Go with
$$f(x) = \sin(2x).$$
Give a clean formula for the function h[x] defined by
$$h(x) = \int_0^x f(s) ds$$
How are $f(x)$ and $\frac{d}{dx}h(x)$ related?
Is the outcome an accident? Why or why not?
My answer:
$$h(x) = (2\cos(2x)) - 1$$
$$h'(x) = -4\sin(2x)$$
When they say $f(s)$ do they mean to use integration by substitution? Are $h'(x)$ and $f(x)$ supposed to both equal $\sin(2x)$?
Your mistake is
$$h(x)= \int_0^x f(s) \, ds= \int_0^x \sin(2s) \, ds$$
Hence $$h(x) = -\frac{\cos(2x)}{2}+\frac12$$