Metamath Proof Explorer


Theorem itgeq1f

Description: Equality theorem for an integral. (Contributed by Mario Carneiro, 28-Jun-2014) Avoid axioms. (Revised by GG, 1-Sep-2025)

Ref Expression
Hypotheses itgeq1f.1 𝑥 𝐴
itgeq1f.2 𝑥 𝐵
Assertion itgeq1f ( 𝐴 = 𝐵 → ∫ 𝐴 𝐶 d 𝑥 = ∫ 𝐵 𝐶 d 𝑥 )

Proof

Step Hyp Ref Expression
1 itgeq1f.1 𝑥 𝐴
2 itgeq1f.2 𝑥 𝐵
3 1 2 nfeq 𝑥 𝐴 = 𝐵
4 eleq2 ( 𝐴 = 𝐵 → ( 𝑥𝐴𝑥𝐵 ) )
5 4 anbi1d ( 𝐴 = 𝐵 → ( ( 𝑥𝐴 ∧ 0 ≤ 𝑦 ) ↔ ( 𝑥𝐵 ∧ 0 ≤ 𝑦 ) ) )
6 5 ifbid ( 𝐴 = 𝐵 → if ( ( 𝑥𝐴 ∧ 0 ≤ 𝑦 ) , 𝑦 , 0 ) = if ( ( 𝑥𝐵 ∧ 0 ≤ 𝑦 ) , 𝑦 , 0 ) )
7 6 csbeq2dv ( 𝐴 = 𝐵 ( ℜ ‘ ( 𝐶 / ( i ↑ 𝑘 ) ) ) / 𝑦 if ( ( 𝑥𝐴 ∧ 0 ≤ 𝑦 ) , 𝑦 , 0 ) = ( ℜ ‘ ( 𝐶 / ( i ↑ 𝑘 ) ) ) / 𝑦 if ( ( 𝑥𝐵 ∧ 0 ≤ 𝑦 ) , 𝑦 , 0 ) )
8 7 adantr ( ( 𝐴 = 𝐵𝑥 ∈ ℝ ) → ( ℜ ‘ ( 𝐶 / ( i ↑ 𝑘 ) ) ) / 𝑦 if ( ( 𝑥𝐴 ∧ 0 ≤ 𝑦 ) , 𝑦 , 0 ) = ( ℜ ‘ ( 𝐶 / ( i ↑ 𝑘 ) ) ) / 𝑦 if ( ( 𝑥𝐵 ∧ 0 ≤ 𝑦 ) , 𝑦 , 0 ) )
9 3 8 mpteq2da ( 𝐴 = 𝐵 → ( 𝑥 ∈ ℝ ↦ ( ℜ ‘ ( 𝐶 / ( i ↑ 𝑘 ) ) ) / 𝑦 if ( ( 𝑥𝐴 ∧ 0 ≤ 𝑦 ) , 𝑦 , 0 ) ) = ( 𝑥 ∈ ℝ ↦ ( ℜ ‘ ( 𝐶 / ( i ↑ 𝑘 ) ) ) / 𝑦 if ( ( 𝑥𝐵 ∧ 0 ≤ 𝑦 ) , 𝑦 , 0 ) ) )
10 9 fveq2d ( 𝐴 = 𝐵 → ( ∫2 ‘ ( 𝑥 ∈ ℝ ↦ ( ℜ ‘ ( 𝐶 / ( i ↑ 𝑘 ) ) ) / 𝑦 if ( ( 𝑥𝐴 ∧ 0 ≤ 𝑦 ) , 𝑦 , 0 ) ) ) = ( ∫2 ‘ ( 𝑥 ∈ ℝ ↦ ( ℜ ‘ ( 𝐶 / ( i ↑ 𝑘 ) ) ) / 𝑦 if ( ( 𝑥𝐵 ∧ 0 ≤ 𝑦 ) , 𝑦 , 0 ) ) ) )
11 10 oveq2d ( 𝐴 = 𝐵 → ( ( i ↑ 𝑘 ) · ( ∫2 ‘ ( 𝑥 ∈ ℝ ↦ ( ℜ ‘ ( 𝐶 / ( i ↑ 𝑘 ) ) ) / 𝑦 if ( ( 𝑥𝐴 ∧ 0 ≤ 𝑦 ) , 𝑦 , 0 ) ) ) ) = ( ( i ↑ 𝑘 ) · ( ∫2 ‘ ( 𝑥 ∈ ℝ ↦ ( ℜ ‘ ( 𝐶 / ( i ↑ 𝑘 ) ) ) / 𝑦 if ( ( 𝑥𝐵 ∧ 0 ≤ 𝑦 ) , 𝑦 , 0 ) ) ) ) )
12 11 sumeq2sdv ( 𝐴 = 𝐵 → Σ 𝑘 ∈ ( 0 ... 3 ) ( ( i ↑ 𝑘 ) · ( ∫2 ‘ ( 𝑥 ∈ ℝ ↦ ( ℜ ‘ ( 𝐶 / ( i ↑ 𝑘 ) ) ) / 𝑦 if ( ( 𝑥𝐴 ∧ 0 ≤ 𝑦 ) , 𝑦 , 0 ) ) ) ) = Σ 𝑘 ∈ ( 0 ... 3 ) ( ( i ↑ 𝑘 ) · ( ∫2 ‘ ( 𝑥 ∈ ℝ ↦ ( ℜ ‘ ( 𝐶 / ( i ↑ 𝑘 ) ) ) / 𝑦 if ( ( 𝑥𝐵 ∧ 0 ≤ 𝑦 ) , 𝑦 , 0 ) ) ) ) )
13 df-itg 𝐴 𝐶 d 𝑥 = Σ 𝑘 ∈ ( 0 ... 3 ) ( ( i ↑ 𝑘 ) · ( ∫2 ‘ ( 𝑥 ∈ ℝ ↦ ( ℜ ‘ ( 𝐶 / ( i ↑ 𝑘 ) ) ) / 𝑦 if ( ( 𝑥𝐴 ∧ 0 ≤ 𝑦 ) , 𝑦 , 0 ) ) ) )
14 df-itg 𝐵 𝐶 d 𝑥 = Σ 𝑘 ∈ ( 0 ... 3 ) ( ( i ↑ 𝑘 ) · ( ∫2 ‘ ( 𝑥 ∈ ℝ ↦ ( ℜ ‘ ( 𝐶 / ( i ↑ 𝑘 ) ) ) / 𝑦 if ( ( 𝑥𝐵 ∧ 0 ≤ 𝑦 ) , 𝑦 , 0 ) ) ) )
15 12 13 14 3eqtr4g ( 𝐴 = 𝐵 → ∫ 𝐴 𝐶 d 𝑥 = ∫ 𝐵 𝐶 d 𝑥 )