I had to compute $$T_2$$ using the trapezoid rule formula:
$$ T_m (f) = \frac{h}{2}(f(a) + f(b)) + h \sum\limits_{i=1}^{2^m -1} f(x_i) $$
Values are the following:
[a, b] = [0, 2]
m = 2
h = 1/2
f(x) = x^5 + x
f(0) = 0
f(1) = 2
f(2) = 34
From my calculations, the value of the summation should be
$$ \frac{1}{2} [ 6(3) + 34(3)] $$
but that is incorrect and I believe the answer should be something like :
$$ \frac{1}{2} [11.7] $$
I am unable to evaluate the summation properly. A step by step example of what computations occur within the summation sign would be really helpful.
We are given
$$\begin{align}f(x) &= x^5 + x\\ [a, b] &= [0, 2]\\ m &= 2\\ h &= \dfrac{1}{2} \end{align}$$
For $n = 2$, we have
$$ T_2 (f) = \dfrac{1}{2}(f(a) + 2 f(x_1)+ f(b)) $$
We get
$$T_2(f) = \dfrac{1}{2}( 0 + 2(2) + 34) = \dfrac{1}{2}(38) = 19$$
The actual value is
$$\int_0^2 (x^5+x)~dx = \dfrac{38}{3} \approx 12.6667$$
If you use three sub-intervals
$$T_3(x) = 15.5637860082305$$
If we use ten sub-intervals
$$T_{10}(x) = 12.9328$$
Update
Using your formula, we have
$$T_m (f) = \frac{h}{2}(f(a) + f(b)) + h \sum\limits_{i=1}^{2^m -1} f(x_i) = \dfrac{1}{4} (f(a) + f(b)) + \dfrac{1}{2} (f(x_1) + f(x_2) + f(x_3))$$
This means we have $x_1 = 1/2, x_2 = 1, x_3 = 3/2$.
This gives
$\dfrac{1}{4} (f(0) + f(2))+ \dfrac{1}{2} (f(1/2) + f(1) + f(3/2))\\ = \dfrac{1}{4}(0 + 34) + \dfrac{1}{2}(17/32 + 2 + 291/32) \\= 229/16 \approx 14.3125$