The area (in sq. unit) bounded by the curve $$4y^2 = x^2(4-x)(x-2)$$ is equal to

The curve is $$4y^2 = x^2(4-x)(x-2)$$, which requires $$(4-x)(x-2) \ge 0$$, so $$2 \le x \le 4$$. Taking the positive square root, $$y = \frac{x}{2}\sqrt{(4-x)(x-2)}$$, and by symmetry the total area is $$A = 2\int_2^4 \frac{x}{2}\sqrt{(4-x)(x-2)}\,dx = \int_2^4 x\sqrt{(4-x)(x-2)}\,dx$$.

Substituting $$x = 3 + t$$ (shifting the midpoint of $$[2,4]$$ to the origin), so $$dx = dt$$, with limits $$t = -1$$ to $$t = 1$$: $$(4-x)(x-2) = (1-t)(1+t) = 1 - t^2$$. The integral becomes $$\int_{-1}^{1}(3+t)\sqrt{1-t^2}\,dt = 3\int_{-1}^{1}\sqrt{1-t^2}\,dt + \int_{-1}^{1}t\sqrt{1-t^2}\,dt$$.

The second integral vanishes by odd symmetry. The first integral is $$3 \cdot \frac{\pi}{2}$$ (the area of a semicircle of radius 1). Therefore $$A = \frac{3\pi}{2}$$.