Question: prove that: $\int_0^a f(x) dx = \int_0^a f(a-x) dx$
1. **State the problem:** We want to prove that $$\int_0^a f(x) dx = \int_0^a f(a-x) dx.$$
2. **Recall the substitution rule for integrals:** If we let $u = g(x)$ be a differentiable function with continuous derivative, then $$\int f(x) dx = \int f(g^{-1}(u)) \frac{d}{du}g^{-1}(u) du.$$
3. **Apply substitution:** Let $$u = a - x.$$ Then, $$du = -dx \implies dx = -du.$$
4. **Change the limits of integration:** When $x=0$, $$u = a - 0 = a.$$ When $x=a$, $$u = a - a = 0.$$
5. **Rewrite the integral:**
$$\int_0^a f(a-x) dx = \int_{x=0}^{x=a} f(a-x) dx = \int_{u=a}^{u=0} f(u) (-du) = \int_0^a f(u) du.$$
6. **Simplify the integral:**
$$\int_0^a f(u) du = \int_0^a f(x) dx$$ (renaming the dummy variable $u$ back to $x$).
7. **Conclusion:** We have shown that
$$\int_0^a f(x) dx = \int_0^a f(a-x) dx,$$ which completes the proof.