Johann Bernoulli published (something like) the following expression in his journal Acta Eruditorum.
$\int_0^x f(t) dt = xf(x)-\frac{x^2}{2!}f'(x)+\frac{x^3}{3!}f''(x)-\frac{x^4}{4!}f'''(x)+...$
Is there a geometric interpretation of this identity? That is can each term on the right side be interpreted as an area of sorts and all of them taken together as being the area under the curve of $f(t)$ from $0$ to $x$?
As Joriki commented, this is just repeated integration by parts. Let us start considering $\int f(t)~dt$ and use $u=f(t)$ and $v'=dt$; these lead to $u'=f'(t)~dt$,$v=t$. So $$\int f(t)~dt= tf(t)-\int t f'(t)~dt$$ Now, for the second integral, integration by parts again with $u=f'(t)$,$v'=t~dt$; these lead to $u'=f''(t)~dt$,$v=\frac{t^2}{2}$. So,$$\int t f'(t)~dt=\frac{t^2}{2}f'(t)-\frac{1}{2}\int t^2 f''(t)~dt$$ Let us repeat once more with $u=f''(t)$,$v'=t^2~dt$; these lead to $u'=f'''(t)~dt$,$v=\frac{t^3}{3}$. So,$$\int t^2 f'(t)~dt=\frac{t^3}{3}f''(t)-\frac{1}{3}\int t^3 f'''(t)~dt$$ and so on.
So, the antiderivative becomes $$\int f(t) dt = tf(t)-\frac{t^2}{2!}f'(t)+\frac{t^3}{3!}f''(t)-\frac{t^4}{4!}f'''(t)+...$$ and then the value of the integral between bounds $0$ and $x$ you wrote after Bernoulli.
Concerning a "geometric" interpretation, since each integral corresponds to the area between a curve and the $x$ axis, the final formula gives a sum of areas (again, as Joriki commented)