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 φ A
ftc2.b φ B
ftc2.le φ A B
ftc2.c φ F : A B cn
ftc2.i φ D F 𝐿 1
ftc2.f φ F : A B cn
Assertion ftc2 φ A B F t dt = F B F A

Proof

Step Hyp Ref Expression
1 ftc2.a φ A
2 ftc2.b φ B
3 ftc2.le φ A B
4 ftc2.c φ F : A B cn
5 ftc2.i φ D F 𝐿 1
6 ftc2.f φ F : A B cn
7 1 rexrd φ A *
8 2 rexrd φ B *
9 ubicc2 A * B * A B B A B
10 7 8 3 9 syl3anc φ B A B
11 fvex x A B A x F t dt F x A V
12 11 fvconst2 B A B A B × x A B A x F t dt F x A B = x A B A x F t dt F x A
13 10 12 syl φ A B × x A B A x F t dt F x A B = x A B A x F t dt F x A
14 eqid TopOpen fld = TopOpen fld
15 14 subcn TopOpen fld × t TopOpen fld Cn TopOpen fld
16 15 a1i φ TopOpen fld × t TopOpen fld Cn TopOpen fld
17 eqid x A B A x F t dt = x A B A x F t dt
18 ssidd φ A B A B
19 ioossre A B
20 19 a1i φ A B
21 cncff F : A B cn F : A B
22 4 21 syl φ F : A B
23 17 1 2 3 18 20 5 22 ftc1a φ x A B A x F t dt : A B cn
24 cncff F : A B cn F : A B
25 6 24 syl φ F : A B
26 25 feqmptd φ F = x A B F x
27 26 6 eqeltrrd φ x A B F x : A B cn
28 14 16 23 27 cncfmpt2f φ x A B A x F t dt F x : A B cn
29 ax-resscn
30 29 a1i φ
31 iccssre A B A B
32 1 2 31 syl2anc φ A B
33 fvex F t V
34 33 a1i φ x A B t A x F t V
35 2 adantr φ x A B B
36 35 rexrd φ x A B B *
37 elicc2 A B x A B x A x x B
38 1 2 37 syl2anc φ x A B x A x x B
39 38 biimpa φ x A B x A x x B
40 39 simp3d φ x A B x B
41 iooss2 B * x B A x A B
42 36 40 41 syl2anc φ x A B A x A B
43 ioombl A x dom vol
44 43 a1i φ x A B A x dom vol
45 33 a1i φ x A B t A B F t V
46 22 feqmptd φ D F = t A B F t
47 46 5 eqeltrrd φ t A B F t 𝐿 1
48 47 adantr φ x A B t A B F t 𝐿 1
49 42 44 45 48 iblss φ x A B t A x F t 𝐿 1
50 34 49 itgcl φ x A B A x F t dt
51 25 ffvelrnda φ x A B F x
52 50 51 subcld φ x A B A x F t dt F x
53 14 tgioo2 topGen ran . = TopOpen fld 𝑡
54 iccntr A B int topGen ran . A B = A B
55 1 2 54 syl2anc φ int topGen ran . A B = A B
56 30 32 52 53 14 55 dvmptntr φ dx A B A x F t dt F x d x = dx A B A x F t dt F x d x
57 reelprrecn
58 57 a1i φ
59 ioossicc A B A B
60 59 sseli x A B x A B
61 60 50 sylan2 φ x A B A x F t dt
62 22 ffvelrnda φ x A B F x
63 17 1 2 3 4 5 ftc1cn φ dx A B A x F t dt d x = D F
64 30 32 50 53 14 55 dvmptntr φ dx A B A x F t dt d x = dx A B A x F t dt d x
65 22 feqmptd φ D F = x A B F x
66 63 64 65 3eqtr3d φ dx A B A x F t dt d x = x A B F x
67 60 51 sylan2 φ x A B F x
68 30 32 51 53 14 55 dvmptntr φ dx A B F x d x = dx A B F x d x
69 26 oveq2d φ D F = dx A B F x d x
70 69 65 eqtr3d φ dx A B F x d x = x A B F x
71 68 70 eqtr3d φ dx A B F x d x = x A B F x
72 58 61 62 66 67 62 71 dvmptsub φ dx A B A x F t dt F x d x = x A B F x F x
73 62 subidd φ x A B F x F x = 0
74 73 mpteq2dva φ x A B F x F x = x A B 0
75 56 72 74 3eqtrd φ dx A B A x F t dt F x d x = x A B 0
76 fconstmpt A B × 0 = x A B 0
77 75 76 eqtr4di φ dx A B A x F t dt F x d x = A B × 0
78 1 2 28 77 dveq0 φ x A B A x F t dt F x = A B × x A B A x F t dt F x A
79 78 fveq1d φ x A B A x F t dt F x B = A B × x A B A x F t dt F x A B
80 oveq2 x = B A x = A B
81 itgeq1 A x = A B A x F t dt = A B F t dt
82 80 81 syl x = B A x F t dt = A B F t dt
83 fveq2 x = B F x = F B
84 82 83 oveq12d x = B A x F t dt F x = A B F t dt F B
85 eqid x A B A x F t dt F x = x A B A x F t dt F x
86 ovex A B F t dt F B V
87 84 85 86 fvmpt B A B x A B A x F t dt F x B = A B F t dt F B
88 10 87 syl φ x A B A x F t dt F x B = A B F t dt F B
89 79 88 eqtr3d φ A B × x A B A x F t dt F x A B = A B F t dt F B
90 lbicc2 A * B * A B A A B
91 7 8 3 90 syl3anc φ A A B
92 oveq2 x = A A x = A A
93 iooid A A =
94 92 93 eqtrdi x = A A x =
95 itgeq1 A x = A x F t dt = F t dt
96 94 95 syl x = A A x F t dt = F t dt
97 itg0 F t dt = 0
98 96 97 eqtrdi x = A A x F t dt = 0
99 fveq2 x = A F x = F A
100 98 99 oveq12d x = A A x F t dt F x = 0 F A
101 df-neg F A = 0 F A
102 100 101 eqtr4di x = A A x F t dt F x = F A
103 negex F A V
104 102 85 103 fvmpt A A B x A B A x F t dt F x A = F A
105 91 104 syl φ x A B A x F t dt F x A = F A
106 13 89 105 3eqtr3d φ A B F t dt F B = F A
107 106 oveq2d φ F B + A B F t dt - F B = F B + F A
108 25 10 ffvelrnd φ F B
109 33 a1i φ t A B F t V
110 109 47 itgcl φ A B F t dt
111 108 110 pncan3d φ F B + A B F t dt - F B = A B F t dt
112 25 91 ffvelrnd φ F A
113 108 112 negsubd φ F B + F A = F B F A
114 107 111 113 3eqtr3d φ A B F t dt = F B F A