Finite element discretization example clarification 
We introduce on the domain $\Omega = (0,1)$ a mesh $0=x_0<x_1<x_2<\dots<x_{n+1}=1$ and let $V_h$ be the space of piecewise linear hat functions $\varphi_i$ such that
$$\varphi_i'=
 \left\{
        \begin{array}{ll}
            \frac{1}{x_i-x_{i-1}}=\frac{1}{h_i}, & \quad x_{i-1}<x<x_i \\
            \frac{-1}{x_{i+1}-x_{i}}=-\frac{1}{h_{i+1}}, & \quad x_{i}<x<x_{i+1} \\
0, & \quad\text{ otherwise}
        \end{array}
    \right.
$$
We can then easily compute the corresponding stiffness matrix $K$ and mass matrix $M$ by evaluating the integrals below,
$$K_{ij}:=(\varphi_i', \varphi_j') = \int_0^1 \varphi'_i\varphi'_jdx$$
$$M_{ij}:=(\varphi_i, \varphi_j) = \int_0^1 \varphi_i\varphi_jdx$$
So that
$$K_{ij} = (\varphi_i', \varphi_j') =
 \left\{
        \begin{array}{ll}
            -\frac{1}{h_i}, & \quad j=i-1 \\
            \frac{1}{h_i} + \frac{1}{h_{i+1}}, & \quad j=i \\
\frac{-1}{h_{i+1}}, & \quad j=i+1 \\
0, & \quad\text{ otherwise}
        \end{array}
    \right.$$
$$M_{ij} = (\varphi_i, \varphi_j) =
 \left\{
        \begin{array}{ll}
            \frac{1}{6}h_i, & \quad j=i-1 \\
            \frac13({h_i} +{h_{i+1}}), & \quad j=i \\
\frac16 h_{i+1}, & \quad j=i+1 \\
0, & \quad\text{ otherwise}
        \end{array}
    \right.$$

What I don't understand in this example is how actually the values $K_{ij}$ and $M_{ij}$ were obtained. Let me explain what I mean:
Let $x\in (x_{i-1}, x_i)$, then
$$K_{ij}=(\varphi_i', \varphi_j')=\int_0^1 \varphi_i'\varphi_j' dx
=\int_0^1 \frac{1}{h_i}\frac{1}{h_j}dx=\frac{1}{h_i h_j}=\frac{1}{h_i^2},$$
which does not coincide with what is in the book.
What is it that I'm misunderstanding?
 A: The piecewise linear hat function $\;\phi_i\;$ itself is:
$$
\phi_i(x)=
 \left\{ \begin{matrix}
            (x-x_{i-1})/(x_i-x_{i-1}) & \quad \text{for} \quad x_{i-1} < x < x_i \\
            (x_{i+1}-x)/(x_{i+1}-x_{i}) & \quad \text{for} \quad x_{i}< x < x_{i+1} \\
             0 & \quad \text{ otherwise}
         \end{matrix} \right.
$$
But I think that the mass matrix is given instead by the integral below:
$$
M_{i,j}:=(\phi_i, \phi_j) = \int_{x_{i-1}}^{x_{i+1}} \phi_i(x)\phi_j(x)\;dx
$$
Three non-zero cases (make a sketch):
$$
M_{i,i-1} = \int_{x_{i-1}}^{x_{i+1}} \phi_i(x)\phi_{i-1}(x)\;dx = \\
\frac{1}{(x_i-x_{i-1})^2} \int_{x_{i-1}}^{x_i} (x-x_{i-1})(x_{i}-x)\;dx = \frac{x_i-x_{i-1}}{6}
$$
$$
M_{i,i+1} = \int_{x_{i-1}}^{x_{i+1}} \phi_i(x)\phi_{i+1}(x)\;dx = \\
\frac{1}{(x_{i+1}-x_{i})^2} \int_{x_{i}}^{x_{i+1}} (x-x_{i+1})(x_{i}-x)\;dx = \frac{x_{i+1}-x_{i}}{6}
$$
$$
M_{i,i} = \int_{x_{i-1}}^{x_{i+1}} \phi_i(x)\phi_{i}(x)\;dx = \\
\frac{1}{(x_{i}-x_{i-1})^2} \int_{x_{i-1}}^{x_{i}} (x-x_{i-1})^2\;dx
+\frac{1}{(x_{i+1}-x_{i})^2} \int_{x_{i}}^{x_{i+1}} (x_{i+1}-x)^2\;dx = \frac{x_{i+1}-x_{i-1}}{3}
$$
Suppose that also the stiffness matrix is given instead by the integral below:
$$
K_{i,j}:=(\phi_i', \phi_j') = \int_{x_{i-1}}^{x_{i+1}} \phi'_i(x)\phi'_j(x)\;dx
$$
Then can you do this part of the exercise yourself?
I wouldn't be much surprised if the confusion is due to a typo in your book: a mix up of local and global coordinates could be the case. The following two references at MSE seem to be related to your question and might be helpful:
*
*
Understanding Galerkin method of weighted residuals
*Are there any two-dimensional quadrature that only uses the values at the vertices of triangles? .
A: Imagine that $x_i$ is an interior node: then $\varphi_i$ is nonzero only in  the interval $(x_{i-1},x_{i+1})$, and the same happens with $\varphi_i'$. The integral is computed over $(0,1)$ but the functions are nonzero in smaller intervals.
