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If I took the definite integral of a velocity graph from 0 to 10 seconds, the answer would be the change in position over those 10 seconds correct? I am told by my teacher the area is change in displacement but that doesn't make sense.

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You’re right. You’ll get the change in position $\Delta x$, which is the displacement. Maybe your teacher is confused about “displacement” ($\Delta x$) vs. “position” ($x$).

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Both are correct as it is just a matter of wording although I would favour change in position and displacement most of the time.

Consider one dimensional motion along the x-axis with the unit vector $\hat x$ defining the positive x-direction.

A body starts at position $+3 \,\hat x$ with velocity $+6\,\hat x$ and after undergoing constant acceleration over a time interval of $3$ it has a velocity $-3\,\hat x$.

The area under the velocity time graph is $+6 +(-3) = +3$ and this is the change in position, so the new position is $+6\,\hat x$.

One could also say that the displacement of the particle during that time interval is $+3 \,\hat x$.

However going back to position $+3 \, \hat x$ that is a displacement of $+3 \,\hat x$ from the origin and position $+6 \,\hat x$ is a displacement of $+6 \,\hat x$ from the origin.

So the change in position is also the change in displacement if the displacements are measured from the origin.

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