let:
$$F(x) = \int_0^x\sqrt{t}\sin(t)dt$$
Say whether $F(x)$ is positive, negative or zero at each of the following points, and give a reason in each case:
$$x = \pi$$ $$x = 2\pi$$
I was trying:
$$F'(x) = \sqrt{x}\sin(x)$$ $$F''(x) = \sqrt{x}\cos(x)+\sin(x)\frac{1}{2\sqrt{x}}$$
If we evaluate the second derivative at $\pi$ we see that we are at a local maximum is that enough reason to say that when we evaluate or function at $\pi$ is positive?
Since derivatives (and second derivatives) only give you the rates of change, we cannot usually directly determine very much about the function values at specific points (apart from certain special cases, such as the function value starting at $0$ and the derivative then always being positive meaning the value will always be positive after the starting point). Instead, here are a couple of alternate methods to solve your problem.
For $x = \pi$, since for $0 \lt t \lt \pi$ we have $\sqrt{t} \gt 0$ and $\sin(t) \gt 0$, then $F(\pi) \gt 0$ (note that, for this case, your result of $F'(x) = \sqrt{x}\sin(x)$ means that $F'(x) \gt 0$ for $0 \lt x \lt \pi$, so with $F(0) = 0$, this also shows that $F(\pi) \gt 0$).
With $x = 2\pi$, the second term in the third line below uses the substitution $y = 2\pi - t \;\;\to\;\; dy = -dt$, and then that $\sin(2\pi - y) = -\sin(y)$, to get that
$$\begin{equation}\begin{aligned} F(2\pi) & = \int_{0}^{2\pi}\sqrt{t}\sin(t)dt \\ & = \int_{0}^{\pi}\sqrt{t}\sin(t)dt + \int_{\pi}^{2\pi}\sqrt{t}\sin(t)dt \\ & = \int_{0}^{\pi}\sqrt{t}\sin(t)dt + \int_{\pi}^{0}\sqrt{2\pi - y}\sin(2\pi - y)(-dy) \\ & = \int_{0}^{\pi}\sqrt{t}\sin(t)dt - \int_{0}^{\pi}\sqrt{2\pi - y}\sin(y)dy \\ & = \int_{0}^{\pi}\sqrt{z}\sin(z)dz - \int_{0}^{\pi}\sqrt{2\pi - z}\sin(z)dz \\ & = \int_{0}^{\pi}(\sqrt{z} - \sqrt{2\pi - z})\sin(z)dz \end{aligned}\end{equation}$$
For $0 \lt z \lt \pi$, we get $z \lt 2\pi - z \;\;\to\;\; \sqrt{z} - \sqrt{2\pi - z} \lt 0$, with $\sin(z) \gt 0$ as mentioned earlier, so $F(2\pi) \lt 0$.