Consider the following operator $A: C[-1,1] \to C[-1,1]$, $(Ax)(t) = \int_{-1}^t x(s) ds - \int_0^1 s x(s) ds$. What's its norm?
It's easy to show that $\Vert Ax \Vert \le \frac52 \Vert x \Vert \,\, \forall x \in C[-1,1]$. So $\Vert A \Vert \le \frac52$. But it seems that it's too inaccurate. I suspect that $\Vert Ax \Vert \le \frac32 \Vert x \Vert$ which turns into equality for $x(t) = 1$.
If you have an operator $T:f\longmapsto \int_{-1}^1g(t,s)f(s)\,ds$, with $g$ good enough (say, so that $g_t$ below is continuous), then $$\tag1 \|T\|=\sup\{g_t:\ t\in[-1,1]\}, $$ where $$ g_t=\int_{-1}^1|g(t,s)|\,ds. $$ Indeed, we have $$ \|Tf\|=\sup_t\left|\int_{-1}^1g(t,s)\,f(s)\,ds \right|\leq\|f\|\,\sup_t\int_{-1}^1|g(t,x)|\,ds. $$ Given $\varepsilon>0$, there exists $t_0$ such that $g_{t_0}+\varepsilon>\sup\{g_t:\ t\}$. Now choose $f\in C[-1,1]$ such that $\|f\|=1$ and $Tf(t_0)+\varepsilon>g_{t_0}$. Then $$ \|T\|+2\varepsilon\geq\|Tf\|+2\varepsilon\geq Tf(t_0)+2\varepsilon>g_{t_0}+\varepsilon>\sup\{g_t:\ t\}. $$ As $\varepsilon$ was arbitrary, we have the reverse inequality, and thus equality.
In the case at hand, we have $$ g(t,s)=1_{[-1,t]}(s)-s\,1_{[0,1]}(s). $$ Then $$ \int_{-1}^1|g(t,s)|\,ds=\int_{-1}^1|1_{[-1,t]}(s)-s\,1_{[0,1]}(s)|\,ds. $$ When $t\leq0$, this is $$ \int_{-1}^1 1_{[-1,t]}(s)+s\,1_{[0,1]}(s)|\,ds =t+1+\int_0^1 s\,ds=t+\frac32. $$ When $t>0$, it is $$ \int_{-1}^1 1_{[-1,0]}(s)+(1-s)\,1_{[0,t]}+s\,1_{[t,1]}(s)|\,ds =1+t-\frac{t^2}2+\frac12-\frac{t^2}2=t+\frac32-t^2. $$ The maximum $t$ will then be $t=1/2$, which gives $$ \|T\|=\frac32+\frac14=\frac74. $$