Evaluating the stochastic integral of Brownian motion without using Ito's formula? I am currently trying to understand some lecture notes (which offer little depth) and the following exercise (without a solution) occurs:

Using the definition
  $$
\int_0^t f_s \hspace{1mm} dW_s := \text{l.i.m}_{\lambda \rightarrow 0} \left[ \sum_i f_{s_i} (W_{s_{i+1}} - W_{s_i}) \right] \hspace{10mm} (*)
$$
  where l.i.m = limit in (square) mean, show that
  $$
\int_0^t W_s \hspace{1mm} dW_s = \frac{W_t^2}{2} + \frac{t}{2}
$$

Can someone please help me to understand how to do this? Plugging $W_s$ into the formula $(*)$ we get
$$
\int_0^t W_s \hspace{1mm} dW_s = \text{l.i.m}_{\lambda \rightarrow 0} \left[ \sum_i W_{s_i} (W_{s_{i+1}} - W_{s_i}) \right]
$$
Unfortunately, I am not sure where to go from here. I am also not certain what $\lambda$ is supposed to denote, since it is not defined in the lecture notes. I suspect (by comparing this formula to the formula for Riemann integration) that
$$
\lambda = s_{i+1} - s_i
$$

The integral value should be:
$$\int_0^tW_s\;dW_s=\frac{W_t^2}{2}-\frac{t}{2}$$
Notice that 
$$\sum_iW_{s_i}(W_{s_{i+1}}-W_{s_i})=\frac{1}{2}\sum_i(W_{s_{i+1}}^2-W_{s_i}^2)-\frac{1}{2}\sum_i(W_{s_{i+1}}-W_{s_i})^2$$
The first sum is a telescoping sum that reduces to $\frac{W_t^2}{2}$ (since $W_0 = 0$). The second sum becomes half the quadratic variation of $W_s$. So your integral is...
