‫ﻣﺘﺮﺟﻢ ﻣﻦ ﺍﻹﻧﺠﻠﻴﺰﻳﺔ ﺇﻟﻰ ﺍﻟﻌﺮﺑﻴﺔ ‪www.onlinedoctranslator.com -‬‬
‫ﻭﺭﻗﺔﺍﻟﻐﺶ ﺣﺴﺎﺏ ﺍﻟﺘﻔﺎﺿﻞ ﻭﺍﻟﺘﻜﺎﻣﻞ‬
‫ﺣﺪﻭﺩ‬
‫ﺗﻌﺮﻳﻒﺩﻗﻴﻖ ‪:‬ﻧﻘﻮﻝ ﻟﻴﻢ‪=(x)F‬ﺇﻝﻟﻮ‬
‫‪→x‬ﺃ‬
‫ﻟﻜﻞ‪ 0 <ε‬ﻳﻮﺟﺪ ﻣﻠﻒ‪ 0 <δ‬ﺑﺤﻴﺚ ﻛﻠﻤﺎ ‪-x> 0‬ﺃ>‪δ‬‬
‫ﺛﻢ‪-(x)F‬ﺇﻝ>‪.ε‬‬
‫"ﺗﻌﺮﻳﻒ ﺍﻟﻌﻤﻞ ‪:‬ﻧﻘﻮﻝ ﻟﻴﻢ‪=(x)F‬ﺇﻝ‬
‫ﺗﻌﺮﻳﻔﺎﺕ‬
‫ﺍﻟﺤﺪﻋﻨﺪ ﺍﻟﻼﻧﻬﺎﻳﺔ‪:‬ﻧﻘﻮﻝ ﻟﻴﻢ‪=(x)F‬ﺇﻝﺍﺫﺍ ﻧﺤﻦ‬
‫‪∞ →x‬‬
‫ﻳﺴﺘﻄﻴﻊﺟﻌﻞ‪(x)F‬ﺃﻗﺮﺏ ﺇﻟﻰﺇﻝﻛﻤﺎ ﻧﺮﻳﺪ ﺑﺄﺧﺬﻫﺎ‪x‬‬
‫ﻛﺒﻴﺮﺓﺑﻤﺎ ﻓﻴﻪ ﺍﻟﻜﻔﺎﻳﺔ ﻭﺇﻳﺠﺎﺑﻴﺔ‪.‬‬
‫ﻳﻮﺟﺪﺗﻌﺮﻳﻒ ﻣﻤﺎﺛﻞ ﻟـ ‪=(x)Flim‬ﺇﻝ‬
‫‪→x‬ﺃ‬
‫ﺇﻻﺃﻧﻨﺎ ﻧﻄﻠﺐ‪x‬ﻛﺒﻴﺮ ﻭﺳﻠﺒﻲ‪.‬‬
‫ﺇﺫﺍﺍﺳﺘﻄﻌﻨﺎ ﺍﻟﻘﻴﺎﻡ ﺑﺬﻟﻚ‪(x)F‬ﺃﻗﺮﺏ ﺇﻟﻰﺇﻝﻛﻤﺎ ﻧﺮﻳﺪ‬
‫ﻋﻦﻃﺮﻳﻖ ﺃﺧﺬ‪x‬ﺑﻤﺎ ﻓﻴﻪ ﺍﻟﻜﻔﺎﻳﺔ ﻗﺮﻳﺒﺔ ﻣﻦﺃ)ﻋﻠﻰ ﺍﻟﺠﺎﻧﺐ‬
‫ﺍﻵﺧﺮﻣﻦﺃ( ﺩﻭﻥ ﺍﻟﺴﻤﺎﺡ‪=x‬ﺃ‪.‬‬
‫‪∞− →x‬‬
‫ﺣﺪﻻﻧﻬﺎﺉﻲ‪:‬ﻧﻘﻮﻝ ﻟﻴﻢ‪∞ =(x)F‬ﺍﺫﺍ ﻧﺤﻦ‬
‫‪→x‬ﺃ‬
‫ﻳﺴﺘﻄﻴﻊﺟﻌﻞ‪(x)F‬ﻛﺒﻴﺮ ﺑﺸﻜﻞ ﺗﻌﺴﻔﻲ )ﻭﺇﻳﺠﺎﺑﻲ(‬
‫ﺣﺪﺍﻟﻴﺪ ﺍﻟﻴﻤﻨﻰ‪:‬ﻟﻴﻢ‪=(x)F‬ﺇﻝ‪.‬ﻫﺬﺍ ﻟﺪﻳﻪ‬
‫ﻋﻦﻃﺮﻳﻖ ﺃﺧﺬ‪x‬ﺑﻤﺎ ﻓﻴﻪ ﺍﻟﻜﻔﺎﻳﺔ ﻗﺮﻳﺒﺔ ﻣﻦﺃ)ﻋﻠﻰ ﺍﻟﺠﺎﻧﺐ‬
‫ﺍﻵﺧﺮﻣﻦﺃ( ﺩﻭﻥ ﺍﻟﺴﻤﺎﺡ‪=x‬ﺃ‪.‬‬
‫‪→x‬ﺃ‪+‬‬
‫ﻧﻔﺲﺗﻌﺮﻳﻒ ﺍﻟﺤﺪ ﺇﻻ ﺃﻧﻪ ﻳﺘﻄﻠﺐ‪<x‬ﺃ‪.‬‬
‫ﻳﻮﺟﺪﺗﻌﺮﻳﻒ ﻣﻤﺎﺛﻞ ﻟـ ‪∞− =(x)Flim‬‬
‫‪→x‬ﺃ‬
‫ﺣﺪﺍﻟﻴﺪ ﺍﻟﻴﺴﺮﻯ‪:‬ﻟﻴﻢ‪=(x)F‬ﺇﻝ‪.‬ﻫﺬﺍ ﻟﺪﻳﻪ‬
‫ﺇﻻﺃﻧﻨﺎ ﻧﺼﻨﻊ‪(x)F‬ﺑﺸﻜﻞ ﺗﻌﺴﻔﻲ ﻛﺒﻴﺮ ﻭﺳﻠﺒﻲ‪.‬‬
‫‪→x‬ﺃ‪-‬‬
‫ﻧﻔﺲﺗﻌﺮﻳﻒ ﺍﻟﺤﺪ ﺇﻻ ﺃﻧﻪ ﻳﺘﻄﻠﺐ ‪>x‬ﺃ‪.‬‬
‫ﺍﻟﻌﻼﻗﺔﺑﻴﻦ ﺍﻟﺤﺪ ﻭﺍﻟﻘﻴﻮﺩ ﻣﻦ ﺟﺎﻧﺐ ﻭﺍﺣﺪ‬
‫ﻟﻴﻢ‪=(x)F‬ﻟﻴﻢ‪=(x)F‬ﺇﻝ⇒ﻟﻴﻢ‪=(x)F‬ﺇﻝ‬
‫ﻟﻴﻢ‪=(x)F‬ﺇﻝ⇒ﻟﻴﻢ‪=(x)F‬ﻟﻴﻢ‪=(x)F‬ﺇﻝ‬
‫‪→x‬ﺃ‬
‫‪→x‬ﺃ‪+‬‬
‫‪→x‬ﺃ‪+‬‬
‫‪→x‬ﺃ‪-‬‬
‫‪→x‬ﺃ‪-‬‬
‫ﻟﻴﻢ‪≠(x)F‬ﻟﻴﻢ‪⇒(x)F‬ﻟﻴﻢ‪(x)F‬ﻏﻴﺮ ﻣﻮﺟﻮﺩ‬
‫‪→x‬ﺃ‪+‬‬
‫‪→x‬ﺃ‬
‫‪→x‬ﺃ‬
‫‪→x‬ﺃ‪-‬‬
‫ﻣﻠﻜﻴﺎﺕ‬
‫ﺍﻓﺘﺮﺽﻟﻴﻢ‪(x)F‬ﻭ ﻟﻴﻢﺯ)‪(x‬ﻛﻼﻫﻤﺎ ﻣﻮﺟﻮﺩ ﻭﺝﻫﻮ ﺃﻱ ﺭﻗﻢ ﺇﺫﻥ ‪،‬‬
‫‪→x‬ﺃ‬
‫‪→x‬ﺃ‬
‫‪-( x)F‬ﻟﻴﻢ‪(x)F‬‬‫‪.4‬ﻟﻴﻢ ‪-‬‬
‫=‪→x‬ﺃ‬‫ﻟﻴﻢﺯ)‪(x‬‬
‫‪→x‬ﺃ‪-‬ﺯ)‪-(x‬‬
‫‪→x‬ﺃ‬
‫‪.1‬ﻟﻴﻢ‪-→x‬ﺃﺭﺍﺟﻊ)‪= -(x‬ﺝﻟﻴﻢ‪→Fx‬ﺃ)‪(x‬‬
‫‪.2‬ﻟﻴﻢ‪-→x‬ﺃ‪±(x)F‬ﺯ)‪= -(x‬ﻟﻴﻢ‪±→x(x)F‬ﺃﻟﻴﻢﺯ)‪(x‬‬
‫‪.3‬ﻟﻴﻢ‪-→x‬ﺃ‪(x)F‬ﺯ)‪= -(x‬ﻟﻴﻢ‪→(xx)F‬ﺃﻟﻴﻢﺯ)‪(x‬‬
‫‪→x‬ﺃ‬
‫‪→x‬ﺃ‬
‫‪.5‬ﻟﻴﻢ ‪F-‬‬
‫)(‪x‬ﻥ‪-‬‬
‫‪.6‬ﻟﻴﻢ ‪-‬ﻥ‬
‫‪→x‬ﺃ‪-‬‬
‫‪= -(x)F‬ﻥﻟﻴﻢ‪F‬‬
‫‪-‬‬
‫‪→x‬ﺃ‬
‫ﻗﺪﻣﺖﻟﻴﻢﺯ)‪0≠(x‬‬
‫‪→x‬ﺃ‬
‫= ‪-‬ﻟﻴﻢ‪-(x)F‬‬
‫‪→x‬ﺃ‬‫‬‫‪→x‬ﺃ‬
‫ﻥ‬
‫)‪(x‬‬
‫ﺗﻘﻴﻴﻤﺎﺕﺍﻟﺤﺪ ﺍﻷﺳﺎﺳﻴﺔ ﻓﻲ‪∞ ±‬‬
‫ﻣﻼﺣﻈﺔ‪)sgn:‬ﺃ(=‪ 1‬ﺇﺫﺍﺃ<‪ 0‬ﻭ ‪)sgn‬ﺃ(= ‪ 1-‬ﺇﺫﺍﺃ>‪.0‬‬
‫‪.1‬ﻟﻴﻢﻩ‪& ∞=x‬‬
‫‪∞ →x‬‬
‫‪.2‬ﻟﻴﻢ ‪∞ =(x)ln‬‬
‫‪∞ →x‬‬
‫‪.3‬ﺇﺫﺍﺹ<‪ 0‬ﺛﻢ ﻟﻴﻢ‬
‫ﻟﻴﻢﻩ‪0=x‬‬
‫‪∞ - →x‬‬
‫& ﻟﻴﻢ ‪ ∞− =(x)ln‬ﺏ‬
‫‪0→x‬‬
‫‪+‬‬
‫‪x∞ →x‬‬
‫ﺹ=‪0‬‬
‫‪.4‬ﺇﺫﺍﺹ<‪ 0‬ﻭ‪x‬ﺹﺣﻘﻴﻘﻲ ﻣﻘﺎﺑﻞ ﺳﻠﺒﻲ‪x‬‬
‫ﺏ‬
‫=‪0‬‬
‫ﺛﻢﻟﻴﻢ‬
‫‪x∞− →x‬ﺹ‬
‫ﻳﺰﻭﺭ‪ http://tutorial.math.lamar.edu‬ﻟﻤﺠﻤﻮﻋﺔ ﻛﺎﻣﻠﺔ ﻣﻦ ﻣﻼﺣﻈﺎﺕ ﺍﻟﺘﻔﺎﺿﻞ ﻭﺍﻟﺘﻜﺎﻣﻞ‪.‬‬
‫‪.5‬ﻥﺣﺘﻰ‪ :‬ﻟﻴﻢ‪x‬ﻥ=∞‬
‫‪∞ ± →x‬‬
‫‪.6‬ﻥﻏﺮﻳﺐ‪ :‬ﻟﻴﻢ‪x‬ﻥ=∞ &‬
‫‪∞ →x‬‬
‫ﻟﻴﻢ‪x‬ﻥ= ‪∞-‬‬
‫‪∞ - →x‬‬
‫‪.7‬ﻥﺣﺘﻰ‪ :‬ﻟﻴﻢﻓﺄﺱﻥ‪+bxL ++‬ﺝ=‪)sgn‬ﺃ(∞‬
‫‪∞ ± →x‬‬
‫‪.8‬ﻥﻏﺮﻳﺐ‪ :‬ﻟﻴﻢﻓﺄﺱﻥ‪+bxL ++‬ﺝ=‪)sgn‬ﺃ(∞‬
‫‪.9‬ﻥﻏﺮﻳﺐ ‪:‬‬
‫‪∞ →x‬‬
‫ﻟﻴﻢﻓﺄﺱﻥ‪+cxL ++‬ﺩ= ‪)sgn-‬ﺃ(∞‬
‫‪∞− →x‬‬
‫©‪ 2005‬ﺑﻮﻝ ﺩﻭﻛﻴﻨﺰ‬
‫ﻭﺭﻗﺔﺍﻟﻐﺶ ﺣﺴﺎﺏ ﺍﻟﺘﻔﺎﺿﻞ ﻭﺍﻟﺘﻜﺎﻣﻞ‬
‫ﺗﻘﻨﻴﺎﺕﺍﻟﺘﻘﻴﻴﻢ‬
‫ﻗﺎﻋﺪﺓﻟﻮﺑﻴﺘﺎﻝ‬
‫ﻭﻇﺎﺉﻒﻣﺴﺘﻤﺮﺓ‬
‫ﻟﻮ‪(x)F‬ﻣﺴﺘﻤﺮ ﻓﻲﺃﺛﻢ ﻟﻴﻢ‪)F=(x)F‬ﺃ(‬
‫‪∞ ± (x)F‬‬
‫‪0 (x)F‬‬
‫=‬
‫= ﺃﻭﻟﻴﻢ‬
‫ﺇﺫﺍ‬
‫‪→x 0‬ﺃﺯ)‪(x‬‬
‫ﻟﻴﻢﺃﺯ)‪(x‬‬
‫‪→x‬‬
‫‪∞±‬‬
‫‪→x‬ﺃ‬
‫ﺍﻟﻮﻇﺎﺉﻒﻭﺍﻟﺘﻜﻮﻳﻦ ﺍﻟﻤﺴﺘﻤﺮ‬
‫‪(x)F‬ﻣﺴﺘﻤﺮ ﻓﻲﺏﻭ ﻟﻴﻢﺯ)‪=(x‬ﺏﺛﻢ‬
‫ﻟﻴﻢ‪)F‬ﺯ)‪ )F=((x‬ﻟﻴﻢﺯ)‪((x‬‬
‫‪→x‬ﺃ‬
‫‪→x‬ﺃ‬
‫‪(x)F‬‬
‫ﻟﻴﻢ‬
‫‪→x‬ﺃﺯ)‪(x‬‬
‫‪→x‬ﺃ‬
‫=‪)F‬ﺏ(‬
‫ﺹ)‪(x‬ﻭﻑ)‪(x‬ﻛﺜﻴﺮﺍﺕ ﺍﻟﺤﺪﻭﺩ‪ .‬ﻟﺤﺴﺎﺏ‬
‫ﺹ)‪(x‬‬
‫ﻟﻴﻢ∞ﻑ)‪(x‬‬
‫‪± →x‬‬
‫‪8 6+x‬‬
‫=‬
‫=ﻟﻴﻢ‬
‫‪2‬‬
‫‪x 2→x‬‬
‫ﻋﻘﻞﺍﻟﺒﺴﻂ ‪ /‬ﺍﻟﻤﻘﺎﻡ‬
‫‪x‬‬
‫‪+ 3x-3‬‬
‫‪x- 3‬‬
‫=ﻟﻴﻢ‬
‫ﻟﻴﻢ‬
‫‪81 -2x9→x‬‬
‫‪x + 3 81 -2x9→x‬‬
‫‪1‬‬‫‪x-9‬‬
‫=ﻟﻴﻢ‬
‫=ﻟﻴﻢ‬
‫‪+ 3 (9+x)9→x‬‬
‫‪x + 3 ( 81 -2x) 9→x‬‬
‫)‬
‫ﻋﺎﻣﻞﺃﻛﺒﺮ ﻗﻮﺓ‪x‬ﻓﻲﻑ)‪(x‬ﺧﺎﺭﺝ‬
‫ﻟﻜﻠﻴﻬﻤﺎﺹ)‪(x‬ﻭﻑ)‪(x‬ﺛﻢ ﺣﺴﺎﺏ ﺍﻟﺤﺪ‪.‬‬
‫=‪4‬‬
‫)‬
‫‪→x‬ﺃ‬
‫ﺯ‪(x)′‬‬
‫ﺃﻫﻮ ﺭﻗﻢ∞ﺃﻭ ‪∞-‬‬
‫ﻛﺜﻴﺮﺍﺕﺍﻟﺤﺪﻭﺩ ﻓﻲ ﺇﻧﻔﻴﻨﻴﺘﻲ‬
‫ﻋﺎﻣﻞﻭﺇﻟﻐﺎء‬
‫)‪(6+x) (2-x‬‬
‫‪12-x4 +2x‬‬
‫=ﻟﻴﻢ‬
‫ﻟﻴﻢ‬
‫‪(2-x)x 2→x‬‬
‫‪x2 -2x 2→x‬‬
‫(‬
‫=ﻟﻴﻢ‬
‫‪( x)′F‬‬
‫ﺛﻢ‪،‬‬
‫‪4 -2x3‬‬
‫ﻟﻴﻢ‬
‫‪2x2-x5∞− →x‬‬
‫=ﻟﻴﻢ‬
‫‪(x‬‬
‫‪1‬‬
‫‪1‬‬‫=‪-‬‬
‫=‬
‫‪108‬‬
‫)‪(6) (18‬‬
‫ﺍﺟﻤﻊﺍﻟﺘﻌﺒﻴﺮﺍﺕ ﺍﻟﻤﻨﻄﻘﻴﺔ‬
‫‪+x)-x-1‬ﺡ(‪-‬‬
‫‪1 -1‬‬
‫‪-1‬‬
‫ﻟﻴﻢ‪-‬‬
‫‬‫‪= -‬ﻟﻴﻢ ﺡ→‪0‬ﺡ‪- -‬‬‫‬‫(‬
‫ﺡ‬
‫‪+‬‬
‫‪x‬‬
‫)‬
‫‪x‬‬
‫‬‫‪x‬‬
‫ﺡ‬
‫‪+‬‬
‫‪x‬‬
‫‬‫ﺡ‬
‫ﺡ→ ‪0‬‬
‫‬‫‬‫‪- -1‬ﺡ ‪-‬‬
‫‪1‬‬
‫‪1‬‬‫=‪-‬‬
‫ ﻟﻴﻢ‬‫=ﻟﻴﻢ ‪-‬‬
‫‪2x‬‬
‫ﺡ→‪0‬ﺡ‪+xx -)-‬ﺡ‪ - (= -‬ﺡ→‪+x)x0‬ﺡ(‬
‫‪)x‬‬
‫‪2 ∞− →x‬‬
‫ﺩﺍﻟﺔﻣﺘﻔﺮﻗﺔ‬
‫)‬
‫‪- 3 2x‬‬
‫ﻟﻴﻢﺯ)‪(x‬ﺃﻳﻦﺯ)‪- =(x‬‬
‫‪2- →x‬‬
‫(‬
‫=ﻟﻴﻢ‬
‫‪2x4‬‬
‫‪(2 -x5‬‬
‫‪∞− →x‬‬
‫‪→x‬‬
‫‪-2-‬‬
‫‪→x‬‬
‫‪→x‬‬
‫‪+2-‬‬
‫‪→x‬‬
‫‪+2-‬‬
‫‪- =x 5‬‬
‫‪2-x‬‬
‫‪ 5 +2x‬ﻟﻮ‪2->x‬‬‫‪x3 −1-‬‬
‫ﺍﺣﺴﺐﺣﺪﻳﻦ ﻣﻦ ﺟﺎﻧﺐ ﻭﺍﺣﺪ ‪،‬‬
‫ﻟﻴﻢﺯ)‪=(x‬ﻟﻴﻢ‪9 = 5 +2x‬‬
‫‪-2-‬‬
‫‪-3‬‬
‫‪2 4‬‬
‫ﻟﻮ‪2- ≤x‬‬
‫ﻟﻴﻢﺯ)‪=(x‬ﻟﻴﻢ ‪7=x3 −1‬‬
‫ﺍﻟﺤﺪﻭﺩﻣﻦ ﺟﺎﻧﺐ ﻭﺍﺣﺪ ﻣﺨﺘﻠﻔﺔ ﺟﺪﺍﺯ)‪(x‬‬
‫‪2- →x‬‬
‫ﻏﻴﺮﻣﻮﺟﻮﺩ‪ .‬ﺇﺫﺍ ﻛﺎﻧﺖ ﺍﻟﺤﺪﻳﻦ ﻋﻠﻰ ﺟﺎﻧﺐ ﻭﺍﺣﺪ‬
‫ﻣﺘﺴﺎﻭﻳﺘﻴﻦ ‪،‬ﻓﻌﻨﺪﺉﺬ ٍ‪lim‬ﺯ)‪(x‬ﻛﺎﻧﺖ ﻣﻮﺟﻮﺩﺓ‬
‫‪2- →x‬‬
‫ﻭﻛﺎﻥﻟﻪ ﻧﻔﺲ ﺍﻟﻘﻴﻤﺔ‪.‬‬
‫ﺑﻌﺾﺍﻟﻮﻇﺎﺉﻒ ﺍﻟﻤﺴﺘﻤﺮﺓ‬
‫ﻗﺎﺉﻤﺔﺟﺰﺉﻴﺔ ﻟﻠﻮﻇﺎﺉﻒ ﺍﻟﻤﺴﺘﻤﺮﺓ ﻭﻗﻴﻢ‪x‬ﺍﻟﺘﻲ ﻫﻢ ﻣﺴﺘﻤﺮﻭﻥ ﻣﻦ ﺃﺟﻠﻬﺎ‪ .1 .‬ﻣﺘﻌﺪﺩﺍﺕ ﺍﻟﺤﺪﻭﺩ‬
‫ﻟﻠﺠﻤﻴﻊ‪.x‬‬
‫‪.7‬ﻛﻮﺱ)‪(x‬ﻭﺍﻟﺨﻄﻴﺉﺔ)‪(x‬ﻟﻠﺠﻤﻴﻊ‪.x‬‬
‫‪.2‬ﻭﻇﻴﻔﺔ ﻋﻘﻼﻧﻴﺔ ‪ ،‬ﺑﺎﺳﺘﺜﻨﺎء‪x‬ﺍﻟﺘﻲ ﺗﻌﻄﻲ ﺍﻟﻘﺴﻤﺔ‬
‫‪.8‬ﺗﺎﻥ)‪(x‬ﻭ ﺛﺎﻧﻴﺔ)‪(x‬ﻣﺘﺎﺡ‬
‫ﻋﻠﻰﺻﻔﺮ‪.‬‬
‫‪π3 π π π3‬‬
‫‪.3‬ﻥ‪)x‬ﻥﻏﺮﻳﺐ( ﻟﻠﺠﻤﻴﻊ‪.x‬‬
‫‪،‬ﻻﻡ‬
‫‪ -،‬ﻭ ﻭ‬
‫‪≠x‬ﻻﻡ ‪- ،‬‬
‫‪2 2 2‬‬
‫‪2‬‬
‫‪.4‬ﻥ‪)x‬ﻥﺣﺘﻰ( ﻟﻠﺠﻤﻴﻊ‪.0≤x‬‬
‫‪.9‬ﺳﺮﻳﺮ)‪(x‬ﻭ ‪(x)CSC‬ﻣﺘﺎﺡ‬
‫‪.5‬ﻩ‪x‬ﻟﻠﺠﻤﻴﻊ‪.x‬‬
‫‪≠x‬ﻻﻡ ‪π- ،π2- ،‬ﻭ‪π، 0‬ﻭ‪π2‬ﻭﺇﻝ‬
‫‪x6. ln‬ﻝ‪.0<x‬‬
‫ﻧﻈﺮﻳﺔﺍﻟﻘﻴﻤﺔ ﺍﻟﻤﺘﻮﺳﻄﺔ‬
‫ﻟﻨﻔﺘﺮﺽﺃﻥ‪(x)F‬ﻣﺴﺘﻤﺮ ﻋﻠﻰ ]ﺃ ‪ ،‬ﺏ[ ﻭﺍﺳﻤﺤﻮﺍﻡﻳﻜﻮﻥ ﺃﻱ ﺭﻗﻢ ﺑﻴﻦ‬
‫ﺛﻢﻳﻮﺟﺪ ﺭﻗﻢﺝﻣﺜﻞ ﺫﻟﻚﺃ>ﺝ>ﺏﻭ‪)F‬ﺝ(=ﻡ‪.‬‬
‫ﻳﺰﻭﺭ‪ http://tutorial.math.lamar.edu‬ﻟﻤﺠﻤﻮﻋﺔ ﻛﺎﻣﻠﺔ ﻣﻦ ﻣﻼﺣﻈﺎﺕ ﺍﻟﺘﻔﺎﺿﻞ ﻭﺍﻟﺘﻜﺎﻣﻞ‪.‬‬
‫‪)F‬ﺃ(ﻭ‪)F‬ﺏ(‪.‬‬
‫©‪ 2005‬ﺑﻮﻝ ﺩﻭﻛﻴﻨﺰ‬
‫‪3‬‬
‫‪2‬‬
‫ﻭﺭﻗﺔﺍﻟﻐﺶ ﺣﺴﺎﺏ ﺍﻟﺘﻔﺎﺿﻞ ﻭﺍﻟﺘﻜﺎﻣﻞ‬
‫ﺍﻟﻤﺸﺘﻘﺎﺕ‬
‫ﻟﻮﺫ=‪(x)F‬‬
‫ﺍﻟﺘﻌﺮﻳﻒﻭﺍﻟﺘﺮﻣﻴﺰ‬
‫ﺛﻢﻳﺘﻢ ﺗﻌﺮﻳﻒ ﺍﻟﻤﺸﺘﻖ ﻟﻴﻜﻮﻥ‪=(x)′F‬ﻟﻴﻢ‬
‫ﺡ→‪0‬‬
‫ﻟﻮﺫ=‪(x)F‬ﺛﻢ ﻛﻞ ﻣﺎ ﻳﻠﻲ‬
‫ﺡ‬
‫ﻟﻮﺫ=‪(x)F‬ﻛﻞ ﻣﺎ ﻳﻠﻲ ﻫﻮ ﺭﻣﻮﺯ ﻣﻜﺎﻓﺉﺔ ﻟﻠﻤﺸﺘﻘﺎﺕ‬
‫ﺍﻟﺘﻲﺗﻢ ﺗﻘﻴﻴﻤﻬﺎ ﻓﻲ‪=x‬ﺃ‪.‬‬
‫ﺍﻟﺮﻣﻮﺯﺍﻟﻤﻜﺎﻓﺉﺔ ﻟﻠﻤﺸﺘﻖ‪.‬‬
‫‪df dy d‬‬
‫‪=(x)′F‬ﺫ‪=((x)F)= = = ′‬ﻣﺪﺍﻓﻊ)‪(x‬‬
‫‪dx dx dx‬‬
‫‪()′‬‬
‫ﻛﺮﺓﺍﻟﻘﺪﻡ‬
‫ﺗﻔﺴﻴﺮﺍﻟﻤﺸﺘﻖ‬
‫ﻟﻮﺫ=‪(x)F‬ﺛﻢ‪،‬‬
‫‪+x ) F‬ﺡ‪) F(-‬‬
‫‪(x‬‬
‫‪.‬‬
‫=ﺫ‪′‬‬
‫‪=x‬ﺃ‬
‫=‬
‫ﻣﺪﺍﻓﻊ‬
‫‪=xdx‬ﺃ‬
‫ﺩﻯ‬
‫=‬
‫‪=xdx‬ﺃ‬
‫=ﻣﺪﺍﻓﻊ)ﺃ(‬
‫‪)′F.2‬ﺃ(ﻫﻮ ﻣﻌﺪﻝ ﺍﻟﺘﻐﻴﻴﺮ ﺍﻟﻔﻮﺭﻱ‪(x)F‬ﻓﻲ‪=x‬‬
‫ﺃ‪.‬‬
‫‪.1‬ﻡ=‪)′F‬ﺃ(ﻫﻮ ﻣﻨﺤﺪﺭ ﺍﻟﻈﻞ‬
‫ﺧﻂﻝﺫ=‪(x)F‬ﻓﻲ‪=x‬ﺃﻭ ﺍﻝ‬
‫‪.3‬ﺇﺫﺍ‪(x)F‬ﻫﻮ ﻣﻮﺿﻊ ﺍﻟﻜﺎﺉﻦ ﻓﻲ‬
‫ﻣﻌﺎﺩﻟﺔﺧﻂ ﺍﻟﻤﻤﺎﺱ ﻋﻨﺪ‪=x‬ﺃﺍﻋﻄﻲ ﻣﻦ ﻗﺒﻞﺫ‬
‫=‪)F‬ﺃ(‪)′F+‬ﺃ( )‪-x‬ﺃ(‪.‬‬
‫ﻭﻗﺖ‪x‬ﺛﻢ‪)′F‬ﺃ(ﻫﻲ ﺳﺮﻋﺔ ﺍﻟﺠﺴﻢ ﻋﻨﺪ‪=x‬ﺃ‪.‬‬
‫ﺍﻟﺨﺼﺎﺉﺺﺍﻷﺳﺎﺳﻴﺔ ﻭﺍﻟﺼﻴﻎ‬
‫ﻟﻮ ‪(x)F‬ﻭﺯ)‪(x‬ﻫﻲ ﻭﻇﺎﺉﻒ ﻗﺎﺑﻠﺔ ﻟﻠﺘﻔﺎﺿﻞ )ﺍﻟﻤﺸﺘﻖ ﻣﻮﺟﻮﺩ( ‪،‬ﺝﻭﻥﻫﻲ ﺃﻱ ﺃﺭﻗﺎﻡ ﺣﻘﻴﻘﻴﺔ ‪،‬‬
‫ﺩ‬
‫‪.5‬‬
‫‪dx‬‬
‫ﺩ‬
‫)‪x‬ﻥ(=‪nx‬ﻥ‪-1−‬ﺣﻜﻢ ﺍﻟﻘﻮﺓ‬
‫‪.6‬‬
‫‪dx‬‬
‫ﺩ‬
‫‪.7‬‬
‫)‪)F‬ﺯ)‪)′F=(((x‬ﺯ)‪((x‬ﺯ‪(x)′‬‬
‫‪dx‬‬
‫‪).1‬ﺭﺍﺟﻊ(‪= ′‬ﺭﺍﺟﻊ‪(x)′‬‬
‫)ﺝ(=‪0‬‬
‫‪±F).2‬ﺯ(‪±(x)′F= ′‬ﺯ‪(x)′‬‬
‫‪′F= ′(fg).3‬ﺯ‪-′fg+‬ﺳﻴﺎﺩﺓ ﺍﻟﻤﻨﺘﺞ‬
‫‪′-F‬‬‫‪= - -.4‬‬
‫‪-‬ﺯ‪-‬‬
‫ﺩ)‪(x‬‬
‫‪dx‬‬
‫‪′F‬ﺯ‪′fg-‬‬
‫ﺯ‬
‫‪2‬‬
‫=‪1‬‬
‫ﺩ)ﺍﻟﺨﻄﻴﺉﺔ‪=(x‬ﻛﻮﺱ‪dx‬‬
‫‪x‬‬
‫ﺩ)ﻛﻮﺱ‪- =(x‬ﺍﻟﺨﻄﻴﺉﺔ‪dx‬‬
‫‪x‬‬
‫ﺩ)ﺗﺎﻥ‪=(x‬ﺛﺎﻧﻴﺔ‪dx2‬‬
‫‪x‬‬
‫ﺩ)ﺛﺎﻧﻴﺔ‪=(x‬ﺛﺎﻧﻴﺔ‪x‬ﺗﺎﻥ‪x dx‬‬
‫‪-‬ﺍﻟﻘﺎﻋﺪﺓ ﺍﻟﺤﺎﺻﻞ‬
‫ﻫﺬﺍﺍﻝﺣﻜﻢ ﺍﻟﺴﻠﺴﻠﺔ‬
‫ﺍﻟﻤﺸﺘﻘﺎﺕﺍﻟﺸﺎﺉﻌﺔ‬
‫ﺩ‬
‫‪dx‬‬
‫ﺩ‬
‫)ﺳﺮﻳﺮ ﻧﻘﺎﻝ‪x CSC- =(x‬‬
‫‪dx‬‬
‫‪1‬‬
‫ﺩ‬
‫)ﺍﻟﺨﻄﻴﺉﺔ ‪=(x‬‬
‫‪dx‬‬
‫‪2x−1‬‬
‫‪1‬‬
‫ﺩ‬
‫‪-(=x‬‬
‫) ﺍ‪-‬‬
‫‪1CS‬‬
‫‪dx‬‬
‫‪2x−1‬‬
‫‪1‬‬
‫ﺩ‬
‫)ﺗﺎﻥ‪=(x1−‬‬
‫‪2x+1‬‬
‫‪dx‬‬
‫)‪xCSC- =(xCSC‬ﺳﺮﻳﺮ ﻧﻘﺎﻝ‪x‬‬
‫‪2‬‬
‫‪1−‬‬
‫ﻳﺰﻭﺭ‪ http://tutorial.math.lamar.edu‬ﻟﻤﺠﻤﻮﻋﺔ ﻛﺎﻣﻠﺔ ﻣﻦ ﻣﻼﺣﻈﺎﺕ ﺍﻟﺘﻔﺎﺿﻞ ﻭﺍﻟﺘﻜﺎﻣﻞ‪.‬‬
‫ﺩ‬
‫‪dx‬‬
‫ﺩ‬
‫)ﻩ‪=(x‬ﻩ‪x‬‬
‫‪dx‬‬
‫‪1‬‬
‫ﺩ‬
‫)‪ =((x)ln‬ﻭ ‪0<x‬‬
‫‪x‬‬
‫‪dx‬‬
‫‪1‬‬
‫ﺩ‬
‫)‪ = ( xln‬ﻭ ‪0≠x‬‬
‫‪x‬‬
‫‪dx‬‬
‫‪1‬‬
‫ﺩ‬
‫ﻭ ‪0< x‬‬
‫)(( =‬
‫)ﺳﺠﻞﺃ ‪x‬‬
‫‪lnx‬ﺃ‬
‫‪dx‬‬
‫)ﺃ‪=(x‬ﺃ‪)lnx‬ﺃ(‬
‫©‪ 2005‬ﺑﻮﻝ ﺩﻭﻛﻴﻨﺰ‬
‫ﻭﺭﻗﺔﺍﻟﻐﺶ ﺣﺴﺎﺏ ﺍﻟﺘﻔﺎﺿﻞ ﻭﺍﻟﺘﻜﺎﻣﻞ‬
‫ﻣﺘﻐﻴﺮﺍﺕﻗﺎﻋﺪﺓ ﺍﻟﺴﻠﺴﻠﺔ‬
‫ﺗﻄﺒﻖﻗﺎﻋﺪﺓ ﺍﻟﺴﻠﺴﻠﺔ ﻋﻠﻰ ﺑﻌﺾ ﺍﻟﻮﻇﺎﺉﻒ ﺍﻟﻤﺤﺪﺩﺓ‪.‬‬
‫‪.1‬‬
‫ﺩ‪)dx‬‬
‫(‬
‫‪- -(x)F--‬ﻥ=ﻥ‪-(x)F-‬‬
‫ﻥ‪1−‬‬
‫‪-‬‬
‫‪(x)′F‬‬
‫ﺩ‬
‫‪.2‬‬
‫‪dx‬‬
‫‪(x) ′F‬‬
‫ﺩ‬
‫)‪=(-(x)Fln-‬‬
‫‪.3‬‬
‫‪(x)F‬‬
‫‪dx‬‬
‫ﺩ‬
‫‪.4‬‬
‫)ﺍﻟﺨﻄﻴﺉﺔ ‪(x)′F=(-(x)F-‬ﻛﻮﺱ ‪(x)F-‬‬
‫‪dx‬‬
‫)ﻩ‪(x)′F=((x)F‬‬
‫‪2‬‬
‫‪=(x)()F=(x)′ ′F‬‬
‫‪.6‬‬
‫ﻩ‪(x)F‬‬
‫ﻳﺘﻢﺍﻹﺷﺎﺭﺓ ﺇﻟﻰ ﺍﻟﻤﺸﺘﻖ ﺍﻟﺜﺎﻧﻲ ﻛـ‬
‫‪2‬‬
‫ﻣﺪﺍﻓﻊ‬
‫‪dx‬‬
‫‪2‬‬
‫ﺩ‬
‫‪)dx .5‬ﻛﻮﺱ ‪(x)′F- =(-(x)F-‬ﺍﻟﺨﻄﻴﺉﺔ ‪- -(x)F-‬‬
‫ﺩ‪)dx‬‬
‫(‬
‫ﺗﺎﻥ ‪(x)′F= - -(x)F-‬ﺛﺎﻧﻴﺔ‬
‫‪2‬‬
‫ﺩ‬
‫‪.7‬‬
‫‪dx‬‬
‫ﺩ‬
‫‬‫‪)dx .8‬ﺗﺎﻥ‪-=(-(x)F-1‬‬
‫‪-(x)F-‬‬
‫)ﺛﺎﻧﻴﺔ]‪(x) ′F=([(x)F‬ﺛﺎﻧﻴﺔ]‪[(x)F‬ﺗﺎﻥ]‪[(x)F‬‬
‫‪-‬‬
‫ﺍﻟﻤﺸﺘﻘﺎﺕﺫﺍﺕ ﺍﻟﺘﺮﺗﻴﺐ ﺍﻷﻋﻠﻰ‬
‫ﺛﻢﺫ‬
‫ﺍﻟﻤﺸﺘﻖﻫﻮ ﺍﻟﻤﺸﺎﺭ ﺇﻟﻴﻪ‬
‫‪)F‬ﻥ()‪= (x‬‬
‫ﻭﻳﺘﻢﺗﻌﺮﻳﻔﻪ ﻋﻠﻰ ﺃﻧﻪ‬
‫‪، ′((x)′F)=(x)′ ′F‬ﺃﻱﻣﺸﺘﻖ ﺍﻟﻤﺸﺘﻖ ﺍﻷﻭﻝ ‪)′F،‬‬
‫‪.(x‬‬
‫‪(x)′F‬‬
‫‪-2-(x)F- +1‬‬
‫ﻥ‬
‫ﻣﺪﺍﻓﻊ‬
‫‪dx‬‬
‫ﻥ‬
‫ﻭﻳﺘﻢﺗﻌﺮﻳﻔﻪ ﻋﻠﻰ ﺃﻧﻪ‬
‫)‬
‫(‬
‫‪)F‬ﻥ()‪)F =(x‬ﻥ‪ (x ′)(1−‬ﻭﺃﻱﻣﺸﺘﻖ ﻣﻦ‬
‫ﺍﻝ)ﻥ‪(1-‬ﺷﺎﺭﻉﺍﻟﻤﺸﺘﻖ‪)F،‬ﻥ‪.(x)(1−‬‬
‫ﺍﻻﺷﺘﻘﺎﻕﺍﻟﻀﻤﻨﻲ‬
‫ﻳﺠﺪﺫ‪′‬ﻟﻮﻩ‪9−x2‬ﺫ‪3x+‬ﺫ‪ =2‬ﺍﻟﺨﻄﻴﺉﺔ)ﺫ(‪.x11+‬ﻳﺘﺬﻛﺮﺫ=ﺫ)‪(x‬ﻫﻨﺎ ‪ ،‬ﻟﺬﺍ ﻓﺈﻥ ﻣﻨﺘﺠﺎﺕ ‪ /‬ﺣﻮﺍﺟﺰ‪x‬ﻭﺫ‬
‫ﺳﻴﺴﺘﺨﺪﻡﺍﻟﻤﻨﺘﺞ ‪ /‬ﻗﺎﻋﺪﺓ ﺣﺎﺻﻞ ﺍﻟﻘﺴﻤﺔ ﻭﻣﺸﺘﻘﺎﺗﻪﺫﺳﻴﺴﺘﺨﺪﻡ ﻗﺎﻋﺪﺓ ﺍﻟﺴﻠﺴﻠﺔ‪" .‬ﺍﻟﺤﻴﻠﺔ" ﻫﻲ ﺍﻟﺘﻔﺮﻳﻖ‬
‫ﻛﺎﻟﻤﻌﺘﺎﺩﻭﻓﻲ ﻛﻞ ﻣﺮﺓ ﺗﻔﺮﻕ ﻓﻴﻬﺎﺫﺃﻧﺖ ﺗﻌﻠﻖ ﻋﻠﻰ ﺃﺫ‪) ′‬ﻣﻦ ﻗﺎﻋﺪﺓ ﺍﻟﺴﻠﺴﻠﺔ(‪.‬‬
‫ﺑﻌﺪﺍﻟﺘﻔﺮﻳﻖ ﻳﺤﻞ ﻝﺫ‪.′‬‬
‫ﻩ‪9−x2‬ﺫ)‪9 - 2‬ﺫ‪2x3+(′‬ﺫ‪3x2 +2‬ﺱ ﺹ‪= ′‬ﻛﻮﺱ)ﺫ(ﺫ‪2 11+ ′‬ﻩ‪9−x2‬ﺫ‪9-‬ﺫ‬
‫‪′‬ﻩ‪9−x2‬ﺫ‪2x3+‬ﺫ‪3x2 +2‬ﺱ ﺹ‪= ′‬ﻛﻮﺱﺱ ﺹ‪11+ ′‬‬
‫)(‬
‫)‪3x2‬ﺫ‪9-‬ﻩ‪9−x2‬ﺫ‪-‬ﻛﻮﺱ)ﺫ((ﺫ‪2 −11= ′‬ﻩ‪9−x2‬ﺫ‪ x3-‬ﺫ‬
‫⇒‬
‫‪2 2‬‬
‫‪2‬‬
‫ﺱ ‪2‬ﺹ‬
‫‪2 −11‬ﻩ‪9−x2‬ﺫ ‪3 -‬‬
‫ﺫ‪= ′‬‬
‫‪3x2‬ﺫ‪9-‬ﻩ‪9−x2‬ﺫ‪-‬ﻛﻮﺱ)ﺫ(‬
‫ﺯﻳﺎﺩﺓ ‪ /‬ﺗﻨﺎﻗﺺ ‪ -‬ﻣﻘﻌﺮ ﻷﻋﻠﻰ ‪ /‬ﻣﻘﻌﺮ ﻷﺳﻔﻞ‬
‫ﻧﻘﺎﻁﺣﺮﺟﺔ‬
‫‪=x‬ﺝﻫﻲ ﻧﻘﻄﺔ ﺣﺮﺟﺔ ﻣﻦ‪(x)F‬ﺍﻟﻤﻘﺪﻣﺔ ﺳﻮﺍء‬
‫‪)′F.1‬ﺝ(=‪ 0‬ﺃﻭ‪)′F.2‬ﺝ(ﻏﻴﺮ ﻣﻮﺟﻮﺩ‪.‬‬
‫ﺯﻳﺎﺩﺓ ‪ /‬ﺗﻨﺎﻗﺺ‬
‫‪.1‬ﺇﺫﺍ‪ 0<(x)′F‬ﻟﻠﺠﻤﻴﻊ‪x‬ﻓﻲ ﻓﺘﺮﺓﺃﻧﺎﺛﻢ‬
‫‪(x)F‬ﻳﺘﺰﺍﻳﺪ ﻓﻲ ﺍﻟﻔﺘﺮﺓﺃﻧﺎ‪.‬‬
‫‪.2‬ﺇﺫﺍ‪ 0>(x)′F‬ﻟﻠﺠﻤﻴﻊ‪x‬ﻓﻲ ﻓﺘﺮﺓﺃﻧﺎﺛﻢ ‪(x)F‬ﻳﺘﻨﺎﻗﺺ‬
‫ﻓﻲﺍﻟﻔﺘﺮﺓﺃﻧﺎ‪.‬‬
‫‪.3‬ﺇﺫﺍ‪ 0=(x)′F‬ﻟﻠﺠﻤﻴﻊ‪x‬ﻓﻲ ﻓﺘﺮﺓﺃﻧﺎﺛﻢ ‪(x)F‬ﺛﺎﺑﺖ ﻓﻲ‬
‫ﻣﻘﻌﺮﻷﻋﻠﻰ ‪ /‬ﻣﻘﻌﺮ ﻟﻸﺳﻔﻞ‬
‫‪.1‬ﺇﺫﺍ‪ 0<(x)′ ′F‬ﻟﻠﺠﻤﻴﻊ‪x‬ﻓﻲ ﻓﺘﺮﺓﺃﻧﺎﺛﻢ‬
‫‪(x)F‬ﻣﻘﻌﺮﺓ ﻓﻲ ﺍﻟﻔﺘﺮﺓ ﺍﻟﻔﺎﺻﻠﺔﺃﻧﺎ‪.‬‬
‫‪.2‬ﺇﺫﺍ‪ 0>(x)′ ′F‬ﻟﻠﺠﻤﻴﻊ‪x‬ﻓﻲ ﻓﺘﺮﺓﺃﻧﺎﺛﻢ ‪(x)F‬ﻣﻘﻌﺮ‬
‫ﻷﺳﻔﻞﻓﻲ ﺍﻟﻔﺘﺮﺓ ﺍﻟﻔﺎﺻﻠﺔﺃﻧﺎ‪.‬‬
‫ﻧﻘﺎﻁﺍﻻﻧﻘﻼﺏ‬
‫‪=x‬ﺝﻫﻲ ﻧﻘﻄﺔ ﺍﻧﻌﻄﺎﻑ‪(x)F‬ﺇﺫﺍ ﻛﺎﻥ‬
‫ﻳﺘﻐﻴﺮﺍﻟﺘﻘﻌﺮ ﻓﻲ‪=x‬ﺝ‪.‬‬
‫ﺍﻟﻔﺘﺮﺓﺃﻧﺎ‪.‬‬
‫ﻳﺰﻭﺭ‪ http://tutorial.math.lamar.edu‬ﻟﻤﺠﻤﻮﻋﺔ ﻛﺎﻣﻠﺔ ﻣﻦ ﻣﻼﺣﻈﺎﺕ ﺍﻟﺘﻔﺎﺿﻞ ﻭﺍﻟﺘﻜﺎﻣﻞ‪.‬‬
‫©‪ 2005‬ﺑﻮﻝ ﺩﻭﻛﻴﻨﺰ‬
‫ﻭﺭﻗﺔﺍﻟﻐﺶ ﺣﺴﺎﺏ ﺍﻟﺘﻔﺎﺿﻞ ﻭﺍﻟﺘﻜﺎﻣﻞ‬
‫ﺇﻛﺴﺘﺮﻳﻤﺎ‬
‫ﻧﺴﺒﻲ)ﻣﺤﻠﻲ( ﺇﻛﺴﺘﺮﻳﻤﺎ‬
‫‪=x.1‬ﺝﻫﻮ ﻧﺴﺒﻲ )ﺃﻭ ﻣﺤﻠﻲ( ﻛﺤﺪ ﺃﻗﺼﻰ‬
‫‪(x)F‬ﻟﻮ‪)F‬ﺝ(≤‪(x)F‬ﻟﻠﺠﻤﻴﻊ‪x‬ﻗﺮﻳﺐﺝ‪.‬‬
‫ﺍﻟﻤﻄﻠﻘﺔﺇﻛﺴﺘﺮﻳﻤﺎ‬
‫‪=x.1‬ﺝﻫﻮ ﺍﻟﺤﺪ ﺍﻷﻗﺼﻰ ﺍﻟﻤﻄﻠﻖ‪(x)F‬‬
‫ﻟﻮ‪)F‬ﺝ(≤‪(x)F‬ﻟﻠﺠﻤﻴﻊ‪x‬ﻓﻲ ﺍﻟﻤﺠﺎﻝ‪.‬‬
‫‪=x.2‬ﺝﻫﻮ ﻧﺴﺒﻲ )ﺃﻭ ﻣﺤﻠﻲ( ﻛﺤﺪ ﺃﺩﻧﻰ ‪(x)F‬ﻟﻮ‪)F‬ﺝ(‬
‫≥‪(x)F‬ﻟﻠﺠﻤﻴﻊ‪x‬ﻗﺮﻳﺐﺝ‪.‬‬
‫‪=x.2‬ﺝﻫﻮ ﺍﻟﺤﺪ ﺍﻷﺩﻧﻰ ﺍﻟﻤﻄﻠﻖ‪(x)F‬‬
‫ﻟﻮ‪)F‬ﺝ(≥‪(x)F‬ﻟﻠﺠﻤﻴﻊ‪x‬ﻓﻲ ﺍﻟﻤﺠﺎﻝ‪.‬‬
‫ﻧﻈﺮﻳﺔﻓﻴﺮﻣﺎﺕ‬
‫ﻟﻮ‪(x)F‬ﻟﺪﻳﻪ ﻗﺮﻳﺐ )ﺃﻭ ﻣﺤﻠﻲ( ﻗﻴﻤﺔ ﻗﺼﻮﻯ ﻋﻨﺪ‬
‫‪=x‬ﺝﻭﺛﻢ‪=x‬ﺝﻫﻲ ﻧﻘﻄﺔ ﺣﺮﺟﺔ ﻣﻦ‪.(x)F‬‬
‫‪1‬ﺷﺎﺭﻉﺍﺧﺘﺒﺎﺭ ﻣﺸﺘﻖ‬
‫ﻟﻮ‪=x‬ﺝﻫﻲ ﻧﻘﻄﺔ ﺣﺮﺟﺔ ﻣﻦ‪(x)F‬ﺛﻢ‪=x‬ﺝﻳﻜﻮﻥ‬
‫‪.1‬ﺫﺍﺕ ﺍﻟﺼﻠﺔ‪ .‬ﺍﻷﻋﻠﻰ‪ .‬ﻝ‪(x)F‬ﻟﻮ‪ 0<(x)′F‬ﻋﻠﻰ ﻳﺴﺎﺭ‪=x‬‬
‫ﺝﻭ‪ 0>(x)′F‬ﻋﻠﻰ ﻳﻤﻴﻦ‪=x‬ﺝ‪.‬‬
‫‪.2‬ﻣﺮﺟﻊ‪ .‬ﺩﻗﻴﻘﺔ‪ .‬ﻝ‪(x)F‬ﻟﻮ‪ 0>(x)′F‬ﻋﻠﻰ ﻳﺴﺎﺭ‪=x‬ﺝﻭ‪F‬‬
‫ﻧﻈﺮﻳﺔﺍﻟﻘﻴﻤﺔ ﺍﻟﻘﺼﻮﻯ‬
‫ﻟﻮ‪(x)F‬ﻣﺴﺘﻤﺮ ﻓﻲ ﺍﻟﻔﺘﺮﺓ ﺍﻟﻤﻐﻠﻘﺔ‬
‫‪ 0<(x)′‬ﻋﻠﻰ ﻳﻤﻴﻦ‪=x‬ﺝ‪.‬‬
‫]ﺃﻭﺏ[ﺛﻢ ﺗﻮﺟﺪ ﺃﺭﻗﺎﻡﺝﻭﺩﻟﻬﺬﺍ ﺍﻟﺴﺒﺐ‪،‬‬
‫‪.1‬ﺃ≥ﺝﻭﺩ≥ﺏﻭ‪)F.2‬ﺝ(ﻫﻮ ﺍﻟﻘﻴﻤﺔ ﺍﻟﻤﻄﻠﻘﺔ‪ .‬ﺍﻷﻋﻠﻰ‪ .‬ﻓﻲ‬
‫]ﺃﻭﺏ[ﻭ‪)F.3‬ﺩ(ﻫﻮ ﺍﻟﻘﻴﻤﺔ ﺍﻟﻤﻄﻠﻘﺔ‪ .‬ﺩﻗﻴﻘﺔ‪ .‬ﻓﻲ]ﺃﻭﺏ[‪.‬‬
‫‪.3‬ﻻ ﻗﻴﻤﺔ ﻗﺼﻮﻯ ﻧﺴﺒﻲ‪(x)F‬ﻟﻮ‪(x)′F‬ﻫﻲ ﻧﻔﺲ‬
‫ﺍﻟﻌﻼﻣﺔﻋﻠﻰ ﺟﺎﻧﺒﻲ‪=x‬ﺝ‪.‬‬
‫‪2‬ﺍﺧﺘﺼﺎﺭ ﺍﻟﺜﺎﻧﻲﺍﺧﺘﺒﺎﺭ ﻣﺸﺘﻖ‬
‫ﻟﻮ‪=x‬ﺝﻫﻲ ﻧﻘﻄﺔ ﺣﺮﺟﺔ ﻣﻦ‪(x)F‬ﻣﺜﻞ ﺫﻟﻚ‬
‫ﺍﻟﺒﺤﺚﻋﻦ ﺍﻟﻤﻄﻠﻘﺔ ﺍﻟﻤﺘﻄﺮﻓﺔ‬
‫‪)′F‬ﺝ(=‪ 0‬ﺑﻌﺪ ﺫﻟﻚ‪=x‬ﺝ‬
‫ﻹﻳﺠﺎﺩﺍﻟﻘﻴﻤﺔ ﺍﻟﻘﺼﻮﻯ ﺍﻟﻤﻄﻠﻘﺔ ﻟﻠﻤﺴﺘﻤﺮ‬
‫ﻭﻇﻴﻔﺔ‪(x)F‬ﻓﻲ ﺍﻟﻔﺘﺮﺓ]ﺃﻭﺏ[ﺍﺳﺘﺨﺪﻡ ﺍﻝ‬
‫‪.1‬ﻫﻮ ﺣﺪ ﺃﻗﺼﻰ ﻧﺴﺒﻲ‪(x)F‬ﻟﻮ‪)′ ′F‬ﺝ(>‪.0‬‬
‫ﺍﻟﻌﻤﻠﻴﺔﺍﻟﺘﺎﻟﻴﺔ‪.‬‬
‫‪.2‬ﻫﻮ ﺣﺪ ﺃﺩﻧﻰ ﻧﺴﺒﻲ ﻣﻦ‪(x)F‬ﻟﻮ‪)′ ′F‬ﺝ(<‪.0‬‬
‫‪.1‬ﺍﻟﺒﺤﺚ ﻋﻦ ﺟﻤﻴﻊ ﺍﻟﻨﻘﺎﻁ ﺍﻟﺤﺮﺟﺔ‪(x)F‬ﻓﻲ]ﺃﻭﺏ[‪.‬‬
‫‪.2‬ﺗﻘﻴﻴﻢ‪(x)F‬ﻓﻲ ﺟﻤﻴﻊ ﺍﻟﻨﻘﺎﻁ ﺍﻟﻤﻮﺟﻮﺩﺓ ﻓﻲ ﺍﻟﺨﻄﻮﺓ ‪.1‬‬
‫‪.3‬ﻗﺪ ﻳﻜﻮﻥ ﺣﺪﺍً ﺃﻗﺼﻰ ﻧﺴﺒﻴﺎً ﺃﻭ ﺣﺪﺍً ﺃﺩﻧﻰ ﻧﺴﺒﻴﺎً‬
‫ﺃﻭﻻ ﻳﻜﻮﻥ ﺃﻳﺎً ﻣﻨﻬﻤﺎ ﺇﺫﺍ‪)′ ′F‬ﺝ(=‪.0‬‬
‫‪.4‬ﺗﺤﺪﻳﺪ ﺍﻟﻘﻴﻤﺔ ﺍﻟﻤﻄﻠﻘﺔ‪ .‬ﺍﻷﻋﻠﻰ‪) .‬ﺃﻛﺒﺮ ﻗﻴﻤﺔ ﺩﺍﻟﺔ(‬
‫ﻭﺍﻟﻘﻴﻤﺔﺍﻟﻤﻄﻠﻘﺔ‪ .‬ﺍﻟﺤﺪ ﺍﻷﺩﻧﻰ )ﺃﺻﻐﺮ ﻗﻴﻤﺔ ﻟﻠﺪﺍﻟﺔ(‬
‫ﻣﻦﺍﻟﺘﻘﻴﻴﻤﺎﺕ ﻓﻲ ﺍﻟﺨﻄﻮﺗﻴﻦ ‪ 2‬ﻭ ‪.3‬‬
‫ﺍﻟﺒﺤﺚﻋﻦ ﺍﻟﻤﺘﻄﺮﻓﻴﻦ ﺍﻟﻨﺴﺒﻴﻴﻦ ﻭ ‪ /‬ﺃﻭ‬
‫ﺗﺼﻨﻴﻒﺍﻟﻨﻘﺎﻁ ﺍﻟﺤﺮﺟﺔ‬
‫‪.1‬ﺍﻟﺒﺤﺚ ﻋﻦ ﺟﻤﻴﻊ ﺍﻟﻨﻘﺎﻁ ﺍﻟﺤﺮﺟﺔ‪.(x)F‬‬
‫‪.3‬ﺗﻘﻴﻴﻢ‪)F‬ﺃ(ﻭ‪)F‬ﺏ(‪.‬‬
‫‪.2‬ﺍﺳﺘﺨﺪﻡ ‪1‬ﺷﺎﺭﻉﺍﺧﺘﺒﺎﺭ ﻣﺸﺘﻖ ﺃﻭ‬
‫‪2‬ﺍﺧﺘﺼﺎﺭ ﺍﻟﺜﺎﻧﻲ‬
‫ﺍﺧﺘﺒﺎﺭﻣﺸﺘﻖ ﻋﻠﻰ ﻛﻞ ﻧﻘﻄﺔ ﺣﺮﺟﺔ‪.‬‬
‫ﻳﻌﻨﻲﻧﻈﺮﻳﺔ ﺍﻟﻘﻴﻤﺔ‬
‫ﻟﻮ ‪(x)F‬ﻣﺴﺘﻤﺮ ﻓﻲ ﺍﻟﻔﺘﺮﺓ ﺍﻟﻤﻐﻠﻘﺔ]ﺃﻭﺏ[ﻭﻗﺎﺑﻞ ﻟﻠﺘﻔﺎﺿﻞ ﻓﻲ ﺍﻟﻔﺘﺮﺓ ﺍﻟﻤﻔﺘﻮﺣﺔ)ﺃﻭﺏ(‬
‫‪)F‬ﺏ(‪)F-‬ﺃ(‬
‫ﺛﻢﻫﻨﺎﻙ ﺭﻗﻢﺃ>ﺝ>ﺏﻣﺜﻞ ﺫﻟﻚ‪)′F‬ﺝ(=‬
‫ﺏ‪-‬ﺃ‬
‫ﻃﺮﻳﻘﺔﻧﻴﻮﺗﻦ‬
‫ﻟﻮ‪x‬ﻥﻫﻞﻥﺫﺗﺨﻤﻴﻦ ﺟﺬﺭ ‪ /‬ﺣﻞ‪ 0=(x)F‬ﺛﻢ )ﻥ‪(1+‬ﺷﺎﺭﻉﺍﻟﺘﺨﻤﻴﻦ ﻫﻮ‪x‬ﻥ‪x=1+‬ﻥ‪-‬‬
‫ﻣﺘﺎﺡ‪x)′F‬ﻥ(ﻣﻮﺟﻮﺩ‪.‬‬
‫ﻳﺰﻭﺭ‪ http://tutorial.math.lamar.edu‬ﻟﻤﺠﻤﻮﻋﺔ ﻛﺎﻣﻠﺔ ﻣﻦ ﻣﻼﺣﻈﺎﺕ ﺍﻟﺘﻔﺎﺿﻞ ﻭﺍﻟﺘﻜﺎﻣﻞ‪.‬‬
‫‪.‬‬
‫‪x)F‬ﻥ(‬
‫‪x)′F‬ﻥ(‬
‫©‪ 2005‬ﺑﻮﻝ ﺩﻭﻛﻴﻨﺰ‬
‫ﻭﺭﻗﺔﺍﻟﻐﺶ ﺣﺴﺎﺏ ﺍﻟﺘﻔﺎﺿﻞ ﻭﺍﻟﺘﻜﺎﻣﻞ‬
‫ﻣﻌﺪﻻﺕﻣﺮﺗﺒﻄﺔ‬
‫ﺭﺳﻢﺻﻮﺭﺓ ﻭﺗﺤﺪﻳﺪ ﺍﻟﻜﻤﻴﺎﺕ ﺍﻟﻤﻌﺮﻭﻓﺔ ‪ /‬ﻏﻴﺮ ﺍﻟﻤﻌﺮﻭﻓﺔ‪ .‬ﺍﻛﺘﺐ ﺍﻟﻤﻌﺎﺩﻟﺔ ﺍﻟﻤﺘﻌﻠﻘﺔ ﺑﺎﻟﻜﻤﻴﺎﺕ ﻭﺍﺷﺘﻖ ﻓﻴﻤﺎ ﻳﺘﻌﻠﻖ‬
‫ﺑﻬﺎﺭﺑﺎﺳﺘﺨﺪﺍﻡ ﺍﻟﺘﻔﺎﺿﻞ ﺍﻟﻀﻤﻨﻲ )ﺃﻱﺃﺿﻒ ﻋﻠﻰ ﻣﺸﺘﻖ ﻓﻲ ﻛﻞ ﻣﺮﺓ ﺗﻔﺮﻕ ﺩﺍﻟﺔﺭ(‪ .‬ﺍﻟﻤﻜﻮﻧﺎﺕ ﻓﻲ ﻣﻌﺮﻓﺔﻥ‬
‫ﻛﻤﻴﺎﺕﻭﺣﻞ ﻟﻠﻜﻤﻴﺔ ﺍﻟﻤﺠﻬﻮﻟﺔ‪.‬‬
‫ﺍﻟﺴﺎﺑﻖ‪.‬ﺳﻠﻢ ‪ 15‬ﻗﺪﻣﺎً ﻳﺴﺘﺮﻳﺢ ﻋﻠﻰ ﺍﻟﺤﺎﺉﻂ‪ .‬ﺍﻟﻘﺎﻉ‬
‫ﺍﻟﺴﺎﺑﻖ‪.‬ﺷﺨﺼﺎﻥ ﻋﻠﻰ ﺑﻌﺪ ‪ 50‬ﻗﺪﻣﺎً ﻋﻨﺪﻣﺎ ﻳﺒﺪﺃ‬
‫ﻓﻲﺍﻟﺒﺪﺍﻳﺔ ﻋﻠﻰ ﺑﻌﺪ ‪ 10‬ﺃﻗﺪﺍﻡ ﻭﻳﺠﺮﻱ‬
‫ﺃﺣﺪﻫﻤﺎﻓﻲ ﺍﻟﻤﺸﻲ ﺷﻤﺎﻻً‪ .‬ﺍﻟﺰﺍﻭﻳﺔ‪θ‬ﻳﺘﻐﻴﺮ ﻋﻨﺪ‬
‫‪0.01‬ﺭﺍﺩ ‪ /‬ﺩﻗﻴﻘﺔ‪ .‬ﺑﺄﻱ ﻣﻌﺪﻝ ﺗﺘﻐﻴﺮ ﺍﻟﻤﺴﺎﻓﺔ‬
‫‪14‬ﻗﺪﻡ ‪ /‬ﺛﺎﻧﻴﺔ‪ .‬ﻛﻴﻒ ﺳﺮﻳﻊ‬
‫ﺩﻓﻌﺖﻧﺤﻮ ﺍﻟﺤﺎﺉﻂ ﻋﻨﺪ‬
‫ﺑﻴﻨﻬﻤﺎﻣﺘﻰ‪ 0.5 =θ‬ﺭﺍﺩ؟‬
‫ﻫﻞﺍﻟﻘﻤﺔ ﺗﺘﺤﺮﻙ ﺑﻌﺪ ‪ 12‬ﺛﺎﻧﻴﺔ؟‬
‫ﻟﺪﻳﻨﺎ‪ 0.01= ′θ‬ﺭﺍﺩ ‪ /‬ﺩﻗﻴﻘﺔ‪ .‬ﻭﺗﺮﻳﺪ ﺃﻥ ﺗﺠﺪ ‪.′x‬ﻳﻤﻜﻨﻨﺎ‬
‫ﺍﺳﺘﺨﺪﺍﻡﻗﻴﻢ ﺣﺴﺎﺏ ﺍﻟﻤﺜﻠﺜﺎﺕ ﺍﻟﻤﺨﺘﻠﻔﺔ ﻭﻟﻜﻦ ﺍﻷﺳﻬﻞ ﻫﻮ‬
‫‪′x‬ﺳﻠﺒﻲ ﻷﻥ‪x‬ﻳﺘﻨﺎﻗﺺ‪ .‬ﺑﺎﺳﺘﺨﺪﺍﻡ ﻧﻈﺮﻳﺔ‬
‫ﻓﻴﺜﺎﻏﻮﺭﺱﻭﺍﻻﺷﺘﻘﺎﻕ ‪،‬‬
‫‪2+ ′xx2‬ﺱ ﺹ‪0= ′‬‬
‫⇒‬
‫‪+2x‬ﺫ‪215 =2‬‬
‫ﺑﻌﺪ‪ 12‬ﺛﺎﻧﻴﺔ ﻟﺪﻳﻨﺎ‪ 7 =(4)12− 10=x‬ﻭ‬
‫ﻟﺬﺍﺫ=‬
‫ﻝﺫ‪.′‬‬
‫‪1‬‬
‫‪ .176 =27 -215‬ﻗﻢ ﺑﺘﻮﺻﻴﻞ ﻭﺣﻞ‬
‫‪176 +(4 1-)7‬ﺫ‪⇒0= ′‬ﺫ‪= ′‬‬
‫‪7‬‬
‫‪416‬‬
‫ﻗﺪﻡ ‪ /‬ﺛﺎﻧﻴﺔ‬
‫‪′x‬‬
‫‪x‬‬
‫⇒ ﺛﺎﻧﻴﺔ‪θ‬ﺗﺎﻥ‪= ′θ θ‬‬
‫ﺛﺎﻧﻴﺔ‪=θ‬‬
‫‪50‬‬
‫‪50‬‬
‫ﻧﻌﻠﻢ‪ 0.5=θ‬ﻟﺬﺍ ﻗﻢ ﺑﺘﻮﺻﻴﻠﻪ‪′θ‬ﻭﺣﻠﻬﺎ‪.‬‬
‫‪′x‬‬
‫ﺗﺎﻥ ‪= 0.01 0.5‬‬
‫ﺛﺎﻧﻴﺔ)(‪() ()0.5‬‬
‫‪50‬‬
‫‪ 0.3112= ′x‬ﻗﺪﻡ ‪ /‬ﺛﺎﻧﻴﺔ‬
‫ﺗﺬﻛﺮﺃﻥ ﻳﻜﻮﻥ ﻟﺪﻳﻚ ﺁﻟﺔ ﺣﺎﺳﺒﺔ ﺑﺎﻟﺘﻘﺪﻳﺮ ﺍﻟﺪﺍﺉﺮﻱ!‬
‫ﺗﺤﺴﻴﻦ‬
‫ﺭﺳﻢﺻﻮﺭﺓ ﺇﺫﺍ ﻟﺰﻡ ﺍﻷﻣﺮ ‪ ،‬ﻗﻢ ﺑﺘﺪﻭﻳﻦ ﺍﻟﻤﻌﺎﺩﻟﺔ ﺍﻟﺘﻲ ﺳﻴﺘﻢ ﺗﺤﺴﻴﻨﻬﺎ ﻭﺗﻘﻴﻴﺪﻫﺎ‪ .‬ﺣﻞ ﺍﻟﻘﻴﺪ ﻷﺣﺪ ﺍﻟﻤﺘﻐﻴﺮﻳﻦ ﻭﻋﻮﺽ‬
‫ﺑﻪﻓﻲ ﺍﻟﻤﻌﺎﺩﻟﺔ ﺍﻷﻭﻟﻰ‪ .‬ﺃﻭﺟﺪ ﺍﻟﻨﻘﺎﻁ ﺍﻟﺤﺮﺟﺔ ﻟﻠﻤﻌﺎﺩﻟﺔ ﻓﻲ ﻧﻄﺎﻕ ﺍﻟﻤﺘﻐﻴﺮﺍﺕ ﻭﺗﺤﻘﻖ ﻣﻦ ﺃﻧﻬﺎ ‪ min / max‬ﻣﺜﻞ ‪n‬‬
‫‪eeded.‬‬
‫ﺍﻟﺴﺎﺑﻖ‪.‬ﻧﺤﻦ ﻧﺮﻓﻖ ﺣﻘﻼ ًﻣﺴﺘﻄﻴﻼ ًﻣﻊ ‪ 500‬ﻗﺪﻡ ﻣﻦ‬
‫ﺍﻟﺴﺎﺑﻖ‪.‬ﺗﺤﺪﻳﺪ ﺍﻟﻨﻘﺎﻁ ﻋﻠﻰﺫ=‪ 1+2x‬ﻫﻲ ﺍﻷﻗﺮﺏ‬
‫ﻣﺎﺩﺓﺍﻟﺴﻴﺎﺝ ﻭﺟﺎﻧﺐ ﻭﺍﺣﺪ ﻣﻦ ﺍﻟﺤﻘﻞ ﻋﺒﺎﺭﺓ ﻋﻦ‬
‫ﺇﻟﻰ)‪.(0،2‬‬
‫ﻣﺒﻨﻰ‪.‬ﺗﺤﺪﻳﺪ ﺍﻷﺑﻌﺎﺩ ﺍﻟﺘﻲ ﻣﻦ ﺷﺄﻧﻬﺎ ﺯﻳﺎﺩﺓ ﺍﻟﻤﺴﺎﺣﺔ‬
‫ﺍﻟﻤﻐﻠﻘﺔ‪.‬‬
‫ﺗﺤﻘﻴﻖﺃﻗﺼﻰ ﻗﺪﺭﺃ=ﺱ ﺹﺗﺨﻀﻊ ﻟﻘﻴﻮﺩ ‪2+x‬ﺫ=‬
‫‪.500‬ﺣﻞ ﺍﻟﻘﻴﺪ ﻝ‪x‬ﻭﻗﻢ ﺑﺘﻮﺻﻴﻠﻪ ﺑﺎﻟﻤﻨﻄﻘﺔ‪.‬‬
‫‪2 - 500=x‬ﺫ⇒‬
‫ﺃ=ﺫ)‪2 - 500‬ﺫ(‬
‫=‪500‬ﺫ‪2-‬ﺫ‬
‫ﻣﻴﺰّﻭﺍﻋﺜﺮ ﻋﻠﻰ ﺍﻟﻨﻘﺎﻁ ﺍﻟﺤﺮﺟﺔ‪.‬‬
‫ﺃ‪4 - 500= ′‬ﺫ⇒ﺫ=‪125‬‬
‫‪2‬‬
‫ﺑﻨﺴﺒﺔ‪2‬ﺍﺧﺘﺼﺎﺭ ﺍﻟﺜﺎﻧﻲﻣﺸﺘﻖ‪ .‬ﺍﺧﺘﺒﺎﺭ ﻫﺬﺍ ﻫﻮ ‪ .rel‬ﺍﻷﻋﻠﻰ‪ .‬ﻭﻛﺬﻟﻚ‬
‫ﺍﻟﺠﻮﺍﺏﺍﻟﺬﻱ ﻧﺴﻌﻰ ﺇﻟﻴﻪ‪ .‬ﺃﺧﻴﺮﺍً ‪ ،‬ﺍﺑﺤﺚ ﻋﻦ‪.x‬‬
‫‪250=(125)2 - 500=x‬‬
‫ﺗﻜﻮﻥﺍﻷﺑﻌﺎﺩ ﺇﺫﻥ ‪.125 × 250‬‬
‫ﻳﺰﻭﺭ‪ http://tutorial.math.lamar.edu‬ﻟﻤﺠﻤﻮﻋﺔ ﻛﺎﻣﻠﺔ ﻣﻦ ﻣﻼﺣﻈﺎﺕ ﺍﻟﺘﻔﺎﺿﻞ ﻭﺍﻟﺘﻜﺎﻣﻞ‪.‬‬
‫‪)+2‬ﺫ‪(2-‬‬
‫ﺗﺼﻐﻴﺮ‪=F‬ﺩ‪(0-x)=2‬‬
‫‪2‬ﻭ ﺍﻝ‬
‫ﺍﻟﻘﻴﺪﺫ=‪ .1+2x‬ﺣﻞ ﺍﻟﻘﻴﺪ ﻝ ‪2x‬ﻭﻗﻢ ﺑﺘﻮﺻﻴﻞ‬
‫ﺍﻟﻮﻇﻴﻔﺔ‪=2x .‬ﺫ‪)+2x=F⇒1-‬ﺫ‪(2-‬‬
‫‪2‬‬
‫‪=2‬ﺫ‪3 -2‬ﺫ‪3+‬‬
‫=ﺫ‪)+1-‬ﺫ‪(2-‬‬
‫ﻣﻴﺰّﻭﺍﻋﺜﺮ ﻋﻠﻰ ﺍﻟﻨﻘﺎﻁ ﺍﻟﺤﺮﺟﺔ‪.‬‬
‫⇒‬
‫‪2= ′F‬ﺫ‪3-‬‬
‫ﺫ=‪2 3‬‬
‫ﺑﻮﺍﺳﻄﺔ‪2‬ﺍﺧﺘﺼﺎﺭ ﺍﻟﺜﺎﻧﻲﺍﺧﺘﺒﺎﺭ ﻣﺸﺘﻖ ﻫﺬﺍ ﻫﻮ ‪ .rel‬ﺩﻗﻴﻘﺔ‪.‬‬
‫ﻭﻟﺬﺍﻛﻞ ﻣﺎ ﻋﻠﻴﻨﺎ ﺍﻟﻘﻴﺎﻡ ﺑﻪ ﻫﻮ ﺇﻳﺠﺎﺩ‪x‬ﻗﻴﻢ(‪.‬‬
‫‪1±=x‬‬
‫‪⇒ 2 1= 12−3=2x‬‬
‫‪2‬‬
‫ﺛﻢ‪ 2‬ﻧﻘﻄﺔ)‬
‫ﻭ( )‬
‫‪3 1‬‬
‫‪22‬‬
‫ﻭ ‪-‬ﻭ‬
‫‪3 1‬‬
‫‪22‬‬
‫(‬
‫©‪ 2005‬ﺑﻮﻝ ﺩﻭﻛﻴﻨﺰ‬
‫‪.‬‬
‫ﻭﺭﻗﺔﺍﻟﻐﺶ ﺣﺴﺎﺏ ﺍﻟﺘﻔﺎﺿﻞ ﻭﺍﻟﺘﻜﺎﻣﻞ‬
‫ﺗﻜﺎﻣﻼﺕ‬
‫ﻻﻳﺘﺠﺰﺃ ﻣﺤﺪﺩ‪:‬ﻳﻔﺘﺮﺽ‪(x)F‬ﻣﺴﺘﻤﺮ‬
‫ﻋﻠﻰ]ﺃﻭﺏ[‪.‬ﻳﻘﺴﻢ]ﺃﻭﺏ[ﺩﺍﺧﻞﻥﻓﺘﺮﺍﺕ ﻓﺮﻋﻴﺔ ﻣﻦ‬
‫ﻋﺮﺽ∆‪x‬ﻭ ﺍﺧﺘﺎﺭ‪x‬‬
‫ﺏ‬
‫ﺛﻢ∫‬
‫ﺃ‬
‫ﻫﻲﻭﻇﻴﻔﺔ ‪(x)F،‬ﻭﻣﺜﻞ ﺫﻟﻚ‪ .(x)F=(x)′F‬ﻻ ﻳﺘﺠﺰﺃ‬
‫ﺇﻟﻰﺃﺟﻞ ﻏﻴﺮ ﻣﺴﻤﻰ‪+(x)F=dx(x)F∫:‬ﺝ ﺃﻳﻦ‪(x)F‬‬
‫ﻫﻮﻣﻀﺎﺩ ﻣﺸﺘﻖ ﻣﻦ‪.(x)F‬‬
‫ﺃﻧﺎﻣﻦ ﻛﻞ ﻓﺘﺮﺓ‪.‬‬
‫*‬
‫∞‬
‫‪=dx(x)F‬ﻟﻴﻢ∑‪x∆(*x)F‬ﺃﻧﺎ‪.‬‬
‫ﻥ→ ∞ﺃﻧﺎ= ‪1‬‬
‫ﺍﻟﻨﻈﺮﻳﺔﺍﻷﺳﺎﺳﻴﺔ ﻟﺤﺴﺎﺏ ﺍﻟﺘﻔﺎﺿﻞ ﻭﺍﻟﺘﻜﺎﻣﻞ‬
‫ﺍﻟﺠﺰءﺍﻷﻭﻝ‪:‬ﻟﻮ‬
‫ﺯ)‪∫=(x‬‬
‫ﺗﻌﺮﻳﻔﺎﺕ‬
‫ﻣﻀﺎﺩﺍﻟﻤﺸﺘﻘﺎﺕ‪:‬ﻣﻀﺎﺩ ﻣﺸﺘﻖ ﻣﻦ‪(x)F‬‬
‫‪x‬‬
‫ﺃ‬
‫‪(x)F‬ﻣﺴﺘﻤﺮ ﻋﻠﻰ]ﺃﻭﺏ[ﺛﻢ‬
‫ﻣﺘﻐﻴﺮﺍﺕﺍﻟﺠﺰء ﺍﻷﻭﻝ‪:‬‬
‫ﺩ‬
‫‪)F‬ﺭ(ﺩ=ﺵ‪-F(x)′‬ﺵ)‪-(x‬‬
‫ﺩ ‪ ∫x‬ﺃ‬
‫ﺏ‬
‫ﺩ‬
‫‪)F‬ﺭ(ﺩ= ‪-‬ﺍﻟﺨﺎﻣﺲ‪-F(x)′‬ﺍﻟﺨﺎﻣﺲ)‪-(x‬‬
‫‪∫ dx‬‬
‫ﺩ ﺵ)‪(x‬‬
‫ﺵ)‪(x‬‬
‫‪) F‬ﺭ(ﺩﻣﺴﺘﻤﺮ ﺃﻳﻀﺎً]ﺃﻭﺏ[‬
‫ﺩ‬
‫‪dx‬‬
‫ﺍﻟﺠﺰءﺍﻟﺜﺎﻧﻲ ‪(x)F:‬ﻣﺴﺘﻤﺮ ﻋﻠﻰ]ﺃﻭﺏ[ﻭ‪(x)F‬ﻫﻮ ﻣﻀﺎﺩ‬
‫ﻭﺯ‪)F∫=(x)′‬ﺭ(ﺩ=‪.(x)F‬‬
‫‪x‬‬
‫ﺃ‬
‫ﺍﻟﺨﺎﻣﺲ)‪(x‬‬
‫‪∫dx‬‬
‫ﻣﺸﺘﻖﻣﻦ‪)(x)F‬ﺃﻱ ‪(dx(x)F∫=(x)F‬‬
‫ﺍﻟﺨﺎﻣﺲ)‪(x‬‬
‫‪)F‬ﺭ(ﺩ=ﺵ‪]F(x)′‬ﺵ)‪-[(x‬ﺍﻟﺨﺎﻣﺲ‪]F(x)′‬ﺍﻟﺨﺎﻣﺲ)‪[(x‬‬
‫ﺏ‬
‫ﺛﻢ∫‪)F=dx(x)F‬ﺏ(‪)F-‬ﺃ(‪.‬‬
‫ﺃ‬
‫ﻣﻠﻜﻴﺎﺕ‬
‫∫‪±(x)F‬ﺯ)‪∫±dx(x)F∫=dx(x‬ﺯ)‪dx(x‬‬
‫∫‬
‫ﺏ‬
‫ﺃ‬
‫ﺃ‬
‫∫‬
‫ﺃ‬
‫ﺏ‬
‫∫‬
‫ﺃ‬
‫ﺃ‬
‫‪0=dx(x)F‬‬
‫ﺃ‬
‫ﺝ‬
‫ﺏ‬
‫∫ﺃ‪F‬ﺏ ) ‪∫≥dx(x‬‬
‫ﺏ‬
‫)(‪dx(x)F∫+dx(x)F∫=fx‬ﻋﻦ ﺃﻱ ﻗﻴﻤﺔﺝ‪.‬‬
‫‪dx‬‬
‫ﺃ‬
‫ﺝ‬
‫ﺏ‬
‫ﻟﻮ‪ 0≤(x)F‬ﻓﻲﺃ≥‪≥x‬ﺏﺛﻢ∫‪0≤dx(x)F‬‬
‫ﻟﻮﻡ≥‪≥(x)F‬ﻡﻋﻠﻰﺃ≥‪≥x‬ﺏﺛﻢﻡ)ﺏ‪-‬ﺃ‬
‫∫ﻙ ﺩﻛﺲ=ﻛﻜﺲ‪+‬ﺝ‬
‫∫‪x‬ﻥ‪x1=dx‬ﻥﻥ‪+11++‬ﺝﻭﻥ≠ ‪1-‬‬
‫‪1‬‬
‫‪+xln=dx‬ﺝ‬
‫∫‪x∫=dx1−x‬‬
‫‪1‬‬
‫ﻓﺄﺱ‪+‬ﺏ‬
‫‪1=dx‬ﺃ‪ln‬ﻓﺄﺱ‪+‬ﺏ‪+‬ﺝ‬
‫∫‪ln‬ﺵ ﺩﻭ=ﺵ‪)ln‬ﺵ(‪-‬ﺵ‪+‬ﺝ‬
‫∫ﻩﺵﺩﻭ=ﻩﺵ‪+‬ﺝ‬
‫ﺃ‬
‫∫‬
‫ﺏ‬
‫ﺃ‬
‫‪dx(x)F‬ﻭﺝﺛﺎﺑﺖ‬
‫ﺏ‬
‫ﺃ∫ ﺝ‪=DX‬ﺝ)ﺏ‪-‬ﺃ(‬
‫ﺃ‬
‫ﻟﻮ ‪≤(x)F‬ﺯ)‪(x‬ﻋﻠﻰﺃ≥‪≥x‬ﺏﺛﻢ‬
‫∫‬
‫ﺏ‬
‫‪gx‬‬
‫)(‪dx∫±x dx‬ﺏ )(‬
‫ﺏ‬
‫∫ﺃﺭﺍﺟﻊ)‪=dx (x‬ﺝ∫‬
‫ﺃ‬
‫‪dx(x)F∫- =dx (x)F‬‬
‫ﺏ‬
‫ﺃ∫‬
‫‪±x() F‬ﺯ )(‪F∫=x dx‬‬
‫ﺏ‬
‫∫ﺭﺍﺟﻊ)‪=dx(x‬ﺝ∫‪dx(x)F‬ﻭﺝﺛﺎﺑﺖ‬
‫ﺏ‬
‫ﺃ‬
‫‪dx(x)F‬‬
‫ﺏ‬
‫‪∫≤dx(x)F‬ﺯ)‪dx(x‬‬
‫ﺃ‬
‫( ≥ ∫ﺃﺏ ‪≥dx(x)F‬ﻡ)ﺏ‪-‬ﺃ(‬
‫ﺍﻟﺘﻜﺎﻣﻼﺕﺍﻟﻤﺸﺘﺮﻛﺔ‬
‫∫ﻛﻮﺱﺵ ﺩﻭ=ﺍﻟﺨﻄﻴﺉﺔﺵ‪+‬ﺝ‬
‫∫ﺍﻟﺨﻄﻴﺉﺔﺵ ﺩﻭ= ‪-‬ﻛﻮﺱﺵ‪+‬ﺝ‬
‫∫ﺛﺎﻧﻴﺔ‪2‬ﺵ ﺩﻭ=ﺗﺎﻥﺵ‪+‬ﺝ‬
‫∫ﺛﺎﻧﻴﺔﺵﺗﺎﻥﺵ ﺩﻭ=ﺛﺎﻧﻴﺔﺵ‪+‬ﺝ‬
‫∫‪CSC‬ﺵﺳﺮﻳﺮ ﻧﻘﺎﻝﺃﻭﺩﻭ= ‪CSC-‬ﺵ‪+‬ﺝ‬
‫∫ﺗﺎﻥﺵ ﺩﻭ=ﻓﻲ ﺛﺎﻧﻴﺔﺵ‪+‬ﺝ‬
‫∫ﺛﺎﻧﻴﺔﺵ ﺩﻭ=ﻓﻲ ﺛﺎﻧﻴﺔﺵ‪+‬ﺗﺎﻥﺵ‪+‬ﺝ‬
‫∫‬
‫∫‬
‫‪1‬‬
‫ﺃ‪+2‬ﺵ‪2‬‬
‫‪1‬‬
‫ﺃ‪-2‬ﺵ‪2‬‬
‫ﺩﻭ=‪ 1‬ﺃﺗﺎ ﻥ)‬
‫‪1-‬‬
‫ﺵ‬
‫ﺃ(‬
‫‪+‬ﺝ‬
‫ﺩﻭ=ﺍﻟﺨﻄﻴﺉﺔ‪ )1−‬ﺃﺵ(‪+‬ﺝ‬
‫∫‪2CSC‬ﺵ ﺩﻭ= ‪-‬ﺳﺮﻳﺮ ﻧﻘﺎﻝﺵ‪+‬ﺝ‬
‫ﻳﺰﻭﺭ‪ http://tutorial.math.lamar.edu‬ﻟﻤﺠﻤﻮﻋﺔ ﻛﺎﻣﻠﺔ ﻣﻦ ﻣﻼﺣﻈﺎﺕ ﺍﻟﺘﻔﺎﺿﻞ ﻭﺍﻟﺘﻜﺎﻣﻞ‪.‬‬
‫©‪ 2005‬ﺑﻮﻝ ﺩﻭﻛﻴﻨﺰ‬
‫ﻭﺭﻗﺔﺍﻟﻐﺶ ﺣﺴﺎﺏ ﺍﻟﺘﻔﺎﺿﻞ ﻭﺍﻟﺘﻜﺎﻣﻞ‬
‫ﺗﻘﻨﻴﺎﺕﺍﻟﺘﻜﺎﻣﻞ ﺍﻟﻘﻴﺎﺳﻴﺔ‬
‫ﻻﺣﻆﺃﻧﻪ ﻓﻲ ﺍﻟﻌﺪﻳﺪ ﻣﻦ ﺍﻟﻤﺪﺍﺭﺱ ‪ ،‬ﻳﺘﻢ ﺗﺪﺭﻳﺲ ﺟﻤﻴﻊ ﺍﻟﻤﺪﺍﺭﺱ ﺑﺎﺳﺘﺜﻨﺎء ﻗﺎﻋﺪﺓ ﺍﻻﺳﺘﺒﺪﺍﻝ ﻓﻲ ﻓﺼﻞ ﺍﻟﺘﻔﺎﺿﻞ ﻭﺍﻟﺘﻜﺎﻣﻞ ﺍﻟﺜﺎﻧﻲ‪.‬‬
‫ﺯ )ﺏ (‬
‫ﺏ‬
‫ﺵ=ﺯ)‪(x‬ﺳﻮﻑ ﻳﺘﺤﻮﻝ∫‪)F‬ﺯﺃ)‪((x‬ﺯ‪(x)′‬‬
‫ﺵﺍﻻﺳﺘﺒﺪﺍﻝ‪:‬ﺍﻻﺳﺘﺒﺪﺍﻝ‬
‫ﺩﻭ=ﺯ‪.dx(x)′‬ﻟﻠﺘﻜﺎﻣﻼﺕ ﻏﻴﺮ ﺍﻟﻤﺤﺪﺩﺓ ﺍﺳﻘﻂ ﺣﺪﻭﺩ ﺍﻟﺘﻜﺎﻣﻞ‪.‬‬
‫‪2‬‬
‫ﺍﻟﺴﺎﺑﻖ‪2x5 ∫.‬ﻛﻮﺱ)‪dx(3x‬‬
‫‪1‬‬
‫ﺵ=‪ ⇒ 3x‬ﺩﻭ=‪=dx2x⇒dx2x3‬‬
‫‪⇒1=x‬ﺵ=‪⇒2=x:: 1 =31‬ﺵ=‪8 =32‬‬
‫‪8‬‬
‫ﻛﻮﺱ‪∫=dx (x‬‬
‫∫‪) x 51‬‬
‫‪1‬‬
‫‪3‬ﺩﻭ‬
‫‪1‬‬
‫‪2‬‬
‫‪=dx‬‬
‫∫ﺯ)ﺃ(‬
‫‪3‬‬
‫‪2‬‬
‫‪)F‬ﺵ(ﺩﻭﺍﺳﺘﺨﺪﺍﻡ‬
‫ﺵ ﺩﻭ‬
‫‪53‬ﻛﻮﺱ)(‬
‫ﻲ)ﺵ(‪3)5= 8‬ﺍﻟﺨﻄﻴﺉﺔ)‪-(8‬ﺍﻟﺨﻄﻴﺉﺔ)‪((1‬‬
‫ﺱ‬
‫=‪ 35‬ﻓ‬
‫‪1‬‬
‫ﺏ‬
‫ﺗﻜﺎﻣﻞﺍﺟﺰﺍء ‪∫:‬ﺵ ﺩ=ﺍﻷﺷﻌﺔ ﻓﻮﻕ ﺍﻟﺒﻨﻔﺴﺠﻴﺔ‪v du∫-‬ﻭ∫ﺵ ﺩ=ﺍﻷﺷﻌﺔ ﻓﻮﻕ‬
‫ﺃ‬
‫ﺍﻟﺒﻨﻔﺴﺠﻴﺔﺏ‬
‫ﺃ‬
‫ﺏ‬
‫ ∫‪.v du‬ﻳﺨﺘﺎﺭﺵﻭﺩﻱ ﻓﻲﻣﻦ‬‫ﺃ‬
‫ﻣﺘﻜﺎﻣﻞﻭﺣﺴﺎﺏﺩﻭﻋﻦ ﻃﺮﻳﻖ ﺍﻟﺘﻔﺮﻳﻖﺵﻭﺍﺣﺴﺐﺍﻟﺨﺎﻣﺲﺍﺳﺘﺨﺪﺍﻡﺍﻟﺨﺎﻣﺲ=∫ﺩﻱ ﻓﻲ‪.‬‬
‫ﺍﻟﺴﺎﺑﻖ‪x∫.‬ﻩ‪dxx-‬‬
‫ﺵ=ﺱ ﺩ=ﻩ‪⇒ x-‬ﺩﻭ=‪=dx v‬‬
‫ﺍﻟﺴﺎﺑﻖ‪.‬‬
‫‪-‬ﻩ ‪x -‬‬
‫∫‪x‬ﻩ‪x- =dxx-‬ﻩ‪∫+x-‬ﻩ‪x- =dxx-‬ﻩ‪-x-‬ﻩ‪+x-‬ﺝ‬
‫ﺍﻟﻤﻨﺘﺠﺎﺕﻭ )ﺑﻌﺾ( ﺍﻗﺘﺒﺎﺳﺎﺕ ﺩﻭﺍﻝ ﺍﻟﻤﺜﻠﺜﺎﺕ ﻝ∫‬
‫ﺍﻟﺨﻄﻴﺉﺔﻥ‪x‬ﻛﻮﺱﻡ‪x dx‬ﻟﺪﻳﻨﺎ ﻣﺎ ﻳﻠﻲ‪:‬‬
‫‪5‬‬
‫∫‪x dxln‬‬
‫‪3‬‬
‫ﺵ=‪ln‬ﺱ ﺩ=‪dx‬‬
‫‪5‬‬
‫∫‪xlnx=x dxln‬‬
‫‪3‬‬
‫‪5‬‬
‫⇒‬
‫‪5‬‬
‫ﺩﻭ=‪dxx 1‬‬
‫‪lnx)=dx 3∫ - 3‬‬
‫=‪2-(3)3ln-(5)5ln‬‬
‫ﺍﻟﺨﺎﻣﺲ=‪x‬‬
‫)‪(x-(x‬‬
‫‪5‬‬
‫‪3‬‬
‫ﻝ∫ﺗﺎﻥﻥ‪x‬ﺛﺎﻧﻴﺔﻡ‪x dx‬ﻟﺪﻳﻨﺎ ﻣﺎ ﻳﻠﻲ‪:‬‬
‫‪.1‬ﻥﻏﺮﻳﺐ‪.‬ﻗﻢ ﺑﺈﺯﺍﻟﺔ ‪ 1‬ﻇﻞ ﻭ ‪ 1‬ﻗﺎﻃﻊ ﺧﺎﺭﺝ ﻭﺗﺤﻮﻳﻞ‬
‫‪.1‬ﻥﻏﺮﻳﺐ‪.‬ﺍﻧﺰﻉ ﺟﻴﺐ ﻭﺍﺣﺪ ﻟﻠﺨﺎﺭﺝ ﻭﻗﻢ ﺑﺘﺤﻮﻳﻞ ﺍﻟﺒﺎﻗﻲ ﺇﻟﻰ‬
‫ﺍﻟﺒﺎﻗﻲﺇﻟﻰ ﻗﻄﻊ ﺑﺎﺳﺘﺨﺪﺍﻡ‬
‫ﺟﻴﺐﺍﻟﺘﻤﺎﻡ ﺑﺎﺳﺘﺨﺪﺍﻡ ﺍﻟﺨﻄﻴﺉﺔ‪ −1=x2‬ﻛﻮﺱ‪x2‬ﻭﺛﻢ‬
‫ﺗﺎﻥ‪=x2‬ﺛﺎﻧﻴﺔ‪ ، 1-x2‬ﺛﻢ ﺍﺳﺘﺨﺪﻡ ﺍﻟﺒﺪﻳﻞ ﺵ=ﺛﺎﻧﻴﺔ‪x‬‬
‫ﺍﺳﺘﺨﺪﻡﺍﻻﺳﺘﺒﺪﺍﻝﺵ=ﻛﻮﺱ‪.x‬‬
‫‪.‬‬
‫‪.2‬ﻡﻏﺮﻳﺐ‪.‬ﺍﻧﺰﻉ ﺟﻴﺐ ﺍﻟﺘﻤﺎﻡ ‪ 1‬ﻭﻗﻢ ﺑﺘﺤﻮﻳﻞ ﺍﻟﺒﺎﻗﻲ‬
‫‪.2‬ﻡﺣﺘﻰ‪.‬ﺍﺧﻠﻊ ﻗﻄﻌﺘﻴﻦ ﻭﻗﻢ ﺑﺘﺤﻮﻳﻞ ﺍﻟﺒﺎﻗﻲ‬
‫ﻟﺠﻴﺐﺑﺎﺳﺘﺨﺪﺍﻡ ﺟﻴﺐ ﺍﻟﺘﻤﺎﻡ‪ −1=x2‬ﺧﻄﻴﺉﺔ‪x2‬ﻭﺛﻢ‬
‫ﺇﻟﻰﺍﻟﻈﻞ ﺑﺎﺳﺘﺨﺪﺍﻡ ﺛﺎﻧﻴﺔ‪ +1=x2‬ﺗﺎﻥ‪x2‬ﻭﺛﻢ‬
‫ﺍﺳﺘﺨﺪﻡﺍﻻﺳﺘﺒﺪﺍﻝﺵ=ﺍﻟﺨﻄﻴﺉﺔ‪.x‬‬
‫ﺍﺳﺘﺨﺪﻡﺍﻻﺳﺘﺒﺪﺍﻝﺵ=ﺗﺎﻥ‪.x‬‬
‫‪.3‬ﻥﻭﻡﻛﻼﻫﻤﺎ ﻏﺮﻳﺐ‪.‬ﺍﺳﺘﺨﺪﻡ ﺇﻣﺎ ‪ .1‬ﺃﻭ ‪.2‬‬
‫‪.3‬ﻥﻏﺮﻳﺐ ﻭﻡﺣﺘﻰ‪.‬ﺍﺳﺘﺨﺪﻡ ﺇﻣﺎ ‪ .1‬ﺃﻭ ‪.2‬‬
‫‪.4‬ﻥﻭﻡﻛﻼﻫﻤﺎ ﺣﺘﻰ‪.‬ﺍﺳﺘﺨﺪﻡ ﺻﻴﻎ ﺍﻟﺰﺍﻭﻳﺔ‬
‫‪.4‬ﻥﺣﺘﻰ ﻭﻡﻏﺮﻳﺐ‪.‬ﺳﻴﺘﻢ ﺍﻟﺘﻌﺎﻣﻞ ﻣﻊ ﻛﻞ ﺟﺰء‬
‫ﺍﻟﻤﺰﺩﻭﺟﺔﻭ ‪ /‬ﺃﻭ ﻧﺼﻒ ﺍﻟﺰﺍﻭﻳﺔ ﻟﺘﻘﻠﻴﻞ ﺍﻟﺘﻜﺎﻣﻞ‬
‫ﺑﺸﻜﻞﻣﺨﺘﻠﻒ‪.‬‬
‫ﺇﻟﻰﺷﻜﻞ ﻳﻤﻜﻦ ﺩﻣﺠﻪ‪.‬‬
‫ﺻﻴﻎ‪:Trig‬ﺍﻟﺨﻄﻴﺉﺔ)‪(x)2sin=(x2‬ﻛﻮﺱ)‪(x‬ﻭﻛﻮﺱ‪ +1)2 1=(x)2‬ﻛﻮﺱ)‪((x2‬ﻭﺍﻟﺨﻄﻴﺉﺔ‪ −1)2 1=(x)2‬ﻛﻮﺱ)‪((x2‬‬
‫ﺍﻟﺴﺎﺑﻖ‪∫.‬ﺗﺎﻥ‪x3‬ﺛﺎﻧﻴﺔ‪x dx5‬‬
‫∫ﺗﺎﻥ‪x3‬ﺛﺎﻧﻴﺔ‪∫=xdx5‬ﺗﺎﻥ‪x2‬ﺛﺎﻧﻴﺔ‪x4‬ﺗﺎﻥ‪x‬ﺛﺎﻧﻴﺔ‪xdx‬‬
‫=∫ )ﺛﺎﻧﻴﺔ‪(1-x2‬ﺛﺎﻧﻴﺔ‪x4‬ﺗﺎﻥ‪x‬ﺛﺎﻧﻴﺔ‪xdx‬‬
‫=∫ )ﺵ‪(1−2‬ﺵ‪4‬ﺩﻭ‬
‫ﺛﺎﻧﻴﺔ‪x7‬‬
‫=‪7 1‬‬
‫‪-‬‬
‫‪15‬ﺛﺎﻧﻴﺔ‬
‫‪5‬‬
‫‪+x‬ﺝ‬
‫)ﺵ=ﺛﺎﻧﻴﺔ‪(x‬‬
‫ﻳﺰﻭﺭ‪ http://tutorial.math.lamar.edu‬ﻟﻤﺠﻤﻮﻋﺔ ﻛﺎﻣﻠﺔ ﻣﻦ ﻣﻼﺣﻈﺎﺕ ﺍﻟﺘﻔﺎﺿﻞ ﻭﺍﻟﺘﻜﺎﻣﻞ‪.‬‬
‫ﺍﻟﺴﺎﺑﻖ‪∫.‬‬
‫∫‬
‫ﺍﻟﺨﻄﻴﺉﺔ‪dxx5‬‬
‫ﻛﻮﺱ‪x3‬‬
‫∫‬
‫=∫‬
‫= ‪∫-‬‬
‫ﺍﻟﺨﻄﻴﺉﺔ‪=dxx5‬‬
‫ﻛﻮﺱ‪x3‬‬
‫ﺍﻟﺨﻄﻴﺉﺔ‪x4‬ﺍﻟﺨﻄﻴﺉﺔ‪=dxx‬‬
‫ﻛﻮﺱ‪x3‬‬
‫∫‬
‫)‪ − 1‬ﻛﻮﺱ‪2(x2‬ﺍﻟﺨﻄﻴﺉﺔ‪dxx‬‬
‫ﻛﻮﺱ‪3x‬‬
‫)‪−1‬ﺵ‪2(2‬‬
‫ﺵ‪3‬‬
‫ﺩﻭ= ‪∫-‬‬
‫)ﺍﻟﺨﻄﻴﺉﺔ‪2(x2‬ﺍﻟﺨﻄﻴﺉﺔ‪dxx‬‬
‫ﻛﻮﺱ‪x3‬‬
‫)ﺵ=ﻛﻮﺱ‪( x‬‬
‫‪1−2‬ﺵ‪+2‬ﺵ‪4‬ﺩﻭ‬
‫ﺵ‪3‬‬
‫ﻛﻮﺱ‬
‫ﺛﺎﻧﻴﺔ‪ 2+x2‬ﻟﻴﻦ ﻛﻮﺱ‪2 1-x‬‬
‫=‪2 1‬‬
‫‪2‬‬
‫‪+x‬ﺝ‬
‫©‪ 2005‬ﺑﻮﻝ ﺩﻭﻛﻴﻨﺰ‬
‫ﻭﺭﻗﺔﺍﻟﻐﺶ ﺣﺴﺎﺏ ﺍﻟﺘﻔﺎﺿﻞ ﻭﺍﻟﺘﻜﺎﻣﻞ‬
‫ﺑﺪﺍﺉﻞﺍﻟﻤﺜﻠﺚ‪:‬ﺇﺫﺍ ﻛﺎﻥ ﺍﻟﺘﻜﺎﻣﻞ ﻳﺤﺘﻮﻱ ﻋﻠﻰ ﺍﻟﺠﺬﺭ ﺍﻟﺘﺎﻟﻲ ‪ ،‬ﻓﺎﺳﺘﺨﺪﻡ ﺍﻟﺘﻌﻮﻳﺾ ﻭﺍﻟﺼﻴﻐﺔ ﺍﻟﻤﻌﻄﺎﺓ‬
‫ﻟﻠﺘﺤﻮﻳﻞﺇﻟﻰ ﺗﻜﺎﻣﻞ ﻳﺘﻀﻤﻦ ﺩﻭﺍﻝ ﺣﺴﺎﺏ ﺍﻟﻤﺜﻠﺜﺎﺕ‪.‬‬
‫ﺏ‪-2x2‬ﺃ‪=x⇒ 2‬ﺃﺛﺎﻧﻴﺔ‪θ‬‬
‫ﺃ‪-2‬ﺏ‪=x⇒ 2x2‬ﺃﺍﻟﺨﻄﻴﺉﺔﺏ‪θ‬‬
‫ﺍﻟﺴﺎﺑﻖ‪∫.‬‬
‫‪16‬‬
‫‪=2x9 -4‬‬
‫ﻳﺘﺬﻛﺮ‬
‫⌡‬
‫‪ 4‬ﺍﻟﺨﻄﻴﺉﺔ ‪θ2cos)θ2‬‬
‫ﻛﻮﺱﺩ‪θ‬‬
‫‪θ 23 =dx‬‬
‫‪=θ24 - 4sin‬‬
‫ﺛﺎﻧﻴﺔ‪ +1 =θ2‬ﺗﺎﻥ‪θ2‬‬
‫‪16‬‬
‫⌠‬
‫‪dx‬‬
‫ﺍﻟﺨﻄﻴﺉﺔ ‪⇒ θ‬‬
‫‪3 2=x‬‬
‫ﺏ‬
‫ﺗﺎﻥ‪ =θ2‬ﺛﺎﻧﻴﺔ‪1−θ2‬‬
‫ﻛﻮﺱ‪ −1 =θ2‬ﺧﻄﻴﺉﺔ‪θ2‬‬
‫‪2x4−9 2x‬‬
‫ﺃ‪+2‬ﺏ‪=x⇒ 2x2‬ﺃﺗﺎﻥ‪θ‬‬
‫‪4‬ﻛﻮﺱ‪ 2 =θ2‬ﻛﻮﺱ‪θ‬‬
‫(‬
‫‪9‬‬
‫) ‪32‬ﻛﻮﺱ‪(θ‬ﺩ‪=θ‬‬
‫∫‬
‫‪12‬‬
‫‪2‬‬
‫ﺍﻟﺨﻄﻴﺉﺔ‪θ‬‬
‫ﺩ‪θ‬‬
‫=∫‪ 12‬ﺳﻢ ﻣﻜﻌﺐ‪2‬ﺩ‪ 12- =θ‬ﺳﺮﻳﺮ ﺃﻃﻔﺎﻝ‪+θ‬ﺝ‬
‫ﺍﺳﺘﺨﺪﻡ‪ Right Triangle Trig‬ﻟﻠﻌﻮﺩﺓ ﺇﻟﻰ‪'x‬ﺱ‪ .‬ﻣﻦ‬
‫ﺍﻻﺳﺘﺒﺪﺍﻝﻟﺪﻳﻨﺎ ﺧﻄﻴﺉﺔ‪x3=θ‬ﻟﺬﺍ‪،‬‬
‫‪2‬‬
‫‪.x=2x‬ﻷﻥ ﻟﺪﻳﻨﺎ ﻷﺟﻞ ﻏﻴﺮ ﻣﺴﻤﻰ‬
‫ﺍﻟﺘﻜﺎﻣﻞﺳﻨﻔﺘﺮﺽ ﻣﻮﺟﺒﺔ ﻭﺇﺳﻘﺎﻁ ﺃﺷﺮﻃﺔ ﺍﻟﻘﻴﻤﺔ‬
‫ﺍﻟﻤﻄﻠﻘﺔ‪.‬ﺇﺫﺍ ﻛﺎﻥ ﻟﺪﻳﻨﺎ ﺗﻜﺎﻣﻞ ﻣﺤﺪﺩ ‪ ،‬ﻓﺴﻨﺤﺘﺎﺝ ﺇﻟﻰ‬
‫ﺣﺴﺎﺑﻪ‪θ‬ﻭﺇﺯﺍﻟﺔ ﺃﺷﺮﻃﺔ ﺍﻟﻘﻴﻤﺔ ﺍﻟﻤﻄﻠﻘﺔ ﺑﻨﺎء ًﻋﻠﻰ ﺫﻟﻚ‬
‫ﻭ‪،‬‬
‫ﻟﻮ‪0≤x‬‬
‫‪x‬‬‫‪- =x‬‬
‫‪ x−-‬ﻟﻮ‪0>x‬‬
‫ﻟﺬﺍ‪،‬‬
‫ﻣﻦﻫﺬﺍ ﻧﺮﻯ ﺳﺮﻳﺮ ﺍﻷﻃﻔﺎﻝ‪.2x4−9=θ‬‬
‫‪x3‬‬
‫∫‬
‫ﻓﻲﻫﺬﻩ ﺍﻟﺤﺎﻟﺔ ﻟﺪﻳﻨﺎ ‪ 2 =2x9 - 4‬ﻛﻮﺱ‪.θ‬‬
‫ﺍﻟﻜﺴﻮﺭﺍﻟﺠﺰﺉﻴﺔ ‪:‬ﻓﻲ ﺣﺎﻟﺔ ﺍﻟﺪﻣﺞ∫‬
‫ﺏ‬
‫‪16‬‬
‫‪2x4−9 2x‬‬
‫‪+2xx4−9 4- =dx‬ﺝ‬
‫ﺹ)‪dx(x‬ﺣﻴﺚ ﺩﺭﺟﺔﺹ)‪(x‬ﺃﺻﻐﺮ ﻣﻦ ﺩﺭﺟﺔ‬
‫ﺱ)‪(x‬‬
‫ﺱ)‪.(x‬ﻋﺎﻣﻞ ﺍﻟﻤﻘﺎﻡ ﺑﺸﻜﻞ ﻛﺎﻣﻞ ﻗﺪﺭ ﺍﻹﻣﻜﺎﻥ ﻭﺇﻳﺠﺎﺩ ﺍﻟﺘﺤﻠﻞ ﺍﻟﺠﺰﺉﻲ ﻟﻠﻜﺴﺮ‬
‫ﺍﻟﺘﻌﺒﻴﺮﺍﻟﻌﻘﻼﻧﻲ‪ .‬ﺍﺩﻣﺞ ﺍﻟﺘﺤﻠﻞ ﺍﻟﺠﺰﺉﻲ ﻟﻠﻜﺴﺮ )‪ .(PFD‬ﻟﻜﻞ ﻋﺎﻣﻞ ﻓﻲ ﺍﻟﻤﻘﺎﻡ ﻧﺤﺼﻞ ﻋﻠﻰ ﻣﺼﻄﻠﺢ )ﻣﺼﻄﻠﺤﺎﺕ(‬
‫ﻓﻲﺍﻟﺘﺤﻠﻞ ﻭﻓﻘﺎً ﻟﻠﺠﺪﻭﻝ ﺍﻟﺘﺎﻟﻲ‪.‬‬
‫ﻋﺎﻣﻞﻓﻲﺱ)‪(x‬‬
‫ﻋﺎﻣﻞﻓﻲﺱ)‪(x‬ﺍﻟﻤﺪﺓ ﻓﻲ ‪PFD‬‬
‫ﺃ‬
‫ﻓﺄﺱ‪+‬ﺏ‬
‫ﻓﺄﺱ‪+‬ﺏ‬
‫ﻓﺄﺱ‪+‬ﺏ‬
‫ﻓﺄﺱ‪+bx+2‬ﺝ‬
‫ﺍﻟﺴﺎﺑﻖ‪∫.‬‬
‫∫‬
‫‪x13+2x7‬‬
‫)‪(4+2x) (1−x‬‬
‫‪x13+2x7‬‬
‫)‪(4+2x) (1−x‬‬
‫ﻓﺄﺱ‪+‬ﺏ‬
‫‪2‬‬
‫ﻓﺄﺱ‪+bx+2‬ﺝ‬
‫)ﻓﺄﺱ‪+‬ﺏ(‬
‫ﺃ‪+x1‬ﺏ‬
‫ﻙ‬
‫)ﻓﺄﺱ ‪ + bx+‬ﺝ(‬
‫‪dx‬‬
‫‪x3+14-x‬‬
‫ﻙ‬
‫‪1‬‬
‫‪+‬‬
‫ﺃ‬
‫‪2‬‬
‫‪1‬‬
‫‪2‬‬
‫ﻓﺄﺱ‪+bx+‬ﺝ‬
‫‪2‬‬
‫ﺃ‬
‫‪+L+‬‬
‫‪+L+‬‬
‫ﻙ‬
‫)ﻓﺄﺱ‪+‬ﺏ(‬
‫ﺃﻙ‪+x‬ﺏ‬
‫ﻙ‬
‫)ﻓﺄﺱ ‪+bx+‬ﺝ(‬
‫‪2‬‬
‫ﻙ‬
‫ﻙ‬
‫‪=x13+2x7‬ﺃ‪+‬ﺑﻜﺲ‪+‬ﺝ=ﺃ)‪) + (4+2x‬ﺑﻜﺲ‪+‬ﺝ( )‪(1−x‬‬
‫‪+ 1−4x ∫=dx‬‬
‫=∫‬
‫)ﻓﺄﺱ‪+‬ﺏ(‬
‫ﺃ‬
‫ﺍﻟﻤﺪﺓﻓﻲ ‪PFD‬‬
‫‪4+2x‬‬
‫‪16+x3‬‬
‫‪4+2x‬‬
‫)‪(4+2x) (1−x‬‬
‫ﺿﻊﺍﻟﺒﺴﻂ ﻋﻠﻰ ﻗﺪﻡ ﺍﻟﻤﺴﺎﻭﺍﺓ ﻭﺍﺟﻤﻊ ﺍﻟﺤﺪﻭﺩ ﺍﻟﻤﺘﺸﺎﺑﻬﺔ‪.‬‬
‫‪dx‬‬
‫‪)=x13+2x7‬ﺃ‪+‬ﺏ(‪)+2x‬ﺝ‪-‬ﺏ(‪4+x‬ﺃ‪-‬ﺝ‬
‫‪dx 416‬‬
‫‪+‬‬
‫‪+2x‬‬
‫=‪ 4‬ﻟﻮ‪ 8+(4 +2x)2ln3+ 1-x‬ﺗﺎﻥ‪)1−‬‬
‫ﻫﻨﺎﺷﻜﻞ ﺍﻟﻜﺴﺮ ﺍﻟﺠﺰﺉﻲ ﻭﺇﻋﺎﺩﺓ ﺗﺠﻤﻴﻌﻪ‪.‬‬
‫‪1−x‬‬
‫‪4+2x‬‬
‫)‪(4+2x) (1−x‬‬
‫‪x‬‬
‫‪(2‬‬
‫ﺿﻊﻣﻌﺎﻣﻼﺕ ﻣﺘﺴﺎﻭﻳﺔ ﻟﻠﺤﺼﻮﻝ ﻋﻠﻰ ﻧﻈﺎﻡ ﻭﺣﻞ ﻟﻠﺤﺼﻮﻝ‬
‫ﻋﻠﻰﺍﻟﺜﻮﺍﺑﺖ‪.‬‬
‫ﺃ‪+‬ﺏ=‪7‬‬
‫ﺃ=‪4‬‬
‫ﺝ‪-‬ﺏ=‪13‬‬
‫ﺏ=‪3‬‬
‫‪4‬ﺃ‪-‬ﺝ=‪0‬‬
‫ﺝ=‪16‬‬
‫ﻃﺮﻳﻘﺔﺑﺪﻳﻠﺔ ﺃﻥﺃﺣﻴﺎﻧﺎﻳﻌﻤﻞ ﻋﻠﻰ ﺇﻳﺠﺎﺩ ﺍﻟﺜﻮﺍﺑﺖ‪ .‬ﺍﺑﺪﺃ ﺑﺠﻌﻞ ﺍﻟﺒﺴﻂ ﻣﺘﺴﺎﻭﻳﺎً ﻓﻲ ﺍﻟﻤﺜﺎﻝ ﺍﻟﺴﺎﺑﻖ‪=x13+2x7 :‬ﺃ)‪+2x‬‬
‫‪)+(4‬ﺑﻜﺲ‪+‬ﺝ( )‪.(1-x‬ﺍﺧﺘﺎﺭﻟﻄﻴﻒ ‪ -‬ﺟﻴﺪﻗﻴﻢ‪x‬ﻭﻗﻢ ﺑﺘﻮﺻﻴﻠﻪ‪.‬‬
‫ﻋﻠﻰﺳﺒﻴﻞ ﺍﻟﻤﺜﺎﻝ ﺇﺫﺍ‪ 1=x‬ﻧﺤﺼﻞ ﻋﻠﻰ ‪5 = 20‬ﺃﺍﻟﺬﻱ ﻳﻌﻄﻲﺃ=‪ .4‬ﻟﻦ ﻳﻌﻤﻞ ﻫﺬﺍ ﺩﺍﺉﻤﺎً ﺑﺴﻬﻮﻟﺔ‪.‬‬
‫ﻳﺰﻭﺭ‪ http://tutorial.math.lamar.edu‬ﻟﻤﺠﻤﻮﻋﺔ ﻛﺎﻣﻠﺔ ﻣﻦ ﻣﻼﺣﻈﺎﺕ ﺍﻟﺘﻔﺎﺿﻞ ﻭﺍﻟﺘﻜﺎﻣﻞ‪.‬‬
‫©‪ 2005‬ﺑﻮﻝ ﺩﻭﻛﻴﻨﺰ‬
‫ﻭﺭﻗﺔﺍﻟﻐﺶ ﺣﺴﺎﺏ ﺍﻟﺘﻔﺎﺿﻞ ﻭﺍﻟﺘﻜﺎﻣﻞ‬
‫ﺍﻟﻤﺴﺎﺣﺔﺍﻟﺼﺎﻓﻴﺔ ‪∫:‬‬
‫ﺏ‬
‫ﺃ‬
‫ﺗﻄﺒﻴﻘﺎﺕﺍﻟﺘﻜﺎﻣﻼﺕ‬
‫‪dx(x)F‬ﻳﻤﺜﻞ ﺻﺎﻓﻲ ﺍﻟﻤﺴﺎﺣﺔ ﺍﻟﻮﺍﻗﻌﺔ ﺑﻴﻦ‪(x)F‬ﻭ ﺍﻝ‬
‫‪-x‬ﻣﺤﻮﺭ ﺑﻤﺴﺎﺣﺔ ﺃﻋﻼﻩ‪-x‬ﺍﻟﻤﺤﻮﺭ ﺍﻹﻳﺠﺎﺑﻲ ﻭﺍﻟﻤﺴﺎﺣﺔ ﺃﺩﻧﺎﻩ‪-x‬ﻣﺤﻮﺭ ﺳﻠﺒﻲ‪.‬‬
‫ﺍﻟﻤﺴﺎﺣﺔﺑﻴﻦ ﺍﻟﻤﻨﺤﻨﻴﺎﺕ‪:‬ﺍﻟﺼﻴﻎ ﺍﻟﻌﺎﻣﺔ ﻟﻠﺤﺎﻟﺘﻴﻦ ﺍﻟﺮﺉﻴﺴﻴﺘﻴﻦ ﻟﻜﻞ ﻣﻨﻬﻤﺎ ‪ ،‬ﺫ=‪fx‬‬
‫)(‬
‫ﺩ‬
‫ﺏ‬
‫ﺃ=∫ﺃ ‪--‬ﻭﻇﻴﻔﺔﺍﻟﻌﻠﻮﻱ‪- ---‬ﻭﻇﻴﻔﺔ ﺃﻗﻞ ‪)F=x&dx-‬ﺫ(⇒ﺃ=∫‬
‫⇒‬
‫ﺝ‬
‫‪--‬ﺍﻟﻮﻇﻴﻔﺔﺍﻟﺼﺤﻴﺤﺔ‪ ----‬ﻭﻇﻴﻔﺔ ﺍﻟﻴﺴﺎﺭ ‪-‬ﺩﻯ‬
‫ﺇﺫﺍﺗﻘﺎﻃﻌﺖ ﺍﻟﻤﻨﺤﻨﻴﺎﺕ ‪ ،‬ﻓﻴﺠﺐ ﺇﻳﺠﺎﺩ ﻣﺴﺎﺣﺔ ﻛﻞ ﺟﺰء ﻋﻠﻰ ﺣﺪﺓ‪ .‬ﻓﻴﻤﺎ ﻳﻠﻲ ﺑﻌﺾ ﺍﻟﺮﺳﻮﻣﺎﺕ ﺍﻟﺘﺨﻄﻴﻄﻴﺔ‬
‫ﻟﺒﻌﺾﺍﻟﻤﻮﺍﻗﻒ ﻭﺍﻟﺼﻴﻎ ﺍﻟﻤﺤﺘﻤﻠﺔ ﻟﺤﺎﻟﺘﻴﻦ ﻣﺤﺘﻤﻠﺘﻴﻦ‪.‬‬
‫ﺃ=∫‬
‫ﺏ‬
‫ﺃ‬
‫ﺃ=∫‬
‫‪-(x)F‬ﺯ)‪dx(x‬‬
‫ﺩ‬
‫‪)F‬ﺫ(‪-‬ﺯ)ﺫ(ﺩﻯ‬
‫ﺝ‬
‫ﺃ =∫‬
‫ﺝ‬
‫ﺃ‬
‫ﺏ‬
‫‪∫+gx dx‬ﺯ)‪dx(x)F-(x‬‬
‫‪()-x() F‬‬
‫ﺝ‬
‫ﺃﺣﺠﺎﻡﺍﻟﺜﻮﺭﺓ‪:‬ﺍﻟﺼﻴﻐﺘﺎﻥ ﺍﻟﺮﺉﻴﺴﻴﺘﺎﻥ ﻫﻤﺎﺍﻟﺨﺎﻣﺲ=∫ﺃ)‪dx(x‬ﻭﺍﻟﺨﺎﻣﺲ=∫ﺃ)ﺫ(ﺩﻯ‪.‬ﻫﻨﺎ‬
‫ﺑﻌﺾﺍﻟﻤﻌﻠﻮﻣﺎﺕ ﺍﻟﻌﺎﻣﺔ ﺣﻮﻝ ﻛﻞ ﻃﺮﻳﻘﺔ ﻣﻦ ﻃﺮﻕ ﺍﻟﺤﻮﺳﺒﺔ ﻭﺑﻌﺾ ﺍﻷﻣﺜﻠﺔ‪.‬‬
‫ﺧﻮﺍﺗﻢ‬
‫)‬
‫ﺃ=‪) π‬‬
‫ﻧﺼﻒ ﺍﻟﻘﻄﺮ ﺍﻟﺨﺎﺭﺟﻲ‬
‫(‪-‬‬
‫)‬
‫ﺃ=‪)π2‬ﻧﺼﻒ ﺍﻟﻘﻄﺮ( )ﻋﺮﺽ ﺍﺭﺗﻔﺎﻉ(‬
‫(‪(2‬‬
‫‪ 2‬ﺩﺍﺉﺮﺓ ﻧﺼﻒ ﻗﻄﺮﻫﺎ ﺍﻟﺪﺍﺧﻠﻲ‬
‫ﺣﺪﻭﺩ‪/x:‬ﺫﻣﻦ ﺣﻖ ‪ bot ring /‬ﻝ‪/x‬ﺫﻣﻦ ﺍﻟﺤﻠﻘﺔ ﺍﻟﻴﺴﺮﻯ ‪ /‬ﺍﻟﻌﻠﻮﻳﺔ‬
‫ﺍﻻﺳﻄﻮﺍﻧﺎﺕ‬
‫ﺣﺪﻭﺩ‪/x:‬ﺫﻣﻦ ﺍﻻﺳﻄﻮﺍﻧﺎﺕ ﺍﻟﺪﺍﺧﻠﻴﺔ‪ .‬ﻝ‪/x‬ﺫﻣﻦ ﺍﻻﺳﻄﻮﺍﻧﺎﺕ ﺍﻟﺨﺎﺭﺟﻴﺔ‪.‬‬
‫ﻫﻮﺭﺯ‪.‬ﺍﺳﺘﺨﺪﺍﻡ ﺍﻟﻤﺤﻮﺭ‪(x)F‬ﻭ‬
‫ﻓﻴﺮﺕ‪.‬ﺍﺳﺘﺨﺪﺍﻡ ﺍﻟﻤﺤﻮﺭ‪)F‬‬
‫ﺫ(ﻭ ﺯ)ﺫ(ﻭﺃ)ﺫ(ﻭﺩﻯ‪.‬‬
‫ﻫﻮﺭﺯ‪.‬ﺍﺳﺘﺨﺪﺍﻡ ﺍﻟﻤﺤﻮﺭ‪)F‬ﺫ(ﻭ‬
‫ﺯ)ﺫ(ﻭﺃ)ﺫ(ﻭﺩﻯ‪.‬‬
‫ﻓﻴﺮﺕ‪.‬ﺍﺳﺘﺨﺪﺍﻡ ﺍﻟﻤﺤﻮﺭ‪)F‬‬
‫‪(x‬ﻭ ﺯ)‪(x‬ﻭﺃ)‪(x‬ﻭ‪.dx‬‬
‫ﺍﻟﺴﺎﺑﻖ‪.‬ﺍﻟﻤﺤﻮﺭ‪:‬ﺫ=ﺃ<‪0‬‬
‫ﺍﻟﺴﺎﺑﻖ‪.‬ﺍﻟﻤﺤﻮﺭ‪:‬ﺫ=ﺃ≥‪0‬‬
‫ﺍﻟﺴﺎﺑﻖ‪.‬ﺍﻟﻤﺤﻮﺭ‪:‬ﺫ=ﺃ<‪0‬‬
‫ﺍﻟﺴﺎﺑﻖ‪.‬ﺍﻟﻤﺤﻮﺭ‪:‬ﺫ=ﺃ≥‪0‬‬
‫ﻧﺼﻒﺍﻟﻘﻄﺮ ﺍﻟﺨﺎﺭﺟﻲ‪:‬ﺃ‪(x)F-‬‬
‫ﻧﺼﻒﺍﻟﻘﻄﺮ ﺍﻟﺨﺎﺭﺟﻲ‪:‬ﺃ‪+‬ﺯ)‪(x‬‬
‫ﻧﺼﻒﺍﻟﻘﻄﺮ ‪:‬ﺃ‪-‬ﺫ‬
‫ﻧﺼﻒﺍﻟﻘﻄﺮ ‪:‬ﺃ‪+‬ﺫ‬
‫ﺩﺍﺉﺮﺓﻧﺼﻒ ﻗﻄﺮﻫﺎ ﺍﻟﺪﺍﺧﻠﻲ ‪:‬ﺃ‪-‬ﺯ)‪(x‬‬
‫ﺩﺍﺉﺮﺓﻧﺼﻒ ﻗﻄﺮﻫﺎ ﺍﻟﺪﺍﺧﻠﻲ‪:‬ﺃ‪(x)F+‬‬
‫ﺯ)‪(x‬ﻭﺃ)‪(x‬ﻭ‪.dx‬‬
‫ﻋﺮﺽ ‪)F:‬ﺫ(‪-‬ﺯ)ﺫ(‬
‫ﻋﺮﺽ ‪)F:‬ﺫ(‪-‬ﺯ)ﺫ(‬
‫ﻫﺬﻩﻟﻴﺴﺖ ﺳﻮﻯ ﺣﺎﻻﺕ ﻗﻠﻴﻠﺔ ﻟﻤﺤﻮﺭ ﺍﻟﺪﻭﺭﺍﻥ ﺍﻷﻓﻘﻲ‪ .‬ﺇﺫﺍ ﻛﺎﻥ ﻣﺤﻮﺭ ﺍﻟﺪﻭﺭﺍﻥ ﻫﻮ‪-x‬ﺍﺳﺘﺨﺪﺍﻡ ﺍﻟﻤﺤﻮﺭ ﺫ=ﺃ≥‪ 0‬ﺣﺎﻟﺔ‬
‫ﻣﻊﺃ=‪ .0‬ﻟﻤﺤﻮﺭ ﺍﻟﺪﻭﺭﺍﻥ ﺍﻟﻌﻤﻮﺩﻱ )‪=x‬ﺃ<‪ 0‬ﻭ‪=x‬ﺃ≥‪ (0‬ﺍﻟﺘﺒﺎﺩﻝ‪x‬ﻭ‬
‫ﺫﻟﻠﺤﺼﻮﻝ ﻋﻠﻰ ﺍﻟﺼﻴﻎ ﺍﻟﻤﻨﺎﺳﺒﺔ‪.‬‬
‫ﻳﺰﻭﺭ‪ http://tutorial.math.lamar.edu‬ﻟﻤﺠﻤﻮﻋﺔ ﻛﺎﻣﻠﺔ ﻣﻦ ﻣﻼﺣﻈﺎﺕ ﺍﻟﺘﻔﺎﺿﻞ ﻭﺍﻟﺘﻜﺎﻣﻞ‪.‬‬
‫©‪ 2005‬ﺑﻮﻝ ﺩﻭﻛﻴﻨﺰ‬
‫ﻭﺭﻗﺔﺍﻟﻐﺶ ﺣﺴﺎﺏ ﺍﻟﺘﻔﺎﺿﻞ ﻭﺍﻟﺘﻜﺎﻣﻞ‬
‫ﻋﻤﻞ ‪:‬ﺇﺫﺍ ﻛﺎﻧﺖ ﻗﻮﺓ‪(x)F‬ﻳﺤﺮﻙ ﺍﻟﺠﺴﻢ‬
‫ﻣﺘﻮﺳﻂﻗﻴﻤﺔ ﺍﻟﻮﻇﻴﻔﺔ‪:‬ﻣﺘﻮﺳﻂ ﺍﻟﻘﻴﻤﺔ‬
‫ﻋﻠﻰﺃ≥‪≥x‬ﺏﻳﻜﻮﻥ‪F‬ﻣﺘﻮﺳﻂ=‬
‫ﻝ‪() fx‬‬
‫ﺏ‬
‫ﻓﻲﺃ≥‪≥x‬ﺏﻭﺍﻟﻌﻤﻞ ﺍﻟﻤﻨﺠﺰ ﻫﻮﺩﺑﻠﻴﻮ=∫‪dx(x)F‬‬
‫ﺃ‬
‫‪1‬‬
‫ﺏ‪-‬ﺃ‬
‫ﺏ‬
‫∫ ﺃ‪dx(x)F‬‬
‫ﻣﺴﺎﺣﺔﺳﻄﺢ ﻃﻮﻝ ﺍﻟﻘﻮﺱ‪:‬ﻻﺣﻆ ﺃﻥ ﻫﺬﺍ ﻏﺎﻟﺒﺎً ﻣﺎ ﻳﻜﻮﻥ ﻣﻮﺿﻮﻉ ‪ .Calc II‬ﺍﻟﺼﻴﻎ ﺍﻷﺳﺎﺳﻴﺔ ﺍﻟﺜﻼﺙ ﻫﻲ ‪ ،‬ﺇﻝ=∫‬
‫ﺏ‬
‫ﺏ‬
‫ﺏ‬
‫‪π2∫=SA‬ﺱ ﺱ)ﺗﻨﺎﻭﺏ ﺣﻮﻝﺫ‪-‬ﻣﺤﻮﺭ(‬
‫‪π2∫=SA‬ﺱ ﺱ)ﺗﻨﺎﻭﺏ ﺣﻮﻝ‪-x‬ﻣﺤﻮﺭ(‬
‫ﺱ‬
‫ﺃ‬
‫ﺃ‬
‫ﺃ‬
‫ﺃﻳﻦﺱﻳﻌﺘﻤﺪ ﻋﻠﻰ ﺷﻜﻞ ﺍﻟﻮﻇﻴﻔﺔ ﺍﻟﺘﻲ ﻳﺘﻢ ﺍﻟﻌﻤﻞ ﺑﻬﺎ ﻋﻠﻰ ﺍﻟﻨﺤﻮ ﺍﻟﺘﺎﻟﻲ‪.‬‬
‫‪dx )+1‬ﺩﻯ(‬
‫ﺩﻯ(ﺩﻯ‬
‫‪dx )+1‬‬
‫ﺱ=‬
‫‪2‬‬
‫ﺱ=‬
‫‪2‬‬
‫ﺩﺱ= ) ﺩ‪ + )(dx‬ﺩ(‬
‫‪ dx‬ﻟﻮﺫ=‪(x)F‬ﻭﺃ≥‪≥x‬ﺏ‬
‫ﺩﻯ‪2‬‬
‫‪2‬‬
‫ﺏ‬
‫ﺩﻟﻮ‪)F=x‬ﺭ(ﻭﺫ=ﺯ)ﺭ(ﻭﺃ≥ﺭ≥‬
‫‪2‬‬
‫ﺩﻛﺘﻮﺭ(ﺩ‪θ‬ﻟﻮﺹ=‪(θ)F‬ﻭﺃ≥‪≥θ‬ﺏ‬
‫ﺱ= ﺹ‪)+2‬‬
‫ﺩ‪θ‬‬
‫ﻟﻮ‪)F=x‬ﺫ(ﻭﺃ≥ﺫ≥ﺏ‬
‫ﻣﻊﻣﺴﺎﺣﺔ ﺍﻟﺴﻄﺢ ﻟﻚﻳﻤﻜﻦﻳﺠﺐ ﺃﻥ ﻳﺤﻞ ﻣﺤﻞ‪x‬ﺃﻭﺫﺣﺴﺐ ﺍﺧﺘﻴﺎﺭﻙ ﻟـﺱﻟﺘﺘﻨﺎﺳﺐ ﻣﻊ ﺍﻟﺘﻔﺎﺿﻞ ﻓﻲﺱ‪ .‬ﻣﻊ‬
‫ﺍﻟﺒﺎﺭﺍﻣﺘﺮﻳﺔﻭﺍﻟﻘﻄﺒﻴﺔ ‪ ،‬ﺳﺘﺤﺘﺎﺝ ﺩﺍﺉﻤﺎً ﺇﻟﻰ ﺍﻻﺳﺘﺒﺪﺍﻝ‪.‬‬
‫ﺗﻜﺎﻣﻞﻏﻴﺮ ﻻﺉﻖ‬
‫ﺍﻟﺘﻜﺎﻣﻞﻏﻴﺮ ﺍﻟﺼﺤﻴﺢ ﻫﻮ ﺍﻟﺘﻜﺎﻣﻞ ﻣﻊ ﻭﺍﺣﺪ ﺃﻭ ﺃﻛﺜﺮ ﻣﻦ ﺍﻟﺤﺪﻭﺩ ﺍﻟﻼﻧﻬﺎﺉﻴﺔ ﻭ ‪ /‬ﺃﻭ ﺍﻟﺘﻜﺎﻣﻞ ﻏﻴﺮ ﺍﻟﻤﺴﺘﻤﺮ‪ .‬ﻳﻄُﻠﻖ ﻋﻠﻰ‬
‫ﺍﻟﺘﻜﺎﻣﻞﺍﺳﻢ ﻣﺘﻘﺎﺭﺏ ﺇﺫﺍ ﻛﺎﻥ ﺍﻟﺤﺪ ﻣﻮﺟﻮﺩﺍً ﻭﻟﻪ ﻗﻴﻤﺔ ﻣﺤﺪﻭﺩﺓ ﻭﻣﺘﺒﺎﻋﺪ ﺇﺫﺍ ﻛﺎﻥ ﺍﻟﺤﺪ ﻏﻴﺮ ﻣﻮﺟﻮﺩ ﺃﻭ ﻟﻪ ﻗﻴﻤﺔ ﻏﻴﺮ‬
‫ﻣﺤﺪﻭﺩﺓ‪.‬ﻫﺬﺍ ﻫﻮ ﻋﺎﺩﺓ ﻣﻮﺿﻮﻉ ‪.Calc II‬‬
‫ﺣﺪﻻﻧﻬﺎﺉﻲ‬
‫‪∫.1‬‬
‫‪∫.3‬‬
‫ﺃ‬
‫‪∞−‬‬
‫= ﻟﻴﻢ∫‬
‫ﺭ‬
‫∞‬
‫‪x() F‬‬
‫∞‬
‫‪∫+dx(x)∞F-∫ =dx (x)F‬‬
‫ﻭﺟﻪﺿﺎﺣﻚ‬
‫ﺭ→ ∞ﺃ‬
‫‪.2‬‬
‫‪x dx() F‬‬
‫ﺝ‬
‫ﻋﺪﺩﺻﺤﻴﺢ ﻣﺘﻘﻄﻊ‬
‫ﺏ‬
‫‪.1‬ﺍﻟﺴﺨﻂ‪ .‬ﻓﻲﺃ‪dx(x)F∫:‬‬
‫ﺃ‬
‫‪.3‬ﺍﻧﻘﻄﺎﻉ ﻓﻲﺃ>ﺝ>ﺏ‪∫:‬‬
‫∞‬
‫ﻟﻴﻢ∫‬
‫=‬
‫ﺭ ﺃ‬
‫→‬
‫ﺏ‬
‫ﺃ‬
‫‪∞-‬‬
‫‪=dx(x)F‬ﻟﻴﻢ‬
‫ﺭ→ ‪∞−‬‬
‫‪dx(x)F‬ﻛﻞ ﻣﻦ ﺍﻟﺘﻜﺎﻣﻼﺕ ﺍﻟﻤﻘﺪﻣﺔ ﻣﺘﻘﺎﺭﺑﺔ‪.‬‬
‫ﺝ‬
‫‪+‬‬
‫∫‬
‫ﺏ‬
‫ﺏ‬
‫ﺭ‬
‫‪.2‬ﺍﻟﺴﺨﻂ‪ .‬ﻓﻲﺏ‪:‬‬
‫‪dx(x)F‬‬
‫‪∫=dx(x)F‬‬
‫ﺝ‬
‫ﺃ‬
‫ﺏ‬
‫∫‬
‫ﺏ‬
‫ﺃ‬
‫∫‬
‫ﺏ‬
‫ﺭ‬
‫‪dx(x)F‬‬
‫ﺭ‬
‫‪=dx(x)F‬‬
‫ﻟﻴﻢ→ﺏ‪ ∫-‬ﺃ‪dx(x)F‬‬
‫ﺭ‬
‫‪dx(x)F∫+dx (x)F‬ﺷﺮﻳﻄﺔ ﺃﻥ ﻳﻜﻮﻥ ﻛﻼﻫﻤﺎ ﻣﺘﻘﺎﺭﺏ‪.‬‬
‫ﺝ‬
‫ﺍﺧﺘﺒﺎﺭﺍﻟﻤﻘﺎﺭﻧﺔ ﻟﻠﺘﻜﺎﻣﻼﺕ ﻏﻴﺮ ﺍﻟﺼﺤﻴﺤﺔ‪:‬ﻟﻮ‪≤(x)F‬ﺯ)‪ 0≤(x‬ﻓﻲ]ﺃ‪(∞ ،‬ﺛﻢ‪،‬‬
‫‪.1‬ﺇﺫﺍ∫‬
‫ﺃ‬
‫∞‬
‫‪dx(x)F‬ﺍﻟﺘﺤﻮﻳﻞ ﺛﻢ∫ﺯ)‪dx(x‬ﺍﻟﺘﺤﻮﻳﻞ‬
‫∞‬
‫ﺣﻘﻴﻘﺔﻣﻔﻴﺪﺓ‪ :‬ﺇﺫﺍﺃ<‪ 0‬ﺑﻌﺪ ﺫﻟﻚ∫‬
‫ﻝ‪inte‬‬
‫∞‬
‫ﺃ‬
‫‪.2‬ﺇﺫﺍ∫‬
‫∞‬
‫ﺃ‬
‫∞)‪dx(x‬‬
‫∫ﺃ‪F‬‬
‫ﺯ)‪dx(x‬ﺛﻢ ‪divg.‬‬
‫‪dx1‬ﻳﺘﻘﺎﺭﺏ ﺇﺫﺍﺹ<‪ 1‬ﻭﻳﺘﺒﺎﻋﺪ ﻝﺹ≥‪.1‬‬
‫‪divg.‬‬
‫ﺃ‪x‬ﺹ‬
‫ﺗﻘﺮﻳﺐﺍﻟﺘﻜﺎﻣﻼﺕ ﺍﻟﻤﺤﺪﺩﺓ‬
‫‪∫ gral‬ﺃ‪F‬ﺏ )‪dx(x‬ﻭ ﺃﻥ)ﻳﺠﺐ ﺃﻥ ﻳﻜﻮﻥ ﺣﺘﻰ ﻟﻘﺎﻋﺪﺓ ﺳﻤﺒﺴﻮﻥ(‬
‫ﺗﺤﺪﻳﺪ∆‪=x‬ﺏ‪-‬ﺃ‬
‫ﻥ‬
‫ﻳﻘﺴﻢ]ﺃﻭﺏ[ﺩﺍﺧﻞﻥﻓﺘﺮﺍﺕ ﻓﺮﻋﻴﺔ]‪0x‬ﻭ‪[1x‬ﻭ]‪1x‬ﻭ‪x]،…،[2x‬ﻥ‪1−‬ﻭ‪x‬ﻥ[ﻣﻊ‪=0x‬ﺃﻭ‪x‬ﻥ=ﺏﺛﻢ‪،‬‬
‫ﺏ‬
‫ﻗﺎﻋﺪﺓﻧﻘﻄﺔ ﺍﻟﻤﻨﺘﺼﻒ‪-(*x)FL +2+(*x)F+(1 *x)F- -x∆ ≈dx(x)F∫:‬‬
‫ﺃ‬
‫ﺏ‬
‫∆‪x‬‬
‫ﻗﺎﻋﺪﺓﺷﺒﻪ ﻣﻨﺤﺮﻑ‪≈dx(x)F ∫ :‬‬
‫ﺃ‬
‫‪2‬‬
‫ﺏ‬
‫∆‪x‬‬
‫ﻗﺎﻋﺪﺓﺳﻤﺒﺴﻮﻥ‪≈dx(x)F ∫ :‬‬
‫‪1+(x)F4 +(0x)F‬‬‫ﺃ‬
‫‪3‬‬
‫ﻥ‪x *، -‬ﺃﻧﺎ ﻫﻲﻧﻘﻄﺔ ﺍﻟﻤﻨﺘﺼﻒ]‪x‬ﺃﻧﺎ‪1−‬ﻭ‪x‬ﺃﻧﺎ[‬
‫‪-(x)F+(-x)F2L ++(2x)F2+ +(1x)F2 +(0x)F-‬‬
‫ﻳﺰﻭﺭ‪ http://tutorial.math.lamar.edu‬ﻟﻤﺠﻤﻮﻋﺔ ﻛﺎﻣﻠﺔ ﻣﻦ ﻣﻼﺣﻈﺎﺕ ﺍﻟﺘﻔﺎﺿﻞ ﻭﺍﻟﺘﻜﺎﻣﻞ‪.‬‬
‫ﻭ‬
‫ﻥ‪1‬‬
‫‪L ++2(x)F 2‬‬
‫‪+(x)F 2‬‬
‫ﻥ‪2−‬‬
‫ﻥ‪-‬‬
‫‪x ) F4‬ﻥ‪)fx +(1−‬‬
‫ﻥ‬
‫©‪ 2005‬ﺑﻮﻝ ﺩﻭﻛﻴﻨﺰ‬
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