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$2$) $\int_{0}^{a} f(x) d x=\int_{0}^{a} f(a-x) d x$$\newline$Proof: Let $\int f(x) d x=g(x)+c$$\newline$Consider R.H.S. : $\int_{0}^{a} f(a-x) \mathrm{d} x$$\newline$put $a-x=t$ i.e. $\quad x=a-t$$\newline$$\therefore \quad-\mathrm{d} x=\mathrm{d} t \Rightarrow \mathrm{d} x=-\mathrm{d} t$$\newline$As $x$ varies from $0$ to $a, t$ varies from a to $0$$\newline$therefore $I=\int_{a}^{0} f(t)(-\mathrm{d} t)$$\newline$$=-\int_{a}^{0} f(t) d t$$\newline$$\int f(x) d x=g(x)+c$$0$$\newline$$\int f(x) d x=g(x)+c$$1$ as definite integration is$\newline$$\int f(x) d x=g(x)+c$$2$ L.H.S. independent of the variable.$\newline$Thus $\int f(x) d x=g(x)+c$$3$