Let $f$ be a bounded, continuous function such that $f(t+17) = f(t)$ for all $t$

Let $f$ be a bounded, continuous function such that $f(t+17) = f(t)$ for all $t$. Show that for all $\alpha$, we have $$\int_0^{17} f(t) \,dt = \int_\alpha^{17+\alpha} f(t) \,dt$$

How can I solve this question?

Edit: Can I take u = t + $\alpha$, t = u - $\alpha$, du = dt $$\int_0^{17} f(t) \,dt = \int_\alpha^{17+\alpha} f(u-\alpha) \,du$$ Then how can I proceed?

Solution 1:

Method 1: Define the function $I(\alpha)$ for $\alpha\in\mathbb{R}$ as

$$I(\alpha)=\int_\alpha^{\alpha+17}f(t)dt\ .$$

By the fundamental theorem of calculus

$$I'(\alpha) = f(\alpha+17)-f(\alpha)=0 $$

by the periodicity of $f$. So $I(\alpha)$ is constant and equal to $I(0)$

$$\int_\alpha^{\alpha+17}f(t)dt=I(\alpha)=I(0)=\int_0^{17} f(t)dt\ .$$

Method 2: Split the integral into three parts and make the substitution $t=s+17$ on the third as follows

$$\begin{align}\int_\alpha^{\alpha+17}f(t)dt &= \int_\alpha^{0}f(t)dt+\int_0^{17}f(t)dt+\int_{17}^{\alpha+17}f(t)dt\\ &= \int_\alpha^{0}f(t)dt+\int_0^{17}f(t)dt+\int_{0}^{\alpha}f(s+17)ds \\ &= \int_\alpha^{0}f(t)dt+\int_0^{17}f(t)dt +\int_{0}^{\alpha}f(s)ds\\ &= \int_\alpha^{0}f(t)dt+\int_0^{17}f(t)dt-\int_{\alpha}^{0}f(s)ds \\ &= \int_0^{17}f(t)dt\ .\end{align}$$