Show that $$ \int_0^1 \left(x^3+7\right)d \alpha(x) = \frac{29}{2}, \quad \text{where } \alpha(x) = \begin{cases} 1+x^3, & 0 < x \le 1\\ 0, & x = 0 \end{cases} $$
I don't know how to handle discontinuous integrators in RS integrals. Please help. I tried to solve it but ended up getting 15/2. Don't know where I went wrong. Which property can be used here? I converted the RS integral to R integral but it did not work. Assuming $f$ and $g$ to be nicely differentiable, I used: $$\int f(x)dg(x)=\int f(x)g'(x)dx,$$ so $$ \int_0^1 \left(x^3+7\right)d (1+x^3) = \int_0^1 \left(3x^2)(x^3+7\right)dx= \int_0^1 \left(3x^5+21x^2\right)dx =\frac{15}{2}, $$
However in my textbook the answer is given to be $\frac{29}{2}$.
Your analysis takes care of the continuous part of the integral over $x \in (0,1]$ but there is a discrete point at $x=0$ which will additionally contribute $f(0)\Delta \alpha(0)$. Note that $f(0)=7$ and $$ \Delta \alpha(0) = \alpha(0^+) - \alpha(0) = \left(1+0^3\right) - 0 = 1. $$ Hence, the net contribution of the discrete point $x=0$ is $$ f(0)\Delta \alpha(0) = 7 \cdot 1 = 7, $$ and the final integral is $$ f(0)\Delta \alpha(0) + \int_0^1 \left(x^3+7\right) 3x^2 dx = 7 + \frac{15}{2} = \frac{29}{2} $$ as desired.