The area under velocity-time graph gives:
A. Acceleration
B. Distance
C. Displacement
D. Velocity

Here, y is usually a function of x. Hence, the area under a curve can be found from any point to another point. Similarly, in case of the velocity-time graph, the area under the curve is given as, $\vec A_{vt}=\int\limits_{t_1}^{t_2}\vec v.dt $
Here, the area $\vec A_{vt}$ is given in vector form because the velocity is a vector quantity.
Now, we know one thing : $\vec v=\dfrac{d\vec r}{dt}$ . Here, $\vec r$ is the displacement from a particular point (say the origin). So, $\vec r$ is not related to how a body reached a position, but it is just the linear distance of the position from the starting point along with a direction. Now, putting the expression of $\vec v$ in the above equation, we get,
$\vec A_{vt}=\int\limits_{t_1}^{t_2}\vec v.dt=\int\limits_{t_1}^{t_2}\dfrac{d\vec r}{dt}.dt \\ =\int\limits_{\vec r_1}^{\vec r_2}d\vec r=\vec r_2-\vec r_1 $
Hence, the required area is basically $\vec r_2-\vec r_1$. In the language of science, it is called displacement i.e. the “ultimate distance” that has been travelled in the given time. So, option C is the correct answer.

Additional information:
The slope of the velocity-time graph is $\dfrac{d\vec v}{dt}$ and it is called acceleration. In a similar way the slope of the displacement-time graph i.e. $\dfrac{\vec r}{dt}$ is called the velocity.

Note: Remember the following points:
1. Distance is the length of ground that has been travelled. But, displacement is the ultimate separation between two positions. Distance is scalar but displacement is vector.
2. The area $A_{vt}$ is not an area in true sense. It’s just the area under a particular curve.