Math bac cours 3

Page 1

‫اﻟــــــﺪوال اﻷﺻـﻠﻴـــــﺔ‬ ‫‪ -1‬ﺗﻌﺮﻳـــﻒ ‪:‬‬ ‫ﻟﺘﻜﻦ ‪ f‬داﻟﺔ ﻋﺪدﻳﺔ ﻣﻌﺮﻓﺔ ﻋﻠﻰ ﻣﺠﺎل ‪ . I‬ﻧﻘﻮل أن اﻟﺪاﻟﺔ ‪ F‬داﻟﺔ أﺻﻠﻴﺔ ﻟﻠﺪاﻟﺔ ‪ f‬ﻋﻠﻰ ‪ I‬إذا وﻓﻘﻂ إذا آﺎن ‪:‬‬ ‫‪ F‬داﻟﺔ ﻗﺎﺑﻠﺔ ﻟﻼﺷﺘﻘﺎق ﻋﻠﻰ اﻟﻤﺠﺎل ‪. I‬‬ ‫)‪F '( x) = f ( x‬‬ ‫وﻟﻜﻞ ‪ x‬ﻣﻦ ‪: I‬‬ ‫ﻣﺜــــﺎل ‪:‬‬ ‫‪ -1‬ﻟﺘﻜﻦ‬ ‫إذن ‪:‬‬

‫‪F ( x ) = x2 + x + 1‬‬ ‫‪F '( x) = 2 x + 1‬‬

‫إذن ‪ :‬اﻟﺪاﻟﺔ ‪ F‬هﻲ داﻟﺔ أﺻﻠﻴﺔ ﻟﻠﺪاﻟﺔ ‪ f‬اﻟﻤﻌﺮﻓﺔ ﺑــ ‪:‬‬

‫‪f ( x) = 2 x + 1‬‬

‫‪ -2‬ﺣﺪد داﻟﺔ أﺻﻠﻴﺔ ﻟﻜﻞ داﻟﺔ ﻣﻦ اﻟﺪوال اﻟﺘﺎﻟﻴﺔ ‪:‬‬ ‫‪f ( x) = 2‬‬ ‫‪-a‬‬

‫\ ∈ ‪F ( x) = 2 x + C / C‬‬ ‫‪-b‬‬

‫‪f ( x) = x‬‬ ‫‪1 2‬‬ ‫‪x + C‬‬ ‫‪2‬‬

‫‪-c‬‬

‫‪f ( x ) = x3‬‬ ‫‪1 4‬‬ ‫‪x + C‬‬ ‫‪4‬‬

‫‪-d‬‬

‫= )‪F ( x‬‬

‫= )‪F ( x‬‬

‫*` ∈ ‪/ n‬‬

‫‪f ( x ) = xn‬‬

‫‪1‬‬ ‫‪x n +1 + C‬‬ ‫‪n +1‬‬

‫= )‪F ( x‬‬

‫‪-e‬‬

‫}‪f ( x ) = x r ; r ∈ `* − {−1‬‬

‫‪-f‬‬

‫‪1‬‬ ‫‪x r +1 + C‬‬ ‫‪r +1‬‬ ‫‪f ( x) = x‬‬

‫= )‪F ( x‬‬

‫‪1‬‬

‫‪= x2‬‬ ‫‪2 32‬‬ ‫‪x + Cte‬‬ ‫‪3‬‬ ‫‪-g‬‬

‫)‪(2 x‬‬ ‫‪+ Cte‬‬

‫‪3‬‬

‫)‪+ 1‬‬ ‫‪4‬‬

‫= )‪F ( x‬‬ ‫‪2‬‬

‫‪(x‬‬

‫)‪+ 1‬‬

‫‪1‬‬ ‫‪u r +1 + C‬‬ ‫‪r +1‬‬

‫‪2‬‬

‫‪(x‬‬

‫= )‪f ( x‬‬ ‫‪1‬‬ ‫‪4‬‬

‫= )‪F ( x‬‬

‫اﻷﺻﻠﻴــﺔ‬

‫‪ur ⋅ u ' :‬‬


‫‪ -2‬ﺧـﺎﺻﻴـــﺔ ‪:‬‬ ‫ﻟﺘﻜﻦ ‪ f‬داﻟﺔ ﻋﺪدﻳﺔ‪.‬‬ ‫إذا آﺎﻧﺖ ‪ F‬داﻟﺔ أﺻﻠﻴﺔ ﻟﻠﺪاﻟﺔ ‪ f‬ﻋﻠﻰ ﻣﺠﺎل ‪ I‬ﻓﺈن ﻣﺠﻤﻮﻋﺔ اﻟﺪاﻟﺔ اﻷﺻﻠﻴﺔ ﻟﻠﺪاﻟﺔ ‪ f‬ﻋﻠﻰ ‪ I‬هــﻲ ‪:‬‬ ‫‪F + λ‬‬ ‫\ ∈ ‪.λ‬‬ ‫ﺣﻴﺚ‬ ‫ﺑـﺮهــــﺎن ‪:‬‬ ‫ﻟﺘﻜﻦ ‪ F‬داﻟﺔ أﺻﻠﻴﺔ ﻟﻠﺪاﻟﺔ ‪ f‬ﻋﻠﻰ ‪ I‬و ‪ λ‬ﻋﺪد ﺣﻘﻴﻘﻲ‪.‬‬

‫‪+ λ)' = F ' = f‬‬

‫ﻟﺪﻳﻨﺎ ‪:‬‬

‫‪(F‬‬

‫‪ F + λ‬هـﻲ أﻳﻀـﺎ داﻟﺔ أﺻﻠﻴﺔ ﻟﻠﺪاﻟﺔ ‪ f‬ﻋﻠﻰ ‪. I‬‬ ‫إذن ‪:‬‬ ‫وﻣﻨﻪ ‪ :‬ﻣﺠﻤﻮﻋﺔ اﻟﺪوال اﻷﺻﻠﻴﺔ ﻟﻠﺪاﻟﺔ ‪ f‬ﻋﻠﻰ ‪ I‬هــﻲ ‪. F + λ‬‬ ‫‪ -3‬ﺧـﺎﺻﻴـــﺔ ‪:‬‬ ‫ﻟﺘﻜﻦ ‪ f‬داﻟﺔ ﻋﺪدﻳﺔ ﺗﻘﺒﻞ داﻟﺔ أﺻﻠﻴﺔ ﻋﻠﻰ ‪. I‬‬ ‫ﻟﻴﻜﻦ ‪ x0‬ﻣﻦ ‪ I‬و ‪ y0‬ﻋﻨﺼﺮ ﺣﻘﻴﻘﻲ \ ∈ ‪. y0‬‬ ‫ﺗﻮﺟﺪ داﻟﺔ أﺻﻠﻴﺔ وﺣﻴﺪة ‪ F‬ﻟﻠﺪاﻟﺔ ‪ f‬ﻋﻠﻰ ‪. I‬‬ ‫ﺣﻴﺚ ‪:‬‬

‫‪F ( x0 ) = y0‬‬

‫أﻣـﺜـﻠـــﺔ ‪:‬‬ ‫ﺣﺪد اﻟﺪاﻟﺔ اﻷﺻﻠﻴﺔ ﻟﻠﺪاﻟﺔ ‪ f‬واﻟﺘﻲ ﺗﺤﻘﻖ اﻟﺸﺮط ‪. F ( x0 ) = y0‬‬

‫‪f ( x) = x + 1‬‬

‫‪-1‬‬

‫‪1 2‬‬ ‫‪x + x + C = 1‬‬ ‫‪2‬‬ ‫‪F ( 2) = 1‬‬

‫ﻟﺪﻳﻨﺎ ‪:‬‬ ‫وﺑﻤﺎ أن ‪:‬‬

‫= )‪F ( x‬‬

‫‪1 2‬‬ ‫‪x + x + C = 1‬‬ ‫‪2‬‬ ‫‪2 + 2 + C = 1‬‬ ‫‪C = −3‬‬

‫ﻓﺈن ‪:‬‬ ‫وﻣﻨﻪ ‪:‬‬ ‫‪2‬‬ ‫‪x + 1‬‬

‫‪-2‬‬

‫‪F ( 2) = 1‬‬

‫‪2‬‬

‫‪F ( 0) = 0‬‬

‫= )‪f ( x‬‬

‫ﻟﺪﻳﻨﺎ ‪:‬‬

‫‪F ( x ) = 2 Arc tan x + C‬‬

‫وﺑﻤﺎ أن ‪:‬‬

‫‪F ( 0) = 0‬‬

‫‪C = 0‬‬ ‫‪= 2 Arc tan x‬‬

‫ﻓﺈن ‪:‬‬ ‫إذن ‪:‬‬

‫)‪F ( x‬‬

‫‪f ( x ) = cos 2 x‬‬

‫‪-3‬‬ ‫ﻟﺪﻳﻨﺎ ‪:‬‬

‫وﺑﻤﺎ أن ‪:‬‬ ‫ﻓﺈن ‪:‬‬

‫‪+ C‬‬

‫)‪(2 x‬‬

‫⎞ ‪⎛π‬‬ ‫‪F⎜ ⎟ = 0‬‬ ‫⎠‪⎝2‬‬ ‫‪C = 0‬‬

‫⎞ ‪⎛π‬‬ ‫‪F⎜ ⎟ = 0‬‬ ‫⎠‪⎝2‬‬ ‫‪1‬‬ ‫= )‪F ( x‬‬ ‫‪sin‬‬ ‫‪2‬‬


‫‪1‬‬ ‫‪sin 2 x‬‬ ‫‪2‬‬

‫= )‪F ( x‬‬

‫‪ -4‬ﺧـﺎﺻﻴـــﺔ ‪:‬‬ ‫إذا آﺎﻧﺖ ‪ F‬داﻟﺔ أﺻﻠﻴﺔ ﻟﻠﺪاﻟﺔ ‪ f‬ﻋﻠﻰ ‪. I‬‬ ‫و ‪ G‬داﻟﺔ أﺻﻠﻴﺔ ﻟﻠﺪاﻟﺔ ‪ g‬ﻋﻠﻰ ‪. I‬‬ ‫ﻓﺈن ‪ :‬اﻟﺪاﻟﺔ ‪ F + G‬داﻟﺔ أﺻﻠﻴﺔ ﻟﻠﺪاﻟﺔ ‪ f + g‬ﻋﻠﻰ ‪. I‬‬

‫‪ -5‬ﺧـﺎﺻﻴـــﺔ ‪:‬‬ ‫آﻞ داﻟﺔ ﻣﺘﺼﻠﺔ ﻋﻠﻰ ﻣﺠﺎل ‪ I‬ﺗﻘﺒﻞ داﻟﺔ أﺻﻠﻴﺔ ‪.‬‬ ‫ﻣﻼﺣـﻈــﺔ وﺧـﺎﺻﻴـــﺔ ‪:‬‬

‫إذا آﺎﻧﺖ ‪ F‬و ‪ G‬داﻟﺘﻴﻦ أﺻﻠﻴﺘﻴﻦ ﻟﻠﺪاﻟﺔ ‪ f‬ﻋﻠﻰ ‪ ، I‬ﻓﺈﻧﻪ ﻳﻮﺟﺪ ﻋﺪد ﺣﻘﻴﻘﻲ ‪λ‬‬ ‫‪F − G = λ‬‬ ‫ﺣﻴﺚ ‪:‬‬ ‫‪ -6‬ﺟــﺪول اﻟــﺪوال اﻷﺻﻠﻴــﺔ اﻻﻋﺘﻴـﺎدﻳـــﺔ ‪:‬‬ ‫اﻟﺪاﻟــﺔ ‪f‬‬

‫اﻟﺪاﻟــﺔ ‪ ) F‬اﻷﺻﻠﻴــﺔ (‬

‫‪1‬‬

‫‪x+C‬‬

‫‪x‬‬

‫‪1 2‬‬ ‫‪x +C‬‬ ‫‪2‬‬

‫‪xn‬‬

‫‪1 n +1‬‬ ‫‪x +C‬‬ ‫‪n +1‬‬

‫`∈‪n‬‬

‫‪xr‬‬

‫‪1 r +1‬‬ ‫‪x +C‬‬ ‫‪r +1‬‬

‫}‪r ∈ _ − {−1‬‬

‫' ‪un ⋅u‬‬

‫‪1 n +1‬‬ ‫‪u +C‬‬ ‫‪n +1‬‬

‫`∈‪n‬‬

‫' ‪ur ⋅u‬‬

‫‪1 r +1‬‬ ‫‪u +C‬‬ ‫‪r +1‬‬

‫}‪r ∈ _ − {−1‬‬

‫‪1‬‬ ‫‪x +1‬‬

‫‪Arc tan x + C‬‬

‫‪cos x‬‬

‫‪sin x + C‬‬

‫‪sin x‬‬

‫‪− cos x + C‬‬

‫) ‪cos ( ax + b‬‬

‫‪1‬‬ ‫‪sin ( ax + b ) + C‬‬ ‫‪a‬‬

‫‪a≠0‬‬

‫) ‪sin ( ax + b‬‬

‫‪−1‬‬ ‫‪cos ( ax + b ) + C‬‬ ‫‪a‬‬

‫‪a≠0‬‬

‫‪2‬‬

‫‪1‬‬ ‫‪cos 2 x‬‬

‫= ) ‪1 + tan 2 ( x‬‬

‫‪tan x + C‬‬

‫ﻣـﻼﺣﻈـــﺎت‬

‫\∈ ‪C‬‬

‫‪+ kπ‬‬

‫‪π‬‬ ‫‪2‬‬

‫≠‪x‬‬


: ‫ﺗﻄﺒﻴـﻘـــــﺎت‬ : ‫ ﻓﻲ اﻟﺤﺎﻻت اﻟﺘﺎﻟﻴﺔ‬f ‫ﺣﺪد داﻟﺔ أﺻﻠﻴﺔ ﻟﻠﺪاﻟﺔ‬ x2 − 1 x2 + 1

f ( x) =

-1

=

x2 + 1 − 2 x2 + 1

=

−2 + 1 x + 1 2

F ( x ) = x − 2 Arc tan x + C f ( x) = x

: ‫إذن‬

x2 + 1

3

-2

1 2 x 3 x2 + 1 2 1 1 2 3 = x + 1 (2 x) ( ) 2

f ( x) =

1 2

F ( x) =

1

( x2 + 1) 3

1 1 +1 3

1 3 × 3 4

(x

=

3 8

(x

+ 1) 3

F ( x) =

3 8

= F ( x) =

4

2

x2 + 1

3

+ 1)

(x

+ x + 3) 2

2 3

4

: ‫إذن‬

x2 + x + 3 1

(2 x

-3

+ 1)

3

( x 2 + x + 3) 2 f ( x) =

sin 3 x cos5 x

F ( x ) = tan 3 x ⋅ = f ( x) =

: ‫وﻣﻨﻪ‬

+ 1) 3

2

(2 x 2

+1

4

=

f ( x) =

: ‫ﻟﺪﻳﻨﺎ‬

3

(x

2

+ 1) x 2

-4 1 cos 2 x

1 tan 4 x 4

: ‫ﻟﺪﻳﻨﺎ‬

-5


2 1 2 f ( x) = ( x + 1) 3 2 x 2 5 1 3 F ( x) = x 2 + 1) 3 + C ( 2 5 5 3 = x 2 + 1) 3 + C ( 10

: ‫إذن‬


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