The discussion centers on the integral of a function defined as $$f(x)=\frac{1}{\sqrt{1-x}}-3$$ over the interval [0, 1). The function is continuous and approaches infinity as x approaches 1 from the left. The integral $$A:=\int_0^1 f(x)\, dx$$ is analyzed to determine if its sign can be negative, given that the function is negative for much of the interval. The conclusion is that while the function diverges, the integral can still be finite, leading to the possibility of a negative sign for A.
PREREQUISITESMathematicians, students of calculus, and anyone interested in advanced integral analysis and the behavior of functions near singularities.
Try $$f(x)=\frac{1}{\sqrt{1-x}}-3.$$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##?