I wanted to solve the following Riemann- Stieltjes problems:
Calculate
(a) $\int_{0}^{\pi/2}x d(\sin x)$
(b) $\int_{-1}^{3}x~dg(x)$ where $g(x)= \begin{cases} 0, & \text{for $x=-1,$}\\ 1, & \text{for $-1<x<2,$}\\ -1, & \text{for $-1\leq x\leq 2.$} \end{cases}$
(c) $\int_{-2}^{2}x~dg(x)$ where $g(x)= \begin{cases} x+2, & \text{for $-2\leq x\leq -1,$}\\ 2, & \text{for $-1<x<0,$}\\ x^2+3, & \text{for $0\leq x\leq 2.$} \end{cases}$
I try to solve (a) as follows: Since $\sin x$ is differentiable on $0<x<\pi/2,$ so that (a) changes to Riemann integral by using the following theorem: $$\int_a^bf(x)dg(x)=\int_a^bf(x)g'(x)dx.$$ So, I obtained $$\int_{0}^{\pi/2} x~d(\sin x)= \int_{0}^{\pi/2} x\cos x dx=1.$$ Please, Is the solution (a) correct?
Part (b), I try to solve it the following way; $$\int_{-1}^{3} x dg(x)= \int_{-1}^{-1} x g'(x)dx+\int_{-1}^{2} x g'(x)dx+\int_{2}^{3} x g'(x)dx.$$ But, I am not sure whether it is correct or not.
I don't know how to solve part (c) at all. I need a help on how to solve it. Thanks in advance!
By parts:
$$(a)\;\;\begin{cases}u=x,&u'=1\\v'=\cos x,&v=\sin x\end{cases}\implies\left.\int_0^{\pi/2}x\cos xdx=x\sin x\right|_0^{\pi/2}-\int_0^{\pi/2}\sin xdx=$$
$$=\left.\frac\pi2+\cos x\right|_0^{\pi/2}=\frac\pi2-1$$
Your idea for (b) is good: you shall get zero at the end. For (c) you can do the same as for (b):
$$\int_{-2}^2xdg(x)=\int_{-2}^{-1}x\cdot1\,dx+\int_{-1}^0x\cdot0\,dx+\int_0^2x\cdot2x\,dx=$$
$$\left.\frac12x^2\right|_{-2}^{-1}+0+\left.\frac23x^3\right|_0^2=\frac12(1-4)+\frac23(8-0)=-\frac32+\frac{16}3=\frac{23}6$$