How to find the ball? i have $E=\mathcal{C}([0,1],\mathbb{R})$ and $$d(f,g)=\int_0^1|f(x)-g(x)|dx$$
i want to find $B_d(2,1)$
$$B_d(2,1)=\{f\in E, \int_0^1 |f(x)-2|dx<1\}$$
In the case where $f(x)\leq 2$ then $\int_0^1 |f(x)-2|dx=\int_0^1 -f(x)+2dx$ so i find that $\int_0^1 f(x) dx>1$ 
In the case where $f(x)\geq 2$ then $\int_0^1 |f(x)-2|dx=\int_0^1 f(x)-2dx$ so i find that $\int_0^1 f(x) dx<3$
what can i say ?
how to continue please.
 A: $f(x)=\frac32$ for $x\in[0,1]$ is in the ball, as is easy to verify.  
$g(x)=\frac52$ for $x\in[0,1]$ too. 
But also $h(x)=3x+\frac12$ for $x\in[0,1]$, is in the ball.
See a plot of this function at WolframAlpha, and
the corresponding integral showing that $h$ belongs to $B_d(2,1)$. 
Also see the difference function $h(x)-2$ for the above $h$ 
at WA.
Note that the difference is negative for $x<\frac12$ which makes the 
absolute value of the difference to be piece-wise defined, and makes necessary to split the integral in two pieces, to evaluate it:
$\displaystyle \int_0^1|3x+\frac12-2|\,dx=$
$\displaystyle \int_0^\frac12|3x+\frac12-2|\,dx+\int_\frac12^1|3x+\frac12-2|\,dx= $
$\displaystyle \int_0^\frac12 (2-(3x+\frac12))\,dx+\int_\frac12^1 (3x+\frac12-2)\,dx = 
\frac38+\frac38=\frac34<1$. 
In general, given any $f\in C([0,1],\mathbb R)$, in order to compute the integral $\int_0^1 |f(x)-2|\,dx$ you may need to split $[0,1]$ into sets $A=\{x\in[0,1]:f(x)<2\}$ and $B=\{x\in[0,1]:f(x)>2\}$.
Each of $A$ and $B$ is (Lebesgue) measurable and you could take integrals  $a=\int_A|f(x)−2|\,dx=\int_A(2−f(x))\,dx$ and $b=\int_B|f(x)−2|dx=\int_B(f(x)−2)\,dx$.
Then $\int_0^1 |f(x)−2|\,dx=a+b$ and you need $a+b<1$.
(For the above function $h$ we have $A=[0,\frac12)$ and $B=(\frac12,1]$.) 
What you did in your question is not exactly the same, note the difference between $\int_0^1$ and $\int_A$. Alternatively (to avoid notation like 
$\int_A\,$) take $\int_0^1 \max(2−f(x),0)\,dx$ (same as $\int_A (2−f(x))\,dx)$, and $\int_0^1 \max(f(x)−2,0)\,dx$ (same as $\int_B (f(x)−2)\,dx$). 
The precise answer is the definition of $B_d(2,1)$ as already stated in your question, and the three functions listed above are only illustrative. There are many more, and the structure of the sets $A$ and $B$ may be complicated. 
For example, let $s(x)=2+3x\sin(\frac{\pi}x)$ for $x\in(0,1]$, and $s(0)=2$. See plot at WA.
Then $s$ is in the ball, see  the corresponding integral (where
I replaced $0$ with $0.01$ to help WA compute it numerically).
Compare with the graphs of 
the positive part $\max(s(x)-2,0)$, and
the (absolute value of) the negative part $\max(2-s(x),0)$,
of the difference function $s(x)-2$. 
