Podés escribir:
$$\int_ {-a}^af(x) \,\mathrm{d}x=\int_ {-a}^0 f(x) \,\mathrm{d}x+\int_ {0}^a f(x) \,\mathrm{d}x$$
Llamando $$u=-x$$, la primer integral queda:
$$\int_ {-a}^af(x) \,\mathrm{d}x=-\int_ {a}^0 f(-u) \,\mathrm{d}u+\int_ {0}^a f(x) \,\mathrm{d}x=\int_ {0}^a f(-u) \,\mathrm{d}u+\int_ {0}^a f(x) \,\mathrm{d}x$$
En el caso que $$f$$ sea par se cumple $$f(-u)=f(u)$$:
$$\int_ {-a}^af(x) \,\mathrm{d}x=\int_ {0}^a f(u) \,\mathrm{d}u+\int_ {0}^a f(x) \,\mathrm{d}x=2\int_ {0}^a f(x) \,\mathrm{d}x$$
Si $$f$$ es impar $$f(-u)=-f(u)$$:
$$\int_ {-a}^af(x) \,\mathrm{d}x=-\int_ {0}^a f(u) \,\mathrm{d}u+\int_ {0}^a f(x) \,\mathrm{d}x=0$$
Saludos