Prove that this functional is a continuous linear functional and find its norm Prove that this functional is continuous linear functional in $C[-1,1]$ and find its norm.
$\int_0^1 tx(t)dt-x(1)$
What I have so far:
1) I think I proved the linearity by changing $x$ to $(x+y)$ and getting $F(x)+F(y)$ as a result and changing $x$ to $(dx)$ which leads to $d*F(x)$. Is it correct though?
And I don't know what to do further with continuity and norm.
I translated it from another language so I could make a term mistake.

Linearity follows from linearity of the integral and evaluation $ x \mapsto x(1)$.
It should be clear that 
$|f(x)| \le \int_0^1 t \|x\| dt + \|x\| = {3 \over 2} \|x\|$.
If we could set $x(t) = 1$ for $t \in [0,1)$ and $x(1) = -1$ then it is clear that $|f(x)|={3 \over 2}\|x\|$, but we need $x$ to be continuous.
So, we see if an approximation works and let $x_n(t) = 1$ for $x \in [-1,1-{1 \over n}]$ and $x_n(t) = 1-2n(t-(1-{1 \over n}))$ otherwise and a little bit of work shows that
$\lim_n |f(x_n)| = {3 \over 2}$ (and, of course, $\|x_n\| = 1$). Hence $\|f\| = {3 \over 2}$.
