I was way overthinking how to calculate calories that you would do performing a pullup, which is a straightforward mgh. But with a sittup, you have some non-uniform torso that's being lifted to various different heights.
So I'm imagining simplifying a torso into a really thin rod. How would I setup the equation given the mass of torso $ m(x) $ and height $h$, and lifting from 0 degrees to d degrees?
My attempt $$ W = g\int_{h= 0}^{h= cos(d)} (m* dh/h)hdh$$
Regardless of the torso's shape, it has a center of gravity. When the person is lying flat, its center of gravity is at some height $h_0$. When the person comes to a stop at the top of the sit-up, the center of gravity is at some height $h_u$. The work done is $mg(h_u - h_0)$.
If you stick with your thin rod model, with the axis at one end, then the center of gravity is halfway through the length $\ell$ (since you didn't specify a value for it in your question). So the distance from axis to CG is $\frac \ell 2$. The height to which the CG raises is $h_u - h_0 = \frac \ell 2\sin d$, and so the work done is $$W = \frac {mg \ell \sin d}2$$ where $m$ is the mass of the upper body only.
If you want a more accurate model, then you can find online with a little searching human modeling information that will tell you the distance of the CG of the torso from the hip axis for various positions and body sizes. This will give you a more accurate calculation.