Let $d_{1}$ be defined by $d_{1}: C\bigl([a,b]\bigr) \times C\bigl([a,b]\bigr) \to [0,\infty)$, where
$$d_{1}(f,g) = \int_{a}^{b}|f(x) - g(x)|~\mathrm dx.$$
Show $d_{1}$ is a metric on $C\bigl([a,b]\bigr)$. (ie: prove the three requirements for being a metric.)
Note that $C\bigl([a,b]\bigr)$ is the set of all continuous functions on $[a,b]$.
A hint is given, but I think the question can be answered without it.
Here it is, anyway: if $d_{1}(f,g) = 0$, then $$d_{1}(f,g) = \int_{a}^{x}|f(t) - g(t)|~\mathrm dt = 0 \qquad\forall x \in[a,b].$$
My professors says in order to use that you must prove it.
Suppose $d_1(f,g)=0$. Since the absolute value function is nonnegative, $$\int_a^x|f(t)-g(t)|\ dt \leq \int_a^b |f(t)-g(t)|\ dt =0,$$ for $x\in [a,b]$. This proves the hint.
By the first part of the fundamental theorem of calculus, we know that $$\frac{d}{dx} \int_a^x|f(t)-g(t)|\ dt= |f(x)-g(x)| =0.$$ This implies that $f(x)=g(x)$ as desired.
The other requirements for a metric are straightforward so I assume you were able to them.