Question: Let $X$ be a Banach space and consider a bounded operator $A:X\to \mathbb R$.

> 
> I want to prove that $\|A\| = \alpha$ for a fixed $\alpha>0$ and I
>  already proved $\|A\|\leq \alpha$. So now ideally I would like to find
>  an $f$ such that $|Af| = \alpha$.
> 
> 
> 

I didn't quite manage to do that but instead I found a sequence $(f\_n)\_n$ such that $\|f\_n\|=1$ and $\lim\_{n\to \infty} |Af\_n| = \alpha$. Is this sufficient to conclude the claim?

Answer: 
Yes, you are correct, by Green's Theorem, you should evaluate
$$\int\_{x=0}^1\int\_{y=0}^{1-x}(y^2-x)dydx=\int\_{x=0}^1\left[\frac{y^3}{3}-xy\right]\_{y=0}^{1-x}dx=\int\_{x=0}^1\left(\frac{(1-x)^3}{3}-x(1-x)\right)dx\\
=\frac{1}{3}\int\_{t=0}^1t^3 dt-\int\_{x=0}^1 x(1-x)dx.$$
Can you take it from here?
\begin{align}\int\_0^1 \int\_0^{1-x} (y^2-x) \, dy\, dx &= \int\_0^1 \left[ \frac{y^3}{3}-xy \right]\_{y=0}^{y=1-x}\, dx \\
&=\int\_0^1 \frac{(1-x)^3}{3}-x(1-x) \, dx \\
&= \int\_0^1 \frac{(1-x)^3}{3} - x+x^2 \, dx \\
&= \left[ -\frac{(1-x)^4}{12}-\frac{x^2}{2}+\frac{x^3}{3}\right]\_{x=0}^{x=1} \\
&= \left[ -\frac12+\frac13+\frac1{12} \right]\\
&=\left[\frac{-6+4+1}{12} \right]\\
&=-\frac1{12}\end{align}

Your method seems correct but most likely you make a careless mistake somewhere.