Find norm of operator I have a linear functional $$A: L_2[0,2] \to \mathbb R, Ax = \int_0^2(t^2+2)x(t)dt$$
I need to find $C$, trying to measure $C$ and $||Ax||$ to find it, but how can I do it in this problem?

Here I'll play the role of Robin to Batman, and emulate my mentor's work in a more elementary manner, as befits a protoge of the master!
First of all, note that $y(t) = t^2 +2 \in L^2[0, 2]$; since the inner product $\langle u, v \rangle$ on $L^2[0, 2]$ is given by
$\langle u, v \rangle = \int_0^2 u(t)v(t) dt, \tag{1}$
and thus
$\langle t^2 + 2, t^2 + 2 \rangle = \int_0^2 (t^2 + 2)^2 dt = \int_0^2 (t^4 + 4t^2 + 4)^2 dt$
$= (\dfrac{t^5}{5} + \dfrac{4t^3}{3} + 4t \vert_0^2 = \dfrac{32}{5} + \dfrac{32}{3} + 8 = \dfrac{96}{15} + \dfrac{160}{15} + \dfrac{120}{15} = \dfrac{376}{15}, \tag{2}$
so that
$\Vert t^2 + 2 \Vert^2 = \langle t^2 + 2, t^2 + 2 \rangle = \dfrac{376}{15}, \tag{3}$
or
$\Vert t^2 + 2 \Vert = \sqrt{\dfrac{376}{15}}; \tag{4}$
since $\Vert t^2 + 2 \Vert$ exists (and is finite!), we have indeed have that $t^2 + 2 \in L_2[0, 2]$; thus we may apply the Cauchy-Schwarz inequality to $t^2 + 2$ and $x(t)$ to conclude that
$\vert \langle t^2 + 2, x(t) \rangle \vert \le \Vert t^2 + 2 \Vert \Vert x(t) \Vert; \tag{5}$
since $\langle t^2 + 2, x(t) \rangle$ is in fact the functional $Ax$, we see that (5) immediately shows that $Ax$ is bounded by $\Vert t^2 + 2 \Vert$; 
$\vert A(x) \vert \le \Vert t^2 + 2 \Vert \Vert x(t) \Vert = \sqrt{\dfrac{376}{15}} \Vert x(t) \Vert, \tag{6}$
i.e., 
$\Vert A \Vert \le \sqrt{\dfrac{376}{15}}. \tag{7}$
Now taking
$x(t) = t^2 + 2 \tag{8}$
(2)-(4) show that
$\vert A(t^2 + 2) \vert = \Vert t^2 + 2 \Vert^2 = \sqrt{\dfrac{376}{15}} \Vert t^2 + 2 \Vert; \tag{9}$
since $\vert Ax(t) \vert$ in fact takes the value $\sqrt{376/15}\Vert x(t) \Vert$ for $x(t) = t^2 + 2$, we have that indeed
$\Vert A \Vert = \sqrt{\dfrac{376}{15}}. \tag{10}$
Hope this helps!  Cheerio,
and as always,
Fiat Lux!!!
