\[dV = A_x dx\]
Using the scaing propert of the shape,
\[\frac{A_x}{A} = \frac{x^2}{h^2}\]
Hence
\[dV = A \frac{x^2}{h^2} dx\]
Integrating gives
\[V = \int^h_0 A \frac{x^2}{h^2} dx = A [ \frac{x^3}{3h^2}]^h_0 = A \frac{h^3}{3 h^2} = \frac{1}{3} A h \]