​$$\text{The area under the velocity-time curve between times } t_1 \text{ and } t_2 \text{ is equal to the } \_\_\_ \text{ of the object during that interval of time.}$$​​

Correct option is D

​$$\bullet \text{The } \textbf{area under a velocity-time (v-t) graph} \text{ represents the } \textbf{displacement} \text{ of an object.}\\\bullet \text{Displacement is calculated as:}\\\text{Displacement} = \int_{t_1}^{t_2} v \, dt\\\bullet \text{If velocity is always positive, the area gives the total displacement.}\\\bullet \text{If velocity changes direction, the } \textbf{net displacement} \text{ (difference between forward and backward motion) is obtained.}$$​​