Area under the displacement-time graph giving time?

This is probably very stupid, but today at a lecture, our professor solved a problem where we had to find the time taken to travel from 0-5m, where $v = 3/x$ (velocity is a function of position.)

Then he integrated, $$\int_0^5 x dx = \int_0^t 3 dt$$

The integral of x represents the area under the position time curve which is giving the time taken. How is this possible? Where am I going wrong here?

• "The integral of x represents the area under the position time curve" This is wrong. – lucas Jun 11 '16 at 12:27
• See also this answer for how to handle non uniform acceleration under various scenarios. You will see that $$t = \int \frac{1}{v(x)}\,{\rm d}x$$ – ja72 Jun 11 '16 at 13:39

The displacement is equal to the area under a velocity time graph so in your example the change in displacement $dx$ in a time $dt$ is equal to $v\; dt$ with $v=\frac 3 x$
So $dx = v\; dt = \frac 3 x \; dt \Rightarrow x\; dx= 3\; dt$ and then you do the integration.