"Under the length L"? Do you mean that the overall length of the figure is L and the height is h?

That figure can best be approximated by a rectangle with a half circle on

each end. In order that the half circle fit on the end of the rectangle its radius must be h/2. (In your picture there appears to be a small oddly region

between the rectangle and half circle that I am ignoring. That's why this is an approximation.) The area of the two half-circles together is [itex]\pi r^2= \pi h^2/4[/itex] and the area of the rectangle is hL. The area of the entire figure is approximately [itex]hL+ \pi h^2/4[/itex].