Fourier Series piecewise

Question

Can someone please explain why the integral gose from -1 to 1 to (0 to 1) how does this happen?

Answer 1

It is a general strategy to break integrals of piecewise functions at the boundaries of the pieces. This means that we have simple recipes for the functions in each resulting integral. In your case, \begin{align*} \int_{-1}^{1} \; &\begin{cases} 0, & -1 \leq x < 0 \\ x^2, & 0 \leq x < 1 \end{cases} \, \mathrm{d}x \\ &= \int_{-1}^{0} \; \begin{cases} 0, & -1 \leq x < 0 \\ x^2, & 0 \leq x < 1 \end{cases} \, \mathrm{d}x + \int_{0}^{1} \; \begin{cases} 0, & -1 \leq x < 0 \\ x^2, & 0 \leq x < 1 \end{cases} \, \mathrm{d}x \\ &= \int_{-1}^{0} \; 0 \, \mathrm{d}x + \int_{0}^{1} \; x^2 \, \mathrm{d}x \\ &= 0 + \int_{0}^{1} \; x^2 \, \mathrm{d}x \\ \end{align*}

Notice that once we split the interval of integration at the boundaries of the pieces of the definition of $f$, we only keep the one recipe in that interval. Then the subsequent integrals are much easier to work with.