I don't know how to draw here but I can describe the graph.
Because of the -x^2, the parabola opens downward. Its axis of symmetry is x = 0 por the y-axis. Its vertex is at (0,h). It intersects the x-axis at (-b/2,0) and (b/2,0)
The area in question:
dA = y*dx
Boundaries for dx are from x = -b/2 to x = b/2
A = INT.(-b/2 --> b/2)[h -(4h/(b^2))(x^2)]dx
A = [h*x -(1/3)(4h /(b^2))(x^3)] | (-b/2 --> b/2)
A = [h(b/2) -(1/3)(4h /b^2)(b^3 / 8)] -[h(-b/2) -(1/3)(4h /b^2)(-b^3 / 8)]
A = bh/2 -(1/3)(bh/2) +bh/2 -(1/3)(bh/2)
A = bh -(1/3)bh
A = (2/3)bh --------------answer.