What does the area under a velocity-time graph indicate

If we consider a velocity-time graph, that means we consider movement in only one direction (there may be more, but in the graph we only consider one), and thus if the particle moves in the same direction all the time, the displacement matches the distance. So, for the case where the velocity is positive (i.e. moving in the same direction), both are correct.

To answer the general case, consider a graph like a sine curve representing the velocity of a particle and integrate over a full period. The integral becomes zero, which matches the displacement (the particle has gone back to its original position).

If you want to consider this more rigurously, let $x(t)$ be the position of the particle at time $t$ and let $v = \frac{dx}{dt}$ be its velocity. Then, the area under the curve of the graph of $v$ over a time interval $[a,b]$ is (note if $v$ is negative we add up negative areas):

$$ \int_a^b v(t) dt = x(b)-x(a) = \Delta x$$

Which is indeed the displacement (this matches the result before).

This is not the full answer, though. If you were to consider the negative areas as positive, then you would slice the interval $[a,b]$ into intervals where $v$ has a constant sign and add up the absolute value of each integral (the areas). But we are back to the first paragraph! This would be adding up the motion of the particle over periods where it keeps a constant direction of motion, and this means you are calculating the distance.

So it all depends on what you define as "area under the curve". Is it the integral itself (displacement) or the sliced up version (distnace). That is probably the reason for the discrepancy of your sources.

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Velocity-time graphs are used to describe the motion of objects which are moving in a straight line. They can be used to show acceleration and to work out displacement.