歸約公式..........................................

1a)

圖片參考：http://i117.photobucket.com/albums/o61/billy_hywung/Redform1.jpg

1b)

圖片參考：http://i117.photobucket.com/albums/o61/billy_hywung/Redform2.jpg

1c)i)

圖片參考：http://i117.photobucket.com/albums/o61/billy_hywung/Redform3.jpg

1c)ii)

圖片參考：http://i117.photobucket.com/albums/o61/billy_hywung/Redform4.jpg

2a)

圖片參考：http://i117.photobucket.com/albums/o61/billy_hywung/Redform5.jpg

2b)

圖片參考：http://i117.photobucket.com/albums/o61/billy_hywung/Redform6.jpg

2c)

圖片參考：http://i117.photobucket.com/albums/o61/billy_hywung/Redform7.jpg

【以上圖片乃本人自製圖片，未經本人同意勿擅自連結或使用】

嘩~~好鬼難睇呀
下次用括號括住d分子分母嘛
1.
(a) y = x^n (1 - x)^(3/2)
　y'　　　　　← y' = dy/dx
= nx^(n - 1) (1 - x)^(3/2) + x^n [3/2 (1 - x)^(1/2)](-1)
= nx^(n - 1) (1 - x)√(1 - x) - 3/2 x^n √(1 - x)
= nx^(n - 1)√(1 - x) - nx^n √(1 - x) - 3/2 x^n √(1 - x)
= nx^(n - 1)√(1 - x) - (n + 3/2)x^n √(1 - x)
= nx^(n - 1)√(1 - x) - [(2n + 3)/2]x^n √(1 - x)
(b) by (a), y' = nx^(n - 1)√(1 - x) - [(2n + 3)/2]x^n √(1 - x)
　　 [(2n + 3)/2]x^n√(1 - x) = nx^(n - 1)√(1 - x) - y'
∫[(2n + 3)/2]x^n√(1 - x) dx = ∫nx^(n - 1)√(1 - x) dx - ∫y' dx
[(2n + 3)/2]∫x^n√(1 - x) dx = n∫x^(n - 1)√(1 - x) dx - [x^n(1 - x)^(3/2)]0→1
[(2n + 3)/2]∫x^n√(1 - x) dx = n∫x^(n - 1)√(1 - x) dx
　　　　　∫x^n√(1 - x) dx = [2n/(2n + 3)]∫x^(n - 1)√(1 - x) dx
上面我唔寫由0→1喇,自己加返上去啦
(c)(i)
　∫√(1 - x) dx
= ∫(1 - x)^(1/2) dx
= [(-1) (1 - x)^(3/2) / (3/2)] 0→1
= [-2/3 (1 - x)^(3/2)] 0→1
= [-2/3 (1 - 1)^(3/2)] - [-2/3 (1 - 0)^(3/2)]
= 2/3
　(ii)
　∫x^4√(1 - x) dx
= [8/(8 + 3)]∫x^3√(1 - x) dx
= (8/11) ∫x^3√(1 - x) dx
= (8/11) [6/(6 + 3)] ∫x^2√(1 - x) dx
= (8/11) (2/3) ∫x^2√(1 - x) dx
= (8/11) (2/3) [4/(4 + 3)] ∫x√(1 - x) dx
= (8/11) (2/3) (4/7) [2/(2 + 3)]∫√(1 - x) dx
= (8/11) (2/3) (4/7) (2/5) ∫√(1 - x) dx
= (8/11) (2/3) (4/7) (2/5) (2/3)
= 256/3465

2.
(a) y = cos^m x sinnx
　y'
= m cos^(m - 1) x (- sinx) sinnx + cos^m x (cosnx)(n)
= -m cos^(m - 1) x sinx sinnx + n cosnx cos^m x
= [cos^(m - 1) x] (- m sinx sinnx + n cosnx cosx)
= [cos^(m - 1) x] (-m sinx sinnx + n cosnx cosx + m cosnx cosx - m cosnx cosx)
= [cos^(m - 1) x] [ - m (sinx sinnx + cosnx cosx) + (m + n) cosnx cosx ]
= [cos^(m - 1) x] [ (m + n) cosnx cosx - m cos(nx - x) ]
= [cos^(m - 1) x] [ (m + n) cosx cosnx - m cos(n - 1)x ]
= (m + n) cos^(m - 1) x cosx cosnx - m cos^(m - 1) x cos(n - 1)x
= (m + n) cos^m x cosnx - m cos^(m - 1) x cos(n - 1)x
(b) y' = (m + n) cos^m x cosnx - m cos^(m - 1) x cos(n - 1)x
　　(m + n) cos^m x cosnx = m cos^(m - 1) x cos(n - 1)x + y'
∫(m + n) cos^m x cosnx dx = ∫m cos^(m - 1) x cos(n - 1)x dx + ∫y' dx
　　　　　　 (m + n)Ｉm,n = mＩm-1,n-1+ [cos^m x sinnx]0→π/2
　　　　　　　　　　Ｉm,n = [m/(m + n)]Ｉm-1,n-1
(c) ∫cos^3 x cos3x dx
= Ｉ3,3
= [3/(3 + 3)]Ｉ2,2
= (1/2) [2/(2 + 2)]Ｉ1,1
= (1/2) (1/2) ∫cosx cosx dx
= (1/4) ∫cos^2 x dx
= (1/4) ∫(1 + cos2x)/2 dx
= (1/8) ∫(1 + cos2x) dx
= (1/8) [x + (1/2)sin2x]0→π/2
= (1/8) { [π/2 + (1/2)sinπ] - [0 + (1/2)sin0] }
= (1/8) (π/2)
= π/16

唔明再問

2007-01-16 09:24:59 補充：
我相信人地未學點用substitution去in數同埋integration by parts

1.a.

dy/dx = n * x^(n-1) * (1 - x)^(3/2) + x^n * 3/2 * (1 - x)^(1/2) * (-1)
= n * x^(n-1) * (1 - x)^(1/2) * (1 - x) - 3/2 * x^n * (1 - x)^(1/2)
= n * x^(n-1) * (1 - x)^(1/2) - n * x^n * (1 - x)^(1/2) - 3/2 * x^n * (1 - x)^(1/2)
= n * x^(n-1) * (1 - x)^(1/2) - (2n + 3)/2 * x^n * (1 - x)^(1/2)

b.

由(a), 可知
d[x^n * (1 - x)^(3/2)]/dx = n * x^(n-1) * (1 - x)^(1/2) - (2n + 3)/2 * x^n * (1 - x)^(1/2)
in(由0至1) {d[x^n * (1 - x)^(3/2)]/dx} dx = in(由0至1) {n * x^(n-1) * (1 - x)^(1/2)}dx - in(由0至1) {(2n + 3)/2 * x^n * (1 - x)^(1/2)}dx
[x^n * (1 - x)^(3/2)] (0至1) = n * [in(由0至1) {x^(n-1) * (1 - x)^(1/2)}dx] - (2n + 3)/2 * [in(由0至1) {x^n * (1 - x)^(1/2)}dx]
0 = n * [in(由0至1) {x^(n-1) * (1 - x)^(1/2)}dx] - (2n + 3)/2 * [in(由0至1) {x^n * (1 - x)^(1/2)}dx]
(2n + 3)/2 * [in(由0至1) {x^n * (1 - x)^(1/2)}dx] = n * [in(由0至1) {x^(n-1) * (1 - x)^(1/2)}dx]
in(由0至1) {x^n * (1 - x)^(1/2)}dx = 2n/(2n + 3) * [in(由0至1) {x^(n-1) * (1 - x)^(1/2)}dx]

c.i.
in(由0至1) {(1 - x)^(1/2)}dx = -in(由0至1) {(1 - x)^(1/2)}d(1 - x)
= -(1 - x)^(3/2) / (3/2) (0至1)
= 2/3

ii. 256/3465

2007-01-15 20:10:10 補充：
第二題的做法相似, 自己試試吧, 學習不能不勞而獲的.