Basic Circuit Question: Understanding the area under the v(t) curve?
No, it indicates the definite integral of v(t) over the interval the graph covers.
And if you connected that to a fixed resistance R it still does not indicate energy because instantaneous power is \$v^2(t)/R\$.
With a fixed resistor it would tell you the coulombs of charge that passed through the resistor though, since i(t) = v(t)/R it would be the area divided by R.
Edit: As far as charge, think of the charge flow at any given instant being the current (coulombs per second). So if the current is 1A and it flows for 1 second, we have one coulomb of charge. If your graph is 1 second long then the area under the i(t) curve (which is a straight line) is the charge. And the area under the v(t) curve looks the same, just multiplied by the load resistance R.
Analogous to calculating the total distance you've moved by integrating your speed over some period of time (or looking at the area under the speed curve over a certain period of time)
Now if i(t) or v(t) is not constant, but varies over time then we can still use the area under the curve because it's a linear function. Mathematically, the area under the i(t) curve is the charge Q:
Q = \$\int_{t=0}^{T_0}i(t)dt = \frac 1 R \int_{t=0}^{T_0}v(t)dt \$