So far as I have learned,

one has to integrate the curve f(x) function only in order to find the area under the curve.

But I Quote as follows:

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The volume $\displaystyle V$ of the liquid in a container leaks out at the rate of $\displaystyle 30t$ $\displaystyle cm^3s^_-1$ where $\displaystyle t$ is the time in seconds.

Find the amount of liquid lost in the third second.

*Solution ...*
$\displaystyle \frac{\mathrm{d}V}{\mathrm{d} x}=-30t $ (decreasing)

The third second is between $\displaystyle t=2$ and $\displaystyle t=3$.

So we find (value of $\displaystyle V$ when $\displaystyle t=3$) $\displaystyle -$ (value of $\displaystyle V$ when $\displaystyle t=2$ using a definite integral.

Change of volume = $\displaystyle \int_{2}^{3}\frac{\mathrm{d}V}{\mathrm{d} t} dt=\int_{2}^{3}(-30t) dt= \left[-15t^2\right]_2^3=-135-(-60)=-75$

Hence $\displaystyle 75$$\displaystyle cm^3 $ of liquid was lost in the third second.

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According to me, one should integrate the gradient function back to the curve function and then integrate again as usual to find area.

So,

Equation of Curve = $\displaystyle \int (-30t)\hspace{1mm}\textup{dt}$

$\displaystyle V=-15t^2+c$

Arbitrary constant C may be ignored as t=0 when V=0 for physical variables.

$\displaystyle V=-15t^2$

Therefore,

Change of Volume between $\displaystyle t=2$ and $\displaystyle t=3$

$\displaystyle \int_{2}^{3}(-15t^2)= \left[-5t^3\right]_2^3=(-135)-(-40)=-95$

The volume lost must be 95$\displaystyle cm^3$, How far do you agree?