Is the sign of the integral of this function negative?

[Rlwe]

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 $\text{sgn}(A)=-1$?

 

[mathman]

f(x) < 0 for much of the interval.

 

[JFerreira]

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 $$f(x)=\frac{1}{\sqrt{1-x}}-3.$$