) − f (x) = 0 for every x ∈ I. Suppose that x1 , x2 ∈ I and x1 < x2 . Since D(x) is diﬀerentiable for every x ∈ [x1 , x2 ], it follows from Proposition 5B that D(x) is continuous in the closed interval [x1 , x2 ]. By the Mean value theorem, there exists ξ ∈ (x1 , x2 ) such that D (ξ ) = Clearly D (ξ ) = 0, so that D(x1 ) = D(x2 ). Note now that this argument is valid for any x1 , x2 ∈ I . It follows that there is some constant C ∈ R such that D(x) = C for every x ∈ I , whence G(x) = F (x) + C for every x ∈ I . We next establish the Fundamental theorems of the integral calculus. Proof of Proposition 7E. By Proposition 7B, the Riemann integral exists. Write
B D(x2 ) − D(x1 ) . x2 − x1 L=
A f (x) dx. It follows that for every > 0, there is a dissection ∆ : A = x0 < x1 < x2 < . . . < xn = B such that (4) Next, note that
n L − < s(∆) ≤ S (∆) < L + . (5) F (B ) − F (A) =
i=1 (F (xi ) − F (xi−1 )). Since F (x) is diﬀerentiable in the closed interval [xi−1 , xi ], it follows from the Mean value theorem that there exists ξi ∈ [xi−1 , xi ] such that (6) F (xi ) − F (xi−1 ) = (xi − xi−1 )F (ξi ) = (xi − xi−1 )f (ξi ).
page 17 of 20 Chapter 7 : The Deﬁnite Integral First Year Calculus c W W L Chen, 1994, 2005 Combining (5) and (6), we have
n F (B ) − F (A) =
i=1 (xi − xi−1 )f (ξi ), a Riemann sum of f (x) corresponding to the dissection ∆. Recall now that every Riemann sum is bounded below by the corresponding lower Riemann sum and bounded above by the corresponding upper Riemann sum, so that (7) Combining (4) and (7), we have L − < F (B ) − F (A) < L + , so that L − (F (B ) − F (A)) < . Since L − (F (B ) − F (A)) is a constant and > 0 is arbitrary, we must have L − (F (B ) − F (A)) = 0. The result follows immediately. Proof of Proposition 7F. Suppose ﬁrst of all that A < x < B . Then F (y ) − F (x) 1 = y−x y−x with the convention that
y x y x s(∆) ≤ F (B ) − F (A) ≤ S (∆). f (t) dt −
A A f (t) dt = 1 y−x y f (t) dt,
x f (t) dt = −
x y f (t) dt if x > y . We need to show that
y →x lim F (y ) − F (x) = f (x). y−x In other words, we need to show that (8) Note that 1 y−x
y 1 y →x y − x lim y f (t) dt = f (x).
x f (t) dt − f (x) =
x 1 y−x y (f (t) − f (x)) dt ≤
x 1 y − x y f (t) − f (x) dt
x (here we have used the inequality
B B g (x) dx ≤
A A g (x) dx; for a proof, see Problem 2). Continuity implies that for every > 0, there exists δ > 0 such that f (t) − f (x) < whenever t − x < δ . It follows that if y − x < δ , then 1 y − x
y f (t) − f (x) dt <
x 1 y − x = . y − x This gives (8), and completes the proof when A < x < B . The cases x = A and x = B can be deduced with minor modiﬁcations.
Chapter 7 : The Deﬁnite Integral page 18 of 20 First Year Calculus c W W L Chen, 1994, 2005 We complete this chapter by establishing the Mean value theorem for Riemann integrals. Proof of Proposition 7K. By Proposition 7F, the function
x F (x) =
A f (t) dt is continuous in the interval [A, B ] and diﬀerentiable for every x ∈ (A, B ), with F (x) = f (x). By the Mean value theorem, there exists c ∈ (A, B ) such that F (B ) − F (A) = (B − A)F (c), so that
B A f (t) dt −
A A f (t) dt = (B − A)f (c). Clearly the second integral vanishes, and the result follows. Problems for Chapter 7
1 1. Calculate the integral
0 x dx by dissecting the interval [0, 1] into equal parts. 2. a) Suppose that the function f (x) is continuous in the closed interval [A, B ]. Suppose further that f (x) ≥ 0 for every x ∈ [A, B ]. Explain why
B f (x) dx ≥ 0.
A b) Suppose that the functions f1 (x) and f2 (x) are continuous in the closed interval [A, B ]. Suppose further that f1 (x) ≤ f2 (x) for every x ∈ [A, B ]. Use part (a) to show that
B B f1 (x) dx ≤
A A f2 (x) dx. c) Suppose that the function g (x) is continuous in the closed interval [A, B ]. Explain why
B B g (x) dx ≤
A A g (x) dx. 3. Diﬀerentiate each of the following integrals with respect to x:
x 4 a)
1 t t2 + 1 dt d dx
sin x 1 b)
x t2 (t + 1)3 dt 1−2x c)
2x+1 1 dt 1 + t2 4. Determine 1 √ dt. t+ t
1+ x 5. Show that for every x ∈ (0, 1), we have
Chapter 7 : The Deﬁnite Integral 1−x t−1 dt = 0. t(2 − t)
page 19 of 20 First Year Calculus c W W L Chen, 1994, 2005 6. a) Suppose that f (x) = sin−1 x + cos−1 x. Find the largest domain of f (x) as a real valued function and show that f (x) = π/2 for all x in this domain. √ 1/ 2 √ −1 b) Diﬀerentiate the function g (x) = x sin x + 1 − x2 . Hence ﬁnd the integral sin−1 x dx.
0 c) Use parts (a) and (b), or otherwise, to ﬁnd the area bounded by the curves y = sin−1 x, y = cos−1 x and the y axis. Harder Problems for Chapter 7 7. Prove Proposition 7L. 8. Prove Proposition 7M. 9. Prove Proposition 7G’. 10. Prove Proposition 7H’. 11. Calculate the integral
B xk dx,
A where k > 0 is ﬁxed, by dissecting the interval [A, B ] into n parts in geometric progression, so that A < Aq < Aq 2 < . . . < Aq n = B . 12. a) By using the method of Problem 11, prove that b) Deduce that lim n
n→∞ 1 1 dx = . x2 2 1 1 1 1 1 =. + + ... + (n + 1)2 (n + 2)2 (2n)2 2
α 2 13. Calculate the integral
0 sin x dx by dissecting the interval [0, α] into equal parts. Chapter 7 : The Deﬁnite Integral page 20 of 20 ...
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