Is There a Proof for Tabular Integration?

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Tabular integration is a shortcut method for integration by parts, but it lacks a formal proof. It essentially serves as a shorthand for the traditional integration by parts process. By using placeholder functions for u and v, one can demonstrate that tabular integration yields the same results as the standard method. The method simplifies the process by organizing calculations visually with diagonal lines and alternating signs. Ultimately, tabular integration is a time-saving technique rather than a rigorously proven approach.
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Today we learned tabular integration as a shortcut method for integration by parts. Is there a proof that legitimizes tabular integration out there, or some general formula? Because there has to be some sort of logic for the process of creating the diagonal lines, or assigning pluses and minuses. Thanks in advance.
 
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There isn't really a proof for tabular integration, it's just shorthand for integration by parts. If you did a integration by parts using placeholder functions for u and v (meaning the actual functions could be whatever you want) you'll see that it expands to the same thing that tabular integration does, it just saves time doing by remembering the tabular way.
 

Thread 'Proving that convexity implies second order derivative being positive'

There are probably loads of proofs of this online, but I do not want to cheat. Here is my attempt: Convexity says that $$f(\lambda a + (1-\lambda)b) \leq \lambda f(a) + (1-\lambda) f(b)$$ $$f(b + \lambda(a-b)) \leq f(b) + \lambda (f(a) - f(b))$$ We know from the intermediate value theorem that there exists a ##c \in (b,a)## such that $$\frac{f(a) - f(b)}{a-b} = f'(c).$$ Hence $$f(b + \lambda(a-b)) \leq f(b) + \lambda (a - b) f'(c))$$ $$\frac{f(b + \lambda(a-b)) - f(b)}{\lambda(a-b)}...
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