AC capacitor computational problem using calculus
\$Q = CV\$ 

Hence \$I = C\dfrac{dv}{dt}\$ 

Rearranging,  \$V = \dfrac{1}{C}\int i~ dt\$

If i is a ramp, it has the form kt where k is a constant and t is time.

So, \$V = \dfrac{k}{C}t^2\$

Voltage rise as a square of time then falls as a square of time etc..