∑
k=0
N
f(k)=∫
0
N
f(x)dx+
2
f(N)+f(0)
+∑
k=1
m
(2k)!
B
2k
(f
(2k−1)
(N)−f
(2k−1)
(0))+R
m
Mathbot Says...
I wasn't able to parse your question, but the HE.NET team is hard at work making me smarter.
∑
k=0
N
f(k)=∫
0
N
f(x)dx+
2
f(N)+f(0)
+∑
k=1
m
(2k)!
B
2k
(f
(2k−1)
(N)−f
(2k−1)
(0))+R
m
I wasn't able to parse your question, but the HE.NET team is hard at work making me smarter.