Why is currying part of the definition of exponential objects? For context: an exponential of objects $B$ and $A$ in a category $\mathcal{C}$ is defined as an object $B^A$ and a morphism $\epsilon: B^A \times A \rightarrow B$ such that for every object $C$ in $\mathcal{C}$ and every morphism $f: C \times A \rightarrow B$ there is a unique morphism $\lambda f: C \rightarrow B^A$ such that $\epsilon \circ (\lambda f \times \mathrm{id}_A) = f$. The $\lambda$ operation in this context is often referred to as currying, or partial application.
If $C=1$ (the terminal object in $C$), we get a bijection between morphisms $A \rightarrow B$ and morphisms $1 \rightarrow B^A$, which allows us to interpret the exponential object as the "space of morphisms" from $A$ to $B$ (aka "internal hom").
The question is, why is it crucial to require this bijection for every object $C$ instead of only for the terminal object? I suppose the definitions are not equivalent (admit I didn't check, though); the question is more in the lines of "why is this definition more interesting/important/natural/obvious/etc.?".
 A: If you want to interpret $B^A$ as a "space of morphisms" you need to know how to map into or out of this space. The definition of exponential object tells you how to map into this space; that is, it tells you what a $C$-parameterized family of maps $A \to B$ is, namely it's a map $C \times A \to B$. This uniquely determines $B^A$, if it exists, by the Yoneda lemma, because it gives a universal property that $B^A$ satisfies. If you don't require a condition for every $C$ then you don't have a universal property and so in particular no guarantee of uniqueness.
Already the definition of $A \times B$ does not refer just to points $1 \to A \times B$ but to arbitrary maps $C \to A \times B$, the point being that the universal property of the product tells you not only what a point in $A \times B$ is but what a $C$-parameterized family of points is (roughly speaking). 
A: Currying has bijection:
   $\mathsf{Hom}(A\times B,C)\cong \mathsf{Hom}(A,C^B)$
This bijection comes from product-hom adjunction.
You get evaluation $\epsilon$ by $A=C^B$. Evaluation wouldnt work if you do $A=1$. Both need to work. Note that to get evaluation working, you need identity function $\mathrm{id}_{C^B}$, this is needed in $\mathsf{Hom}(C^B \times B,C)\cong \mathsf{Hom}(C^B,C^B)$. Identity function $\mathrm{id}_K$ requires that $K$ is an object, so $C^B$ is an object, instead of arrow.
