Question about the accuracy of integrated resistors
\$R\$ depends on \$L\$, \$W\$, \$\overline{\rho}\$ and \$x_j\$.

If the errors of those 4 parameters are statistically independent, then the variance of \$R\$ will be the sum of the variances of those 4 parameters, hence the summation.

Another way to look at it is by taking partial derivatives of \$R\$ with respect to each parameter, then dividing by \$R\$ and squaring. Then you can add it up all together because all variations are statistically independent. Otherwise you couldn't just add them up, you should take into account any cross-correlation between parameters.

And just in case you're wondering how it's possible to add errors that are in the denominator of the equation (parameters \$W\$ and \$x_j\$): we can do it because there is a first order approximation involved.

$$
\frac{1}{W+\Delta W} = \frac{\frac{1}{W}}{1+\frac{\Delta W}{W}} \approx \frac{1}{W} \left( {1-\frac{\Delta W}{W}}  \right) \text{   if   } \vert{\frac{\Delta W}{W}}\rvert \ll 1 \text{ thus}\\
R+\Delta R \big\vert_{\text{due to }\Delta W} = \frac{L \overline{\rho}}{(W+\Delta W)x_j}= \frac{L \overline{\rho}}{Wx_j}\left( {1-\frac{\Delta W}{W}}  \right) = R - R\frac{\Delta W}{W} \\
\Delta R \big\vert_{\text{due to }\Delta W} = - R\frac{\Delta W}{W} \\
{\frac{\Delta R}{R}} \Bigg\vert_{\text{due to }\Delta W} = - \frac{\Delta W}{W}
$$

After squaring, the minus sign disappears. The same applies to the error contributed by \$x_j\$.