Double Integrals of Characteristic Functions

Question

I was working on the double integral of a characteristic function and was wondering if I had properly set it up. I want to integrate $$\int_0^{b_2}\int_0^{s_2}(s_2-s_1)\chi_{[0,b_1]}(s_1)\chi_{[0,b_2]}(s_2) \ \mathrm{d}s_1\mathrm{d}s_2,$$ where $\chi_{[0,b_1]}$ is the characteristic function with respect to $s_1$ and $\chi_{[0,b_2]}$ is the characteristic function with respect to $s_2$. Note also that $b_1\leq b_2$. I was thinking of first dragging out the $\chi_{[0,b_2]}(s_2)$ term as it's constant with respect to $s_1$: $$\int_0^{b_2} \chi_{[0,b_2]}(s_2)\left(\int_0^{s_2}(s_2-s_1)\chi_{[0,b_1]}(s_1)\ \mathrm{d}s_1\right)\mathrm{d}s_2.$$ Then, noting that $b_1\leq b_2$, I can split the integral into: $$\int_0^{b_1}\chi_{[0,b_2]}(s_2)\left(\int_0^{s_2}(s_2-s_1)\chi_{[0,b_1]}(s_1)\ \mathrm{d}s_1\right)\mathrm{d}s_2 + \int_{b_1}^{b_2}\chi_{[0,b_2]}(s_2)\left(\int_0^{s_2}(s_2-s_1)\chi_{[0,b_1]}(s_1)\ \mathrm{d}s_1\right)\mathrm{d}s_2.$$ Now, in the first term, we have $s_2$ ranging from $0$ to $b_1$, thus $s_2\leq b_1$ in the inner integral and we can drop the characteristic function. On the second term, $s_2\geq b_1$ as it ranges from $b_1$ to $b_2$, thus we only integrate up to $b_1$. We then have the following: $$\int_0^{b_1}\chi_{[0,b_2]}(s_2) \left(\int_0^{s_2}s_2-s_1\ \mathrm{d}s_1\right)\mathrm{d}s_2 + \int_{b_1}^{b_2}\chi_{[0,b_2]}(s_2) \left(\int_0^{b_1}s_2-s_1\ \mathrm{d}s_1\right)\mathrm{d}s_2.$$ We can then evaluate from there on. I was wondering if the set-up up to this point is correct. Thanks!