Answer
$-3$
Work Step by Step
Suppose that $G$ is an antiderivative of $g$ so:
$$G'(t)=g(t)$$
Usin the $FTC$ it follows:
$$\int_{9}^{16}g(t)dt=G(16)-G(9)$$
$$\int_{9}^{16}G'(t)dt=G(16)-G(9)$$
$$\int_{9}^{16}t^{-\frac{1}{2}}dt=G(16)-G(9)$$
$$[2t^{\frac{1}{2}}]_{9}^{16}=G(16)-G(9)$$
$$2\cdot 16^{\frac{1}{2}}-2\cdot 9^{\frac{1}{2}}=G(16)-G(9)$$
$$2\cdot 16^{\frac{1}{2}}-2\cdot 9^{\frac{1}{2}}=G(16)-(-5)$$
$$2\cdot 4-2\cdot 3=G(16)+5$$
$$2\cdot 4-2\cdot 3-5=G(16)$$
$$8-6-5=G(16)$$
$$-3=G(16)$$
