Breaking Stick

As given in the question, the point where the stick will break is uniformly random.

Now, lets consider, what is the probability, that a particular element (very small) point will be chosen (which is at a distance $x$ from one end)?

Very small, right? Yes!

We must quantify this using calculus.

We can say that the probability will be:

$$\frac{dx}{L}$$

where $L$ is the length of the rod.

Now, if this is the case, then the rod will break into 2 parts, one with length $x$, other with $L-x$.

Using the fundamental definition of Expectation, we can say that

$$E[l]=\sum (l.P(l)),l\text{ varies from }0\text{ to }\frac{L}{2}$$

(as we are looking at the smaller edge)

Here, $P(l)$ is:

$$\frac{dl}{L},\text{right? No!}$$

$P(l)$ means what's the probability that the smaller part will have a length $l$.

Note that there are 2 points where a rod can break, for the smaller length to be a particular $l$
(One is at $l$ distance from one end, the other is at $l$ distance from the other end).

Thus,

$$P(l)=\frac{2dl}{L}$$

Now, the sigma becomes an integration as we have elemental quantities involved.

Evaluating the integral, we get

$$E[l]={\int }_0^{L/2}l\cdot \frac{2dl}{L}$$ $$=\frac{2}{L}{\int }_0^{L/2}ldl$$ $$=\frac{2}{L}\times {\left[\frac{l^2}{2}\right]}_0^{L/2}$$ $$=\frac{2}{L}\times \frac{(L/2{)}^2}{2}$$ $$=\frac{2}{L}\times \frac{L^2}{8}$$ $$=\frac{L}{4}$$

Final Answer

Thus, the expected length of the smaller part is:

$$\boxed{\frac{L}{4}}$$