| f ( x ) = 12√ x f ( x ) = x0 ⋅ x−0,52 ⋅ x0,5 ⋅ x−0,5 f ( x ) = x0 − 0,52 ⋅ x0,5 − 0,5 f ( x ) = x−0,52 ⋅ x0 f ( x ) = 12 ⋅ x−0,5 F ( x ) = 12 ⋅ 2 ⋅ x−0,5 + 1 + k F ( x ) = x0,5 + k F ( x ) = √ x + k |
tirsdag den 30. juni 2015
Integralregning 15
mandag den 29. juni 2015
Integralregning 14
Web-matematik
Regneregler for ubestemt integral
Sumreglen:
∫ ( f ( x ) + g ( x ) ) dx =
∫ f ( x ) dx + ∫ g ( x ) dx
Eksempel:
∫ x3 − 1x2 dx =
∫ x3 dx + ∫ − 1x2 dx =
14 ⋅ x4 + 1x + k
Differentialkvotienter
Differensreglen:
∫ ( f ( x ) − g ( x ) ) dx =
∫ f ( x ) dx − ∫ g ( x ) dx
Eksempel:
∫ 2x5 − 1x dx =
∫ 2x5 dx − ∫ 1x dx =
2 ⋅ 16 ⋅ x6 − ln ( x ) + k =
13 ⋅ x6 − ln ( x ) + k
Differentialregning (8)
Konstant-faktor-reglen:
∫ ( c ⋅ f ( x ) ) dx =
c ⋅ ∫ f ( x ) dx
Eksempel:
∫ 5e2x dx =
5 ⋅ ∫ e2x dx =
5 ⋅ 12 ⋅ e2x + k =
52 ⋅ e2x + k
Integralregning 13
Web-matematik
Integraler
∫ x dx = 12 ⋅ x2
∫ kx dx = k2 ⋅ x2
∫ k dx = k ⋅ x
∫ xn dx = 1n + 1 ⋅ xn + 1
∫ 1x dx = ln ( | x | )
Differentialregning (8)
∫ ax dx = axln ( a )
Integralregning (16)
∫ ex dx = ex
∫ ekx dx = 1k ⋅ ekx
∫ √ x dx = 23 ⋅ x1,5 = 23 ⋅ ( √ x )3
∫ ln ( x ) dx = x ⋅ ln ( x ) − x
Bevis:
( x ⋅ ln x − x )’ =
( x ⋅ ln x )’ − ( x )’ =
( x )’ ⋅ ln x + x ⋅ ( ln x )’ − 1 =
1 ⋅ ln x + x ⋅ 1x − 1 = ln x
søndag den 28. juni 2015
Integralregning 12
| f ( x ) = x220 + 1 g ( x ) = 1 = x0 F ( x ) = x320 ⋅ 3 + x G ( x ) = x / 2 F ( 6 ) − F ( 1 ) − G ( 6 ) + G ( 1 ) = 3,583 A1 = ∑M − 1n = 0 q ⋅ f ( s ( n ) ) = 8,583 A2 = ∑M − 1n = 0 q ⋅ g ( s ( n ) ) = 5 A = A1 − A2 = 3,583 |
Integralregning 11
| f ( x ) = 2 x + e−x ∫ 2 x + e−x dx = 2 ⋅ 12 ⋅ x2 + 1 − 1 ⋅ e−x = F ( x ) = x2 − e−x A = ∫30 2 x + e−x dx = [ x2 − e−x ]30 = F ( 3 ) − F ( 0 ) = ( 32 − e−3 ) − ( 02 − e−0 ) = 9,95 |
Abonner på:
Opslag (Atom)