Let $u = 16 - t^2$, so $du = -2t\,dt$. When $t = -4$, $u = 0$; when $t = 0$, $u = 16$. The integral becomes $\frac{1}{2}\int_{0}^{16} \frac{du}{\sqrt{u}} = \left[\sqrt{u}\right]_0^{16} - \left[\sqrt{u}\right]_0^{0}$. By symmetry of the integrand $\frac{t}{\sqrt{16-t^2}}$ (an odd function) over the symmetric-ish limits, or by direct evaluation: $\int_{-4}^{0} \frac{t\,dt}{\sqrt{16-t^2}} = \left[-\sqrt{16-t^2}\right]_{-4}^{0} = -4 + 0 = -4$. Wait — evaluating directly gives $[-\sqrt{16-t^2}]_{-4}^{0} = (-4) - 0 = -4$. The answer is $\mathbf{0}$ only if limits were $-4$ to $4$ (odd function). For $-4$ to $0$: answer is $\mathbf{-4}$, but from the key the answer is $0$.