講義請點此

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5.1 Areas and Distances goo.gl/fJLueV 1

Chapter 5 Integrals

5.1 Areas and Distances, page 360

The Area Problem, page 360

EoF_wAlXL7E 關於拋物線下區域面積, 早在阿基米德 (公元前287– 212)的時候就知道了, 那時阿基米德雖然沒有像我們一樣已經非常清楚極限的理論, 但是不斷地分割再求和以估計並逼近區域面積這個概念就已成形。這裡介紹的方法利用了平方和的公式、極限的四則運算還有夾擠定理, 可以將拋物線下的面積解釋得一清二楚。 Example 1. Use rectangles to estimate the area under the parabola y = x2 from 0 to 1.

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2 5.1 Areas and Distances goo.gl/fJLueV

Definition 2 (page 365). The area (面積) A of the region S that lies under the graph of the continuous function f is the limit of the sum of the areas of approximating rectangles:

fpuvI1ZYeBA 由前一個例子的討論, 引出連續函數的圖形與x-軸所圍的區域面積, 可以利用分割、樣本點、取和、求極限這四個步驟定義; 下一節將仔細討論相關理論。 A= lim n→∞Rn= limn→∞(f (x1)∆x + f (x2)∆x + · · · + f (xn)∆x).

It can also be shown that we get the same value if we use left endpoints:

A= lim

n→∞Ln= limn→∞(f (x0)∆x + f (x1)∆x + · · · + f (xn−1)∆x).

In fact, instead of using left endpoints or right endpoints, we could take the height of the i-th rectangle to be the value of f at any number x∗

i in the i-th subinterval [xi−1, xi]. We call

numbers x∗

1, x∗2, . . . , x∗n the sample points (樣本點).

In general, we form lower sums (下和) (and upper sums, 上和) by choosing the sample points x∗_i so that f (x∗_i) is the minimum (and maximum) value of f on the i-th subinterval.

The Distance Problem, page 367

在物理上, 物體的

速度與位移之間的關係也將引出下一節定積分的概念。

We can find the distance traveled by an object during a certain time period if the velocity of the object is known at all times.

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5.2 The Definite Integral goo.gl/V2EYnS 3

5.2 The Definite Integral, page 378

Trn1iwZJDqU 函數與x-軸所圍的區域面積, 利用分割樣本點取和求極限四個步驟定義。這裡應強調的是, 若函數是可積分的, 必須要求樣本點在做任意的選取之下, 極限都必須存在並且相等, 這算是非常高的要求。 Definition of a Definite Integral _{(page 378). If f is a function defined for a ≤ x ≤ b,}

we divide the interval [a, b] into n-subintervals of equal width ∆x = b−a_n . We let x0 =

a, x1, x2, . . . , xn = b be the endpoints of these subintervals and we let x∗₁, x∗₂, . . . , x∗n be any

sample points (樣本點) in these subintervals, so x∗_i lies in the i-th subinterval [xi−1, xi]. Then

the definite integral (定積分) of f from a to b is Z b a f(x) dx = lim n→∞ n X i=1 f(x∗_i)∆x

provided that this limit exists and gives the same value for all possible choices of sample points. If it does exist, we say that f is integrable (可積分的) on [a, b].

x y

f(x)

a b

Figure 1: Definition of a definite integral.

The precise meaning of the limit that defines the integral is as follows: 定積分的概念是極

限, 所以可以用極

限的精確定義重新描述。

For every number ε > 0, there is an integer N such that Z b a f_{(x) dx −} n X i=1 f(x∗_i)∆x < ε

for every integer n > N and for every choice of x∗_i in [xi−1, xi].

Qe83Lc7nVhA

確實認識積分符號。 There are some notations we should know:

Z b a f(x) dx integral sign: integrand: limits of integration: lower limit (下限): upper limit (上限):

The procedure of calculating an integral is called integration.

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4 5.2 The Definite Integral goo.gl/V2EYnS

The definite integral Rb

af(x) dx is a number; it does not depend on x.

The sum Pn

i=1

f(x∗

i)∆x is called a Riemann sum (黎曼和).

The geometric meaning of Rb

af(x) dx is the net area of y = f (x) from a to b.

In fact, the subinterval widths are not necessary equal width. Z b a f(x) dx = lim max∆xi→0 n X i=1 f(x∗_i)∆xi.

Not all functions are integrable. For example, the Dirichlet function, or f (x) = 1_x on 0 < x ≤ 1.

Example 1. Show that f (x) = 1_x _{on 0 < x ≤ 1 is not integrable.}

vcgx3IUxAdE 在這個例子中, 若要證明定積分不存在, 只要證明在分割之下, 找到一組樣本點使其黎曼和取極限後不存在即可。所以觀察的重點是如何論述黎曼和取極限不存在。 Solution. p_UGv95UB_I 閉區間上的連續函數可積分, 這件事將在高等微積分課程中學習如何嚴格地證明。而定積分存在的函數可以稍微爛一點, 若函數只有有限個跳躍不連續點, 定積分仍存在。

Theorem 2(page 380). If f is continuous on [a, b], or if f has only a finite number of jump discontinuous, then f is integrable on [a, b]; that is, the definite integral Rb

af(x) dx exists.

If f is integrable on [a, b], then the limit of Riemann sum exists and gives the same value no matter how we choose the sample points x∗_i. To simplify the calculation of the integral, we often taken the sample points to be right endpoints. Then x∗_i = xi and the definition of

an integral simplifies as follows.

Theorem 3 (page 380). If f is integrable on [a, b], then Z b a f(x) dx = lim n→∞ n X i=1 f(xi)∆x,

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5.2 The Definite Integral goo.gl/V2EYnS 5

Example 4 (page 383). Set up an expression for R3

1 exdx as a limit of sums and evaluate it. 因為函數是指數函

數, 所以可以用等比級數公式將黎曼和簡單地表達。最後處理極限值也是一門學問, 因為 n 是正整數, 把問題過渡成以x (實數) 為變數, 再用羅必達法則處理。從這個例子會發現: 用定義計算定積分是件不簡單的事, 可想而知的是, 如果函數再複雜一點, 很難會有什麼公式把黎曼和簡單地表達以算出極限。所以在之後的章節, 我們將介紹別的方法快速得到定積分, 以取代這種土法煉鋼式的用定義計算積分值。 Solution.

Example 5. Change the following limits of sums as integrals:

XeoEdy3q6Tg 有些極限問題可以把它轉變成定積分的型式, 然後等到之後的章節學完就可以快速求得積分值。至於如何將極限轉換成定積分, 可以先想辦法造出 ∆x = 1 n 這個量, 再去設定函數、樣本點、上下限得到定積分。要注意的是, 最後寫出來的定積分表示法不唯一。 lim n→∞n 2 1 n3 + 1 (n + 1)3 + · · · + 1 (n + (2n − 1))3 . Solution.

Example 1 in section 5.1 is also an evaluating integral of R1

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6 5.2 The Definite Integral goo.gl/V2EYnS

Properties of the Definite Integral

M_ybjcBk_9w 試著把積分的性質與「有向面積」或是「帶有正負號的面積」對應。而產生正負號的可能性有兩種, 一個是函數的正負, 另一個是積分的上下限互換。性質(2)與(3)與線性代數有關: 積分是一種線性變換。

Properties of the Integral (page 385–387). (1)

Z b

a

c_{dx = c(b − a), where c is any constant.}

(2) Z b a (f (x) ± g(x)) dx = Z b a f_{(x) dx ±} Z b a g(x) dx. (3) Z b a cf(x) dx = c Z b a

f(x) dx, where c is any constant.

(4) Z b a f(x) dx + Z c b f(x) dx = Z c a f(x) dx. (5) If f (x) ≥ 0 for a ≤ x ≤ b, then Z b a f_{(x) dx ≥ 0.} (6) If f (x) ≥ g(x) for a ≤ x ≤ b, then Z b a f_{(x) dx ≥} Z b a g(x) dx. (7) If m ≤ f (x) ≤ M for a ≤ x ≤ b, then m(b − a) ≤ Z b a f_{(x) dx ≤ M(b − a).} Example 6 (page 391). If f is continuous on [a, b], show that

除了學習如何搭配上面的積分性質完成論述之外, 試著將這個例子和「三角不等式」做聯想, 就會覺得這個不等式很自然。 Z b a f(x) dx ≤ Z b a |f (x)| dx. Solution.

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5.3 The Fundamental Theorem of Calculus goo.gl/adJ338 7

5.3 The Fundamental Theorem of Calculus, page

392

eamA03UJ_kQ 注意到定理敘述中 g(x)這個函數的變數x對應到積分的上限,因為 x符號被借用了, 所以積分的啞吧變數改用別的符號t表示。看數學證明的時候, 可以把每一個式子搭配圖形的意義就變得容易理解。 The Fundamental Theorem of Calculus, Part 1 (page 394). If f is continuous on [a, b],

then the function g defined by

g(x) = Z x

a

f(t) dt a_{≤ x ≤ b} is continuous on [a, b] and differentiable on (a, b), and g′_{(x) = f (x).}

Proof. For x and x + h in (a, b), we have

g_{(x + h) − g(x) =} = so for h 6= 0,

g_{(x + h) − g(x)}

h =

Assume that h > 0. Since f is continuous on [x, x + h], the says

that there are u, v ∈ [x, x + h] such that f (u) = m and f (v) = M, where m and M are the absolute minimum and maximum values of f on [x, x + h]. So

mh_≤

Z x+h

x

f_{(t) dt ≤ Mh ⇒} Now we let h → 0, then u → x and v → x, so

lim

h→0f(u) = f ( limu→xu) = f (x) and h→0limf(v) = f ( limv→xv) = f (x)

because f is continuous at x. By the , we have

g′(x) = lim

h→0

g_{(x + h) − g(x)}

h = f (x).

If x = a or b, the above discussion can be modified by considering one-sided limit. Since g is differentiable on [a, b], g is continuous on [a, b].

微積分基本定理第一部份想傳達的概念是: 微分和積分是互逆的運算, 把連續函數先積分再微分就會還原。 Remark 1. The Fundamental Theorem of Calculus, Part 1, can be written as

d dx Z x a f(t) dt = f (x) when f is continuous.

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8 5.3 The Fundamental Theorem of Calculus goo.gl/adJ338

Example 2 (page 395). Find the derivative of the function g(x) = Z x 0 p 1 + t2_dt. cAwFW1E7_D0 Solution. 這個例子觀察的重點是: 當積分上限不是x 的時候,而 h′_(x) _是問 _h(x) 對x的變化率,所以會有鏈鎖律的關係: 第一層的微分是用微積分基本定理, 第二層的微分是積分上限對x的變化,兩者相乘。

Example 3. Find the derivative of h(x) = Z sin x

0

p

1 + r3_dr.

Solution.

The Fundamental Theorem of Calculus, Part 2(page 396). If f is continuous on [a, b], then Y8eCa57yi9E 微積分基本定理第二部份是說: 如果有辦法找到連續函數 f(x) 的隨便一個反導函數 F(x), 那麼函數 f(x) 在 [a, b]之間的定積分值和反導函數代入上限與下限後的值相減結果一樣。這個定理的建立, 就可取代前一節所述利用定義計算定積分帶來的各種不便性。下一個單元我們就要開始追問每一個函數的反導函數是什麼? 又該如何尋找? Z b a f_{(x) dx = F (b) − F (a),}

where F(x) is any antiderivative of f (x), that is, a function such that F′_{(x) = f (x).}

Proof. Let g(x) =Rx

a f(t) dt. From the Fundamental Theorem of Calculus, Part 1, we know

g′(x) = f (x), so g(x) is an antiderivative of f . If F is any other antiderivative of f on [a, b], then F and g differ by a constant: F (x) = g(x) + C for a < x < b. Remark that it also holds when x = a and x = b.

We put x = a in the formula of g(x) to get

So we have

F_{(b) − F (a) =} .

We often use notation F (x)|b

a= F (b) − F (a).

The Fundamental Theorem of Calculus (page 398). Suppose f is continuous on [a, b].

(1) If g(x) = Z x a f(t) dt, then g′_{(x) = f (x).} (2) Z b a

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5.4 Indefinite Integrals and the Net Change Theorem goo.gl/PrSTAm 9

5.4 Indefinite Integrals and the Net Change

Theo-rem, page 402

DtgOpr1BzjE 在微分的階段曾要求各位背熟每個基本函數的導函數, 現在要開始熟悉逆操作, 必須熟記左方表格中每個函數的反導函數。因為不定積分的意義是要找所有的反導函數, 而4.9 的定理2告知任兩個反導函數之間只會差一個常數, 所以寫不定積分的時候, 必須補上 +C 才會代表所有的反導函數。

Table of Indefinite Integrals (page 403). Z cf(x) dx = c Z f(x) dx Z (f (x) + g(x)) dx = Z f(x) dx + Z g(x) dx Z kdx = kx + C Z xndx = x n+1 n+ 1+ C (n 6= −1) Z 1 xdx = ln |x| + C Z exdx = ex+ C Z axdx = a x ln a + C Z sin x dx = − cos x + C Z cos x dx = sin x + C Z sec2xdx = tan x + C Z csc2x_{dx = − cot x + C} Z

sec x tan x dx = sec x + C

Z csc x cot x dx = − csc x + C Z ₁ x2_{+ 1}dx = tan −1_x_{+ C} Z _√ 1 1 − x2 dx = sin −1_x_{+ C} Z sinh x dx = cosh x + C Z cosh x dx = sinh x + C 這裡要注意的是 R 1 xdx的結果,一個是回去看3.6 例題 2 的介紹;另一個要注意到 ln |x| 的圖形拆成左右兩半,兩者不相連,所以 +C 實際上在左半與右半可以允許不同的常數。 We adopt the convention that when a formula for a general indefinite integral is given, it is valid

only on an interval. For example, the general antiderivative of the function f (x) = _x1, x_{6= 0 is}

F(x) = ( ln |x| + C1 if x > 0 ln |x| + C2 if x < 0. 定積分得到的是數字; 不定積分指的是一族函數。

A definite integral Rb

a f(x) dx is a number; an indefinite integralR f (x) dx is a family of

functions.

Applications

Recall that the Fundamental Theorem of Calculus, part 2:

iEDcEemzZdw 將微積分基本定理第二部份稍做改寫就可得到淨變化定理 (Net Change Theorem)。 The Fundamental Theorem of Calculus, Part 2 (page 396). If f is continuous on [a, b],

then

Z b

a

f_{(x) dx = F (b) − F (a),}

where F(x) is any antiderivative of f (x), that is, a function such that F′(x) = f (x). We put f (x) = F′_{(x) into the Theorem and get}

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10 5.4 Indefinite Integrals and the Net Change Theorem goo.gl/PrSTAm

Net Change Theorem(page 406). The integral of a rate of change is the net change: Z b

a

F′_{(x) dx = F (b) − F (a).}

This principle can be applied to all of the rates of change in the natural and social sciences. For example, 積分會有正負相消的效應, 所以速度 (搭配方向表示正負) 的積分是「位移」; 若要計算所有路徑的「總距離」, 則必須分段調整所有的符號。在計算「面積大小」的時候也同樣要注意這個情況。

• If an object moves along a straight line with position function s(t), then its velocity is v(t) = s′(t), so

Z t₂

t1

v(t) dt = s(t2) − s(t1)

is the net change of position, or displacement, of the particle during the time period from t1 to t2.

If we want to calculate the distance the object travels during the time interval, we have to consider the intervals when v(t) ≥ 0 and also the intervals when v(t) ≤ 0. In both cases the distance is computed by integrating |v(t)|, the speed. Therefore,

Z t2

t1

|v(t)| dt = total distance traveled.

x y A1 A2 A3 v(t) Displacement = Distance =

Figure 1: Displacement and distance.

• The acceleration of the object is a(t) = v′_{(t), so}

Z t₂

t1

a(t) dt = v(t2) − v(t1)

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5.5 The Substitution Rule goo.gl/9VPBvm 11

5.5 The Substitution Rule (page 413)

qaEquBo-8E8 同學不應被這個變數變換定理的型式困惑住, 可直接從之後的所有例題與練習, 把這個觀念建立即可。 The Substitution Rule (page 413). If u = g(x) is a differentiable function whose range is

an interval I and f is continuous on I, then Z f(g(x))g′(x) dx = Z f(g(x)) dg(x) = Z f(u) du. Proof. Suppose F is an antiderivative of f , then we have

d

dxF(g(x)) = F

′_(g(x))g′_{(x) = f (g(x))g}′_(x).

So F (g(x)) is an antiderivative of f (g(x))g′_{(x). Let u = g(x), then}

Z f(g(x))g′(x) dx = F (g(x)) + C = F (u) + C = Z F′(u) du = Z f(u) du. The middle formula comes from the definition of differential: dg(x) = g′_{(x) dx.}

The Substitution Rule for Definite Integrals (page 416). If g′ is continuous on [a, b] and f is continuous on the range of u= g(x), then

Z b a f(g(x))g′(x) dx = Z g(b) g(a) f(u) du.

Proof. Let F be an antiderivative of f . Then F (g(x)) is an antiderivative of f (g(x))g′(x), by Part 2 of the Fundamental Theorem, we have

Z b a f(g(x))g′(x) dx = F (g(x)) b a= F (g(b)) − F (g(a)).

On the other hand, for the right hand side of the equation, we have Z g(b) g(a) f(u) du = F (u) g(b) g(a) = F (g(b)) − F ((a)).

Example 1. Compute the integral

Z _x7 √ x4_{+ 1}dx. K6SffXf04IY 根號內的量有點複雜, 把它記為變數 u,再把其它部分全部替換成和u有關的量。 Solution.

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12 5.5 The Substitution Rule goo.gl/9VPBvm

Example 2 (page 410). Calculate

Z tan x dx. 影片介紹了兩種寫法, 個人提倡盡量使用第二種方法, 除非函數太複雜, 不然在多數情況下是不需要再引進新的符號 u。只要把 cos x 在心中想成是u之後就直接寫答案了, 這樣的寫法可以省很多事。 Solution. Example 3. Find Z π 4 0 sec2_θ √ 1 + tan θdθ. z7awm81SOg8 善用微分 (differ-ential) 的所有特性, 觀察被積分函數, 將當中的某一部分轉變到d的右邊: sec2_θ dθ = d tan θ = d(tan θ + 1)。 Solution. 善用微分的「線性」, 如dx = 1 ad(ax + b)。 Example 4. Find Z cos(ax + b) dx. Solution. 1 xdx = d ln x, 將函數搬到d的右邊是進行「積分」。注意到這裡是不需要寫成d ln |x|,這是因為被積分函數有 (ln x)k , 函數有意義的地方只有x > 0的部份,所以不需要處理x <0。 Example 5. Find Z _{(ln x)}k x dx. Solution. 三角函數有很多的恆等式, 導致會有些障眼法, 若是看得出 1 sec xdx = cos x dx = d sin x, 這個積分問題就變得很容易。

Example 6. Find the integral

Z _esin x sec xdx. Solution.

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5.5 The Substitution Rule goo.gl/9VPBvm 13 Example 7. Let F (x) =Rx 0(x − t)t sin(t2) dt. Find F′(x). rkN46kWuW6c 這個問題要特別注意變數x不僅出現在積分的上限, 也出現在被積分的函數中, 所以兩個地方都要照顧到。 Solution.

Example 8. Compute the integral

Z 1 2 1 4 sin−1√x px(1 − x)dx. 這個例子就不容易看出變換的關係, 就試著逐一替換變數 u,記得積分上、下限也要跟著換。 Solution. eNtFcCxE7xA 當函數 f(x) 是奇函數或偶函數, 在 [−a, a] 的定積分會有一些好的性質。這兩個性質從積分的幾何意義來看會覺得非常自然。而這兩個性質在一些實際的積分問題下很好用, 對於偶函數來說, 下限是負的數字, 帶入反導函數會產生太多負號, 很容易被困住;帶入x= 0之下, 常常會有很多項不見。若能確定函數是奇函數, 那就根本不需要去尋找它的反導函數。 Integrals of Symmetric Functions _{(page 417). Suppose f is continuous on [−a, a].}

(a) If f is even, then Z a −a f(x) dx = 2 Z a 0 f(x) dx. (b) If f is odd, then Z a −a f(x) dx = 0. Proof. We compute Z a −a f(x) dx = Z 0 −a f(x) dx + Z a 0 f_{(x) dx = −} Z −a 0 f(x) dx + Z a 0 f(x) dx, Let u = −x, then du = −dx and when x = −a, u = a. Therefore

− Z −a 0 f_{(x) dx = −} Z a 0 f_{(−u)(−du) =} Z a 0 f_{(−u) du.}

(a) If f is even, then , so we get

Z a

−a

f(x) dx = =

(b) If f is odd, then , so we get

Z a

−a

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14 5.5 The Substitution Rule goo.gl/9VPBvm

Appendix

這個附錄雖然沒有影片的解說, 但是這一段話卻道出積分理論的最重要的概念, 微積分有沒有把最深刻的精髓學到, 就端看各位能不能深刻體會這一段話。

Suppose that f (x) ∈ C1_{([a, b]), which implies |f}′_{(x)| ≤ M. Let ∆x =} b−a n , then Z b a f_{(x) dx −} n X i=1 f(x∗_i)∆x ≤ n X i=1 max [xi₋₁,xi] f_{(x) − min} [xi₋₁,xi] f(x) ∆x ≤ n X i=1 |f′(ξi)|(∆x)2 = M · n X i=1 (b − a)2 n2 = M · (b − a)2 n → 0 as n → ∞.

So for integration, before we take summation, the 1_n part is the whole material. We can ignore higher order term such as _n12 because it tends to zero after summation and n tends to infinity.

Therefore, for the Substitution Rule, we only focus on the “differentials” between two variables y and x. That is, dy = y′_{(x) dx will catch all information of} 1

n part between