Calculus107A1Chapter07.dvi

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1 7. Integrtion by Prts goo.gl/qriic5 Chpter 7 Techniques of Integrtion 7. Integrtion by Prts, pge 472 The rule tht corresponds to the Product Rule for differentition is clled the rule for integrtion by prts ( 分佈積分 ). The Product Rule sttes tht if f nd g re differentible functions, then d dx (f(x)g(x)) = f(x)g (x)+g(x)f (x). In the nottion for indefinite integrls this eqution becomes (f(x)g (x)+g(x)f (x))dx = f(x)g(x), or f(x)g (x)dx+ g(x)f (x)dx = f(x)g(x). So we cn rerrnge this eqution s f(x)g (x)dx = f(x)g(x) Formul () is clled the formul for integrtion by prts. g(x)f (x)dx. () Let u = f(x) nd v = g(x), then the differentils re du = f (x)dx nd dv = g (x)dx. By the Substitution Rule, the formul for integrtion by prts becomes udv = uv vdu. (2) FX9qNzLurjY 第七章前四節是在介紹積分技巧 ; 而分佈積分法是探討兩類型函數相乘的積分處理法你會看到這個公式的特色在於等式的左邊和右邊微分的函數不同換言之, 這個公式可以把微分轉移到另一個函數身上接下來會仔細討論任兩類型函數的積分, 到底該把誰看成 u, 該把誰看成 v, 這個方法是有邏輯可循的, 所以重點是要把當中的邏輯學起來 Exmple (pge 472). Evlute Exmple 2 (pge 473). Evlute xsinxdx. lnxdx. 若遇到多項式與三角函數相乘的積分, 因為多項式微分之後會降次, 降到變成常數之後就只剩下三角函數的積分 ; 所以使用分佈積分時, 先把三角函數積分後放到 d 的右邊, 利用分佈積分就可以把多項式微分 3RH0T-cTxFs 對數函數的積分, 直接使用分佈積分法, 它已經分配成 udv 的型式了

2 2 7. Integrtion by Prts goo.gl/qriic5 若遇到多項式與對數相乘的積分, 因為 lnx 微分是 x, 所以使用分佈積分時, 先把多項式積分後放到 d 的右邊, 分佈積分後就可以將對數函數微分注意到, 先把多項式積分將增加多項式的次方, 似乎不利積分的處理, 但是分佈積分後對數微分產生會把次數 x 降一次, 所以情況並沒有變糟 e Exmple 3. Find the integrl Exmple 4. Evlute xe x dx. x(lnx) 2 dx. AcbN6_o3g 若遇到多項式與指數函數相乘的積分, 因為多項式微分之後會降次, 降到變成常數之後就只剩下指數函數的積分 ; 所以使用分佈積分時, 先把指數函數積分後放到 d 的右邊, 在分佈積分後, 就可以把多項式微分若遇到指數函數與三角函數 (sinx,cosx) 相乘的積分, 因為這兩類函數的微分都有週期性 ( 和原函數相關 ), 所以使用分佈積分的時候, 這兩種函數都可以當成 u 或 v 建議兩種變換的方式都確實地操作一次以清楚了解該原理 Exmple 5 (pge 474). Evlute e x sinxdx. sinx 和 cosx 的微分有週期性, 利用兩次分佈積分之後再將整個等式處理

3 7. Integrtion by Prts goo.gl/qriic5 3 Exmple 6 (pge 475). Evlute tn xdx. 6Qqv9nlUx4 反三角函數的積分, 直接使用分佈積分法, 它已經分配成 udv 的型式了反三角函數的積分, 直接用分佈積分法 Exmple 7. Consider the region R enclosed by the curves y = cosx nd y = sinx, nd 0 x π 4. Find the volume of the solid obtined by rotting the region bout the y-xis. HJUoK6h5LYk 學過分佈積分法之後, 就可以處理更多的應用問題, 特別是用柱殼法計算旋轉體體積的時候, 因為公式本身會帶有 x, 只要函數是不同於多項式的函數, 確實處理積分時就會用到分佈積分法 Exmple 8. Find the verge of the horizontl chords in y = sinx,0 x π. FRFcIUbVis 這個例題看似平凡無奇, 但是它和著名的布豐投針 (Buffon s Needle) 相關, 有興趣者可以繼續網路搜尋相關資訊注意函數與水平弦相交時, x 的表示法要會寫

4 4 7.2 Trigonometric Integrls goo.gl/yaj6qb 7.2 Trigonometric Integrls, pge 479 qpidtzqdbwy 這個單元要學習的是三角積分, 有兩種類型, 第一類要處理的是 sinx 與 cosx 各種次方的相乘之積分這裡雖然用非常一般的記號直接計算其結果, 但是各位應該要理解其原理 : 只要次方有一者是奇數, 那就把一個丟到 ( 積分 ) d 的右手邊, 剩下的偶數次方, 透過三角恆等式 sin 2 x + cos 2 x = 就可以變成以 sinx 或 cosx 為變數的多項式積分若兩者的次方都是偶數, 那就用半角公式讓次方減半, 不斷減半的情形下, 終究會讓某個函數的次方變成奇數 In this section we use trigonometric identities to integrte certin combintions of trigonometric functions. Exmple (pge 48). Evlute () If m = 2k +, then = = i=0 sin m xcos n xdx = sin m xcos n xdx, where m,n 0 re integers. sin 2k+ xcos n xdx = k ( cos 2 x) k cos n xdcosx = Ci k k i ( ) i cos 2i xcos n xdcosx k ( ) i+ Ci k (b) If n = 2k +, then = = i=0 sin m xcos n xdx = i=0 cos n+2i xd(cosx) =. sin m x( sin 2 x) k dsinx = k ( ) i Ci k sin m xcos 2k+ xdx = sin m x k Ci k k i ( ) i sin 2i xdsinx i=0 sin m+2i xdsinx =. (c) If m = 2k,n = 2l, then using the hlf-ngle identities sin 2 x = nd cos 2 x =, we hve sin m xcos n xdx = sin 2k xcos 2l xdx ( ) cos2x k ( ) +cos2x l = dx = 2 2 k l i=0 j=0 ( ) i C k i Cl j 2 k+l cos i+j 2xdx. If i+j is odd, we reduce the integrl to cse (b). If i+j is even, we use hlf-ngle identities gin.

5 7.2 Trigonometric Integrls goo.gl/yaj6qb 5 Exmple 2. Evlute sin 4 xdx. bhypcl8ipfa 以實例操作三角積分 Exmple 3 (pge 482). Compute the integrls tnxdx nd secxdx. 重新複習 tnx 的積分 ; 另外也學習 secx 的積分, 這裡介紹的技巧非常高超, 但式子很短, 一下就得到結果 ; 若你想要問是否有一個不像這裡介紹神來一筆的方法, 會在單元 7.4 介紹上述方法太過技巧 ( 誰知道要乘什麼量 ), 但之後會學別的方式處理它 ( 想法比較自然 ) Exmple 4 (pge 482). Evlute tn m xsec n xdx, where m,n N. () If n = 2k,k N, then = i=0 tn m xsec n xdx = tn m xsec 2k xdx = tn m x(tn 2 x+) k dtnx = k = Ci k (b) If m = 2k +,k N, then tn m+2i xdtnx = tn m xsec n xdx = k i=0 tn 2k+ xsec n xdx = k tn m x Ci k tn 2i xdtnx i=0 C k i m+2i+ tnm+2i+ x+c. IeVPjbssxgA 第二類的三角積分要處理的是 tnx 與 secx 各種次方的相乘之積分這裡雖然用一般的記號直接計算其結果, 但是各位應該要理解其原理 : 若 sec n x 的次方是偶數, 或是 tn m x 的次方是奇數, 那就用變數變換以及三角恆等式 sec 2 x = + tn 2 x 處理若 sec n x 的次方是奇數且 tn m x 的次方是偶數, 那就要用分佈積分法處理 k = (sec 2 x ) k sec n xdsecx = Ci k sec2(k i) ( ) i sec n xdsecx = k ( ) i Ci k i=0 i=0 sec 2(k i)+n xdsecx = k i=0 ( ) i C k i 2(k i)+n sec2(k i)+n x+c.

6 6 7.2 Trigonometric Integrls goo.gl/yaj6qb (c) If m = 2k,n = 2l+, then I m = tn m xsec n xdx = tn 2k xsec 2l+ xdx = tn 2k xsec 2l xdsecx = = tn 2k xsec 2l+ x secx(2k )tn 2k 2 sec 2 xsec 2l xdx secxtn 2k x(2l)sec 2l xsecxtnxdx = tn 2k xsec 2l+ x (2k ) tn 2k 2 sec 2l+3 xdx 2l tn 2k xsec 2l+ xdx = tn 2k xsec 2l+ x (2k ) tn 2k 2 (tn 2 x+)sec 2l+ xdx 2l tn 2k xsec 2l+ xdx = tn m xsec n x (m )I m (m )I m 2 (n )I m. Hence we get I m = ( tn m xsec n ) x (m )I m 2, m+n nd the reduction formul will reduce the integrl to sec n xdx = sec 2l+ xdx, l N. Exercise (pge 48). Evlute tn 6 xsec 4 xdx. Exercise (pge 482). Evlute tn 5 xsec 7 xdx. Exmple 5 (pge 484). Evlute the following integrls: o5vhifo9q0u 這裡還要介紹另一種型式的三角積分, 是 sinmx 與 cosnx 相乘的積分, 只要用積化和差公式就能立刻得到結果這裡得到的結果實際上和線性代數大有關係, 它告知向量空間 C([ π,π]) 中會有一組單位正交基底, 而這是傅利葉級數理論的開端 sinmxcosnxdx; cosmxcosnxdx; sinmxsinnxdx. Recll the following identities: sinxcosy = 2 (sin(x+y)+sin(x y)) cosxcosy = 2 (cos(x+y)+cos(x y)) sinxsiny = 2 (cos(x+y) cos(x y)).

7 7.2 Trigonometric Integrls goo.gl/yaj6qb 7 If m+n 0 nd m n 0, then sinmxcosnxdx = sin((m+n)x)+sin((m n)x)dx 2 = cosmxcosnxdx = 2 = cos((m+n)x)+cos((m n)x)dx sinmxsinnxdx = 2 = cos((m+n)x) cos((m n)x)dx In prticulr, π sinmxcosnxdx = π π π cosmxcosnxdx = π π π sinmxsinnxdx = π π Hence {sinmx,m N nd cosnx,n Z,n 0} form n orthonorml bsis in the function spce C([ π,π]).

8 8 7.3 Trigonometric Substitution goo.gl/rvgdn8 7.3 Trigonometric Substitution (pge 486) Trigonometric identities re lso useful to mke substitutions for some rdicl functions. F7vn2FskLjw 根號內呈現變數平方與數字平方的加或減之積分都是屬於三角代換的範疇, 只要把它和三角恆等式做完整地對應即可注意到三角代換 θ 的範圍, 是為了要去掉絕對值而設定 ; 若選取別的範圍, 去絕對值時必須補上負號, 這樣做並不是不行, 但是之後的計算會容易把自己困住 Tble of Trigonometric Substitutions. Expression Substitution Identity 2 x 2 x = sinθ, π 2 θ π 2 sin 2 θ = cos 2 θ 2 +x 2 x = tnθ, π 2 < θ < π 2 +tn 2 θ = sec 2 θ x 2 2 x = secθ, 0 θ < π 2 or π θ < 3π 2 sec 2 θ = tn 2 θ Exmple (pge 487). Find the re enclosed by the ellipse x2 2 + y2 b 2 =. Exmple 2 (pge 490). Find x 3 2x x 2 dx. SddSCojKi-I 遇到根號內是二次三項式的時候, 要和配方法聯想, 配方法在代數上的目的是可以把一次項消掉 9 x 2 Exercise (pge 486). Evlute dx. x 2

9 7.3 Trigonometric Substitution goo.gl/rvgdn8 9 Exmple 3 (pge 488). Find x 2 x 2 +4 dx. Exercise. Evlute the integrl Exmple 4 (pge 489). Find 2 x 2 +x 2 dx. dx, where > 0. x 2 2 ZBcB6VZLB3w 變數平方減常數平方, 與 sec 2 θ = tn 2 θ 聯想 2 Exercise. Find the integrl x 3 dx, x >. x 2 x Exercise. Evlute the integrl x 2 +2x+2 dx. Exmple 5 (pge 490). Find x 3 (4x 2 +9) 3 2 dx.

10 0 7.4 Integrtion of Rtionl Functions by Prtil Frctions goo.gl/obeqat 7.4 Integrtion of Rtionl Functions by Prtil Frctions (pge 493) emee69427sk 部份分式法是要處理有理函數的積分, 它會有標準程序這裡將介紹基本模型的積分該如何處理這裡引用了最一般的記號, 只是為了論述的完整, 看影片時, 必須抓住的是處理積分的原則, 不要被太多符號困住 In this section we show how to integrte ny rtionl function ( rtio of polynomils) by expressing it s sum of simpler frctions, clled prtil frctions ( 部份分式 ). Exmple. Discuss the integrl dx, where k N. (x+b) k Exmple 2. Discuss the integrl Ax+B (x 2 +bx+c) k dx, where b2 4c < 0,k N.

11 7.4 Integrtion of Rtionl Functions by Prtil Frctions goo.gl/obeqat Integrte rtionl functions Step : Perform the long division ( 長除法 ) Definition 3 (pge 494). Consider rtionl function f(x) = P(x) Q(x), where P(x) nd Q(x) re polynomils. () If the degree of P(x) is less thn the degree of Q(x), such rtionl function f(x) is clled proper. (b) If the degree of P(x) is greter or equl to the degree of Q(x), such rtionl function f(x) is clled improper. If f(x) is improper, then we use the long division to get nd R(x) Q(x) is proper. f(x) = P(x) R(x) = S(x)+ Q(x) Q(x), Step 2: Fctor the denomintor Q(x) s product of liner fctors (x+b) nd irreducible qudrtic fctors x 2 +bx+c, b 2 4c < 0. ( 因式分解 ) Step3: Expressthe proper rtionl function R(x) s sum ofprtil frctions. Q(x) ( 拆成部份分式, 不同類型有不同的拆解法 ) Definition 4 (pge 494). A rtionl function is clled prtil frction if it is of the form A (x+b) n or Ax+B (x 2 +bx+c) n. () Q(x) is product of distinct liner fctors. Tht is, 7VboqkOquZc 部份分式法的第一步驟是長除法, 若分子的次數大於分母, 則長除法過後會有一部份是多項式, 而多項式的積分就很容易處理剩下的部份是真分式, 之後的三個步驟都是在處理真分式的積分第二步驟是將分母因式分解, 由代數的理論得知, 多項式一定可以分解成一些一次式與一些二次三項式的乘積第三步驟就是部份分式法的重點, 根據不同類型有不同的拆解方式這裡必須好好體會最後一步驟就是按照前一個影片的方法逐一積分 Q(x) = ( x+b )( 2 x+b 2 ) ( k x+b k ), where no fctor is repeted, then there exist constnts A,A 2,...,A k such tht R(x) Q(x) = A x+b + A 2 2 x+b A k k x+b k. For exmple, 2x 2 3x x(x+)(x ) =, then

12 2 7.4 Integrtion of Rtionl Functions by Prtil Frctions goo.gl/obeqat YQViTvWxu28 根據因式的次方在做部份分式時, 拆解的假設都不同 (2) Q(x) is product of liner fctors, some of which re repeted. Supposethe first liner fctor ( x+b ) is repeted r times, then insted of the single term A x+b, we would use A A 2 + x+b ( x+b ) A r ( x+b ) r. For exmple, 2x2 x+3 (x ) 3 =. (3) Q(x) contins irreducible qudrtic fctors, none of which is repeted. Tht is, Q(x) hs the fctor x 2 +bx+c, where b 2 4c < 0, then the expression for R(x) Q(x) will hve term of the form Ax+B x 2 +bx+c, nd then we will use the formul x dx = tn ( x ) +C. (4) Q(x) contins repeted irreducible qudrtic fctor. If Q(x) hs the fctor (x 2 +bx+ c) r, where b 2 4c < 0, then insted of the single prtil frction, the sum occurs in the R(x) Q(x). A x+b x 2 +bx+c + A 2 x+b 2 (x 2 +bx+c) A r x+b r (x 2 +bx+c) r For exmple, x3 x 2 +2x+2 (x 2 +) 2 =.

13 7.4 Integrtion of Rtionl Functions by Prtil Frctions goo.gl/obeqat 3 Exmple 5 (Cse ()). Show tht secθdθ = ln secθ+tnθ +C. X_cw9yGJspA 我並沒有放錯例題, 它的確是三角函數的積分, 但它可用部份分式法得到結果這裡的想法會比 7.2 介紹的方法容易理解, 但是計算量稍大 Exmple 6 (Cse (2)). Find the integrl x 2 +3x+2 x 3 3x+2 dx. SztwZ7Uyio 例題介紹第二類型的部份分式該如何處理 Exmple 7 (Cse (3)). Evlute the integrl 2x 2 +5x+3 (x 2 +2x+2)(x ) dx. _EPjrXUpV8 例題介紹第三類型的部份分式該如何處理

14 4 7.4 Integrtion of Rtionl Functions by Prtil Frctions goo.gl/obeqat Exmple 8. Find the integrl 2x+ (x 2 +) 2 dx. 3QtcrPI0zZk 例題介紹第四類型的部份分式該如何處理 Rtionlizing substitutions, pge 500 Some nonrtionl functions cn be chnged into rtionl functions by mens of pproprite substitutions. In prticulr, when n integrnd contins n expression of the form n g(x), then the substitution u = n g(x) my be effective. 3_cTz440zNM x+4 Exmple 9 (pge 500). Evlute dx. x 當積分的函數內部有根號, 而且內部不能用三角替換法處理的時候, 可以考慮直接把整個根號令成變數 u, 有時候它可以轉變成對 u 而言的有理函數, 那就再用部份分式法處理

15 7.4 Integrtion of Rtionl Functions by Prtil Frctions goo.gl/obeqat 5 Convert rtionl functions of sinx nd cosx, pge 502 The Germn mthemticin Krl Weierstrss (85-897) noticed tht the substitution t = tn( x 2 ) will convert ny rtionl function of sinx nd cosx into n ordinry rtionl function of t. If t = tn( x 2 ), π < x < π, then we hve ( x cos = 2) By double-ngle formul, we get Furthermore, we cn compute +t 2 ( x nd sin = 2) cosx = t2 +t 2 nd sinx = 2t +t 2. dx = 2 +t 2 dt. t +t 2. 被積函數是 sinx 與 cosx 組成的有理函數, 可透過變數代換 t = tn( x 2 ) 處理 Exmple 0. Find the integrl π cosx dx. 0AOC8TAz7HI 所有以 sinx 與 cosx 為變數的有理函數積分, 都可以透過萬能公式處理, 只要令 t = tn( x 2 ), 就可以把積分變換成以 t 為變數的有理函數雖然這招叫做萬能公式, 但是處理過程非常繁瑣, 所以不建議直接使用, 在其他的積分方法都想不出來的情況下, 不得已再使用這個大絕

16 6 7.5 Strtegy for Integrtion goo.gl/wro3rv 7.5 Strtegy for Integrtion (pge 503) We hve lerned the following techniques to integrte function: DqJHt2gnHsc 這一節回顧到目前為止所學到的五種積分技巧, 必須熟悉關於這裡所列的積分表, 我並沒有背它, 我都是在遇到問題時, 直接透過變數變換的原則很快速地改寫就能得到結果有的時候一個函數的積分處理方法有很多種, 所以平時應多看多試, 培養起你對積分的各種直覺等 Substitution rule, section 5.5. Integrtion prts, section 7.. Trigonometric Integrls 7.2. Trigonometric substitution, section 7.3. Prtil frctions, section 7.4. In this section, we present collection of miscellneous integrls in rndom nd the min chllenge is to recognize which technique or formul to use. Tble of Indefinite Integrls (pge 503). x dx = tn ( x ) +C x 2 2 dx = 2 ln x x+ +C 2 x dx = sin ( x ) +C, > 0 2 x+ +C x x 2 ± dx = ln 2 ± 2 2 Once you re rmed with these bsic integrtion formule, if you don t immeditely see how to ttck given integrl, you might try the following four-steps strtegy. () Simplify the integrnd if possible. Use lgebric mnipultion or trigonometric identities to simplify the integrnd. x(+ x)dx = tnθ sec 2 θ dθ = (sinθ +cosθ) 2 dθ = jijv4j7a0hm 有時候遇到有理函數的積分, 不要急著用部份分式法, 雖然它有標準流程, 但是過程較為繁瑣當分子與分母之間的關聯互為微分與積分的時候, 那就可以直接使用變數變換法而分母是變數平方減常數平方的形式, 也可以嘗試用三角代換處理 (2) Look for n obvious substitution. x x 2 dx = (3) Clssify the integrnd ccording to its form. () Trigonometric function: product of powers of sinx nd cosx, of tnx nd secx, or cotx nd cscx. (b) Rtionl functions: P(x) Q(x), where P(x) nd Q(x) re polynomils. (c) Integrtion by prts: product of power of polynomil nd trnscendentl function (trigonometric, exponentil, or logrithmic).

17 7.5 Strtegy for Integrtion goo.gl/wro3rv 7 (d) Rdicls: ±x 2 ± 2, n x+b, or n g(x). (4) Try gin: remember tht there re bsiclly only two methods of integrtion: substitution nd prts. () Try substitution: inspirtion, ingenuity, despertion. (b) Try prts: it is sometimes effective on single function, such s sin x, tn x, ln x (inverse functions). (c) Mnipulte the integrnd: cosx dx = = (d) Relte the problem to previous problems. tn 2 xsecxdx = (e) Use severl methods: substitution, integrtion by prts, etc. tn 3 x Exmple (pge 505). Compute cos 3 x dx. I3Tl-0lJeqI Solution 2. 積分技巧千變萬化, 微積分課程中只學了其中五種技巧, 而這些技巧可以涵蓋的類型就非常多了所以各位現在開始應該盡量著手思考如何處理每個積分 Exmple 2 (pge 506). Compute e x dx.

18 8 7.5 Strtegy for Integrtion goo.gl/wro3rv Exmple 3 (pge 506). Compute x +x dx. Cn we integrte ll continuous functions? 5MMnkhcWxA 想清楚不可積不會積與積不出來的差別不可積分是定積分之黎曼和極限不存在 ; 不會積指的是個人的積分能力不足 ; 積不出來的意思是反導函數無法表示成基本函數的型式這裡列出幾種積不出來的函數對於積不出來的函數, 有時我們還是可以問它特定的積分值, 因為它積不出來, 所以不可能把反導函數明確表示出, 這時就要透過其他的數學理論去處理 Definition 4 (pge 506). Elementry functions re ll polynomils, rtionl functions, power functions, exponentil functions, logrithmic functions, trigonometric nd inverse trigonometric functions, hyperbolic nd inverse hyperbolic functions, nd ll functions tht cn be obtined from these by the five opertions of ddition, substrction, multipliction, division, nd composition. If f(x) is n elementry function, then f (x) is n elementry function, but f(x)dx need not be n elementry function. For exmple, () 2sin 2 xdx: elliptic integrl ( 橢圓積分 ), 它是計算橢圓弧長的積分表達式 (8. 會介紹如何計算曲線的長度 ) 2 x (2) e t2 dt: error function ( 誤差函數 ), 常用於機率統計 ( 常態分布 ) 與工程 π 0 (3) sin(x 2 )dx, cos(x 2 )dx: Fresnel integrl ( 菲涅耳積分 ), 與誤差函數有關聯 sinx (4) x dx, cos(e x )dx: sine integrl function (cosine integrl function) e x (5) x dx: exponentil integrl ( 指數積分 ) (6) lnx dx: logrithmic integrl ( 對數積分 ) (7) x 3 +dx: 此積分可以化簡成和橢圓積分有關雖然上述函數的積分無法表示成基本函數的型式, 但是有時代入特殊的上下限可以透過其他分析方式 ( 多變數微積分複變函數論微分方程 ) 等求得明確的數值例如 : e x2 dx = π, sin(x 2 )dx = π 2, sinx x dx = π.

19 7.8 Improper Integrls goo.gl/ktcsbm Improper Integrls (pge 527) In this section we extend the concept of definite integrl to the cse where the intervl is infinite nd lso to the cse where f hs n infinite discontinuity in [,b]. In either cse the integrl is clled n improper integrl ( 瑕積分 ). nyqsfbc-b_i Type : Infinite Intervls Definition of n Improper Integrl of Type (pge 528). () If t f(x)dx exists for every number t, then f(x)dx = lim f(x)dx provided this limits exists (s finite number). (b) If b t f(x)dx exists for every number t b, then b provided this limits exists (s finite number). t t f(x)dx = lim t b t f(x)dx The improper integrls f(x)dx nd b f(x)dx re clled convergent ( 收斂 ) if the corresponding limits exists nd divergent ( 發散 ) if the limit does not exist. (c) If both f(x)dx nd f(x)dx convergent, then we define f(x)dx = f(x)dx+ f(x)dx 這一節要處理瑕積分瑕積分有兩類, 首先要觀察的是積分範圍是無窮的情況, 因為在任何有限的範圍內都可以求定積分, 所以這類瑕積分的定義就是對範圍再取一次極限若積分範圍是整個實數軸 R, 則將它分成兩半形成兩個瑕積分, 當確定各別的瑕積分都存在時, 才規定定義在 R 上的函數瑕積分收斂 ; 否則稱為發散 In prt (c) ny rel number cn be used. Exmple (pge 530). Discuss the res of the infinite region R under the curve y = x p,p > 0 nd to the right x =. vtufup2cz-8 這個基本模型必須要徹底理解若用面積的角度理解它的話, p 的次數要大, 區域面積才可能有限而瑕積分好壞的臨界點是 p =

20 Improper Integrls goo.gl/ktcsbm swyjgvfbkoa 第二類瑕積分要處理的是函數在某一點衝到無限大的情況這類瑕積分, 定義方式是在瑕點的附近選擇一點, 那麼定積分就可以處理, 然後再追問這個點趨近於瑕點的時候, 定積分的極限是否存在這裡先定義單邊的瑕積分, 而對於不連續點的函數之瑕積分, 必須確定左極限與右極限瑕積分都必須存在下, 才規定整體的瑕積分收斂, 否則稱為發散對於第二類瑕積分, 也是必須想清楚這個基本模型, 若用面積的角度理解它的話, p 的次數要小, 區域面積才可能有限而瑕積分好壞的臨界點是 p = Type 2: Discontinuous Integrnds Definition of n Improper Integrl of Type 2 (pge 53). () If f is continuous on [,b) nd is discontinuous t b, then if this limits exists (s finite number). (b) If f is continuous on (,b] nd is discontinuous t, then if this limits exists (s finite number). b b f(x)dx = lim t b f(x)dx = lim t + t b t f(x)dx f(x)dx The improper integrls b f(x)dx is clled convergent ( 收斂 ) if the corresponding limits exists nd divergent ( 發散 ) if the limit does not exist. (c) If f hs discontinuity t c, where < c < b, nd both c f(x)dx nd b c f(x)dx convergent, then we define b f(x)dx = c f(x)dx+ b c f(x)dx. Exmple 2 (pge 535). Discuss the res of the region R under the curve y = x p,p > 0, nd between x = 0 nd x =. Exmple 3. Compre Exmple with Exmple 2. Rvet_Spm_r4 多數同學在學習這個單元的時候, 會被怎麼一下 p > 一下又 p < 困住, 實際上這兩類的瑕積分與標準模型彼此是互相等價的只要透過圖形的理解, 就能清楚地知道兩者的關係, 這樣就不會被瑕積分的收斂條件困惑住

21 7.8 Improper Integrls goo.gl/ktcsbm 2 A Comprison Test for Improper Integrls Comprison Theorem (pge 533). Suppose tht f nd g re continuous functions with f(x) g(x) 0 for x. () If f(x)dx is convergent, then g(x)dx is convergent. (b) If g(x)dx is divergent, then f(x)dx is divergent. 定理的條件 f(x) g(x) 0, 函數非負是必要的定理敘述中 for x 可以改成 for some x b, b y y 7wJB_9smHU4 對於複雜函數的瑕積分, 我們在意的是它是否收斂或發散, 而不是追問其明確的積分值判定瑕積分的好壞, 可以用比較判別法處理 : 設法找到一個較為容易的函數, 兩者有大小關係, 而且簡單的函數的瑕積分可以知道收斂或發散, 那就可以得到複雜函數的瑕積分收斂或發散 x x Figure : Comprison Theorem. Exmple 4. () Find the vlues of α for which the improper integrl (b) Evlute the integrl π 0 sec 2 xdx. x α (+ dx converges. x)

22 Improper Integrls goo.gl/ktcsbm cpjncihouz8 Exmple 5 (pge 535). Let I n = 0 x n e x dx. Find the reduction formul. 這個函數與伽瑪函數相關 (Gmm function), 它是一個把階乘連續化的過程 Exmple 6 (pge 535). MoU_SKXztCQ 當 x 很大的時候對數函數比任何冪次函數都還要跑得慢 ; 所以跟冪次函數借一點點去吸收對數, 就可以得知其收斂的條件 () Determine the vlues of α > 0 such tht (b) Find the integrl lnx x 3 dx. lnx dx is convergent. xα PtYMeviy30 Exmple 7. Evlute the improper integrl 2 0 x(2 x) dx. x 瑕積分的問題, 首先要確定瑕點在哪裡, 是無窮遠或是某個特別的點, 再針對它是第一類或第二類瑕積分處理它是收斂或發散這個例題是可以確實透過積分的方式處理瑕積分的值

7. 基本積分公式 (8) sec u tn udu = sec u + C (9) csc u cot udu = csc u + C () tn udu = ln cos u + C = ln sec u + C () cot udu = ln sin u + C = ln csc u + C

7. 基本積分公式 (8) sec u tn udu = sec u + C (9) csc u cot udu = csc u + C () tn udu = ln cos u + C = ln sec u + C () cot udu = ln sin u + C = ln csc u + C 第 7 章積分技巧目錄 7. 基本積分公式............................... 7 7.2 分部積分................................. 72 7.3 遞迴公式................................. 73 7.4 三角函數的冪次.............................. 74 7.5 有理函數的積分..............................