Answer
$$T_1=6,\ \ \ T_2=7 $$
Work Step by Step
Given $$f(0)=3,\ f( 1)= 4,\ f(2)=3 $$
Since $n=1$, $\Delta x=\frac{2-0}{1}=2$, then
\begin{align*}
T_{n}&=\dfrac{1}{2}\left[f(x_0)+2f(x_1)+2f(x_2)+.....+2f(x_{n-1})+f(x_n)\right]\Delta x\\
T_{1}&=\dfrac{1}{2}\left[f(x_0) +f(x_1)\right]\Delta x\\
&=\left[f(0) +f(2)\right] \\
&\approx 6
\end{align*}
and $n=2 $ $\Delta x=\frac{2-0}{2}=1$, so
\begin{align*}
T_{n}&=\dfrac{1}{2}\left[f(x_0)+2f(x_1)+2f(x_2)+.....+2f(x_{n-1})+f(x_n)\right]\Delta x\\
T_{1}&= \dfrac{1}{2}\left[f(x_0)+2f(x_1)+f(x_2) \right]\Delta x\\
&=\dfrac{1}{2}\left[f(0) +2f(1)+f(2)\right] \\
&\approx7
\end{align*}
