I Is the sign of the integral of this function negative?

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The discussion centers on the integral of a function f defined on the interval [0,1) with specific properties, including that it approaches infinity as x approaches 1. The key question is whether the sign of the integral A can be negative, given that f(x) is less than zero for much of the interval. A proposed example function is f(x) = 1/sqrt(1-x) - 3, which illustrates the behavior of f near the boundary. The conversation explores the implications of these conditions on the existence and sign of the integral A. Ultimately, the discussion seeks to clarify the conditions under which the integral can yield a negative value.
Rlwe
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Let ##f:[0;1)\to\mathbb{R}## and ##f\in C^1([0;1))## and ##\lim_{x\to1^-}f(x)=+\infty## and ##\forall_{x\in[0;1)}-\infty<f(x)<+\infty##. Define $$A:=\int_0^1f(x)\, dx\,.$$ Assuming ##A## exists and is finite, is it possible that ##\text{sgn}(A)=-1##?
 
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f(x) < 0 for much of the interval.
 
Rlwe said:
Let ##f:[0;1)\to\mathbb{R}## and ##f\in C^1([0;1))## and ##\lim_{x\to1^-}f(x)=+\infty## and ##\forall_{x\in[0;1)}-\infty<f(x)<+\infty##. Define $$A:=\int_0^1f(x)\, dx\,.$$ Assuming ##A## exists and is finite, is it possible that ##\text{sgn}(A)=-1##?
Try $$f(x)=\frac{1}{\sqrt{1-x}}-3.$$
 

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