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| 2017/05/25 02:35:15瀏覽57｜回應0｜推薦0 | |
| 標題: ［數學］積分面積 發問: 1. y=(x-2)(2x-1)(x+4),y=0 2.y=f(x)=(x-1)^3,x=-1,x=2,y=0 3.y=x^3-2x^2-1,y=x^2+6x-9 如果可以的話,可以畫個圖 謝謝 ：） 最佳解答:
此文章來自奇摩知識+如有不便請留言告知 y = ( x – 2 ) ( 2x – 1 ) ( x + 4 ), y = 0 積分範圍 ~ x = - 4 ~ 1/2, 1/2 ~ 2 ∫(-4,1/2) ( x – 2 ) ( 2x – 1 ) ( x + 4 ) dx - ∫(1/2,2) ( x – 2 ) ( 2x – 1 ) ( x + 4 ) dx = ∫(-4,1/2) 2x3 + 3x2 - 18x + 8 dx - ∫(1/2,2) 2x3 + 3x2 - 18x + 8 dx = ( x^4/2 + x3 - 9x2 + 8x ) |(-4,1/2) - ( x^4/2 + x3 - 9x2 + 8x ) |(1/2,2) = [( 1/32 + 1/8 – 9/4 + 4 ) – ( 128 – 64 – 144 – 32 )] – [( 8 + 8 – 36 + 16 ) – ( 1/32 + 1/8 – 9/4 + 4 )] = ( 61/32 + 48 ) – ( - 4 – 61/32 ) = 1597/32 + 189/32 = 893/16 --------------------- y = f(x) = ( x – 1 )3, x = - 1, x = 2, y = 0 積分範圍 ~ x = - 1 ~ 1, 1 ~ 2 - ∫(-1,1) ( x – 1 )3 dx + ∫(1,2) ( x – 1 )3 dx = - ( x – 1 )^4/4 |(-1,1) + ( x – 1 )^4/4 |(1,2) = - ( 0 – 4 ) + ( 1/4 – 0 ) = 4 + 1/4 = 17/4 --------------------- y = x3 - 2x2 - 1, y = x2 + 6x – 9 得二曲線交點分別為 (-2,-17), (1,-2), (4,31) 積分範圍 ~ x = - 2 ~ 1, 1 ~ 4 ∫(-2,1) ( x3 - 2x2 - 1 ) – ( x2 + 6x – 9 ) dx + ∫(1,4) ( x2 + 6x – 9 ) - ( x3 - 2x2 - 1 ) dx = ∫(-2,1) ( x3 - 3x2 - 6x + 8 ) dx - ∫(1,4) ( x3 - 3x2 - 6x + 8 ) dx = ( x^4/4 - x3 - 3x2 + 8x ) |(-2,1) – ( x^4/4 - x3 - 3x2 + 8x ) |(1,4) = [( 1/4 – 1 – 3 + 8 ) – ( 4 + 8 – 12 – 16 )] – [( 64 – 64 – 48 + 32 ) – ( 1/4 – 1 – 3 + 8 )] = ( 17/4 + 16 ) – ( - 16 – 17/4 ) = 81/4 + 81/4 = 81/2 P.S 圖形請自行參閱部落格, 若有任何問題或解題有誤請留言 http://tw.myblog.yahoo.com/jw!JCVbjQyaBRbXTWOakincl1.Wpxbobg--/article?mid=7934 2009-04-20 23:44:22 補充： 第一題倒數三行更正如下: = ( 61/32 + 112 ) – ( - 4 – 61/32 ) = 3645/32 + 189/32 = 1917/16 謝謝 linch 老師至部落格留言其他解答: |
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