Metamath Proof Explorer


Theorem ftc2

Description: The Fundamental Theorem of Calculus, part two. If F is a function continuous on [ A , B ] and continuously differentiable on ( A , B ) , then the integral of the derivative of F is equal to F ( B ) - F ( A ) . This is part of Metamath 100 proof #15. (Contributed by Mario Carneiro, 2-Sep-2014)

Ref Expression
Hypotheses ftc2.a φA
ftc2.b φB
ftc2.le φAB
ftc2.c φF:ABcn
ftc2.i φDF𝐿1
ftc2.f φF:ABcn
Assertion ftc2 φABFtdt=FBFA

Proof

Step Hyp Ref Expression
1 ftc2.a φA
2 ftc2.b φB
3 ftc2.le φAB
4 ftc2.c φF:ABcn
5 ftc2.i φDF𝐿1
6 ftc2.f φF:ABcn
7 1 rexrd φA*
8 2 rexrd φB*
9 ubicc2 A*B*ABBAB
10 7 8 3 9 syl3anc φBAB
11 fvex xABAxFtdtFxAV
12 11 fvconst2 BABAB×xABAxFtdtFxAB=xABAxFtdtFxA
13 10 12 syl φAB×xABAxFtdtFxAB=xABAxFtdtFxA
14 eqid TopOpenfld=TopOpenfld
15 14 subcn TopOpenfld×tTopOpenfldCnTopOpenfld
16 15 a1i φTopOpenfld×tTopOpenfldCnTopOpenfld
17 eqid xABAxFtdt=xABAxFtdt
18 ssidd φABAB
19 ioossre AB
20 19 a1i φAB
21 cncff F:ABcnF:AB
22 4 21 syl φF:AB
23 17 1 2 3 18 20 5 22 ftc1a φxABAxFtdt:ABcn
24 cncff F:ABcnF:AB
25 6 24 syl φF:AB
26 25 feqmptd φF=xABFx
27 26 6 eqeltrrd φxABFx:ABcn
28 14 16 23 27 cncfmpt2f φxABAxFtdtFx:ABcn
29 ax-resscn
30 29 a1i φ
31 iccssre ABAB
32 1 2 31 syl2anc φAB
33 fvex FtV
34 33 a1i φxABtAxFtV
35 2 adantr φxABB
36 35 rexrd φxABB*
37 elicc2 ABxABxAxxB
38 1 2 37 syl2anc φxABxAxxB
39 38 biimpa φxABxAxxB
40 39 simp3d φxABxB
41 iooss2 B*xBAxAB
42 36 40 41 syl2anc φxABAxAB
43 ioombl Axdomvol
44 43 a1i φxABAxdomvol
45 33 a1i φxABtABFtV
46 22 feqmptd φDF=tABFt
47 46 5 eqeltrrd φtABFt𝐿1
48 47 adantr φxABtABFt𝐿1
49 42 44 45 48 iblss φxABtAxFt𝐿1
50 34 49 itgcl φxABAxFtdt
51 25 ffvelcdmda φxABFx
52 50 51 subcld φxABAxFtdtFx
53 14 tgioo2 topGenran.=TopOpenfld𝑡
54 iccntr ABinttopGenran.AB=AB
55 1 2 54 syl2anc φinttopGenran.AB=AB
56 30 32 52 53 14 55 dvmptntr φdxABAxFtdtFxdx=dxABAxFtdtFxdx
57 reelprrecn
58 57 a1i φ
59 ioossicc ABAB
60 59 sseli xABxAB
61 60 50 sylan2 φxABAxFtdt
62 22 ffvelcdmda φxABFx
63 17 1 2 3 4 5 ftc1cn φdxABAxFtdtdx=DF
64 30 32 50 53 14 55 dvmptntr φdxABAxFtdtdx=dxABAxFtdtdx
65 22 feqmptd φDF=xABFx
66 63 64 65 3eqtr3d φdxABAxFtdtdx=xABFx
67 60 51 sylan2 φxABFx
68 30 32 51 53 14 55 dvmptntr φdxABFxdx=dxABFxdx
69 26 oveq2d φDF=dxABFxdx
70 69 65 eqtr3d φdxABFxdx=xABFx
71 68 70 eqtr3d φdxABFxdx=xABFx
72 58 61 62 66 67 62 71 dvmptsub φdxABAxFtdtFxdx=xABFxFx
73 62 subidd φxABFxFx=0
74 73 mpteq2dva φxABFxFx=xAB0
75 56 72 74 3eqtrd φdxABAxFtdtFxdx=xAB0
76 fconstmpt AB×0=xAB0
77 75 76 eqtr4di φdxABAxFtdtFxdx=AB×0
78 1 2 28 77 dveq0 φxABAxFtdtFx=AB×xABAxFtdtFxA
79 78 fveq1d φxABAxFtdtFxB=AB×xABAxFtdtFxAB
80 oveq2 x=BAx=AB
81 itgeq1 Ax=ABAxFtdt=ABFtdt
82 80 81 syl x=BAxFtdt=ABFtdt
83 fveq2 x=BFx=FB
84 82 83 oveq12d x=BAxFtdtFx=ABFtdtFB
85 eqid xABAxFtdtFx=xABAxFtdtFx
86 ovex ABFtdtFBV
87 84 85 86 fvmpt BABxABAxFtdtFxB=ABFtdtFB
88 10 87 syl φxABAxFtdtFxB=ABFtdtFB
89 79 88 eqtr3d φAB×xABAxFtdtFxAB=ABFtdtFB
90 lbicc2 A*B*ABAAB
91 7 8 3 90 syl3anc φAAB
92 oveq2 x=AAx=AA
93 iooid AA=
94 92 93 eqtrdi x=AAx=
95 itgeq1 Ax=AxFtdt=Ftdt
96 94 95 syl x=AAxFtdt=Ftdt
97 itg0 Ftdt=0
98 96 97 eqtrdi x=AAxFtdt=0
99 fveq2 x=AFx=FA
100 98 99 oveq12d x=AAxFtdtFx=0FA
101 df-neg FA=0FA
102 100 101 eqtr4di x=AAxFtdtFx=FA
103 negex FAV
104 102 85 103 fvmpt AABxABAxFtdtFxA=FA
105 91 104 syl φxABAxFtdtFxA=FA
106 13 89 105 3eqtr3d φABFtdtFB=FA
107 106 oveq2d φFB+ABFtdt-FB=FB+FA
108 25 10 ffvelcdmd φFB
109 33 a1i φtABFtV
110 109 47 itgcl φABFtdt
111 108 110 pncan3d φFB+ABFtdt-FB=ABFtdt
112 25 91 ffvelcdmd φFA
113 108 112 negsubd φFB+FA=FBFA
114 107 111 113 3eqtr3d φABFtdt=FBFA