y = 4x – x²

This is a parabola with negative co-efficient of x², i.e., it’s a downward parabola.

So, area A enclosed is the area of the peak of the parabola above the x – axis.

We need to find the bounds of this peak.

Now, at the point where the peak starts/ends, y = 0,

i.e., 4x – x² = 0

⇒ x(4 – x) = 0

⇒ x = 0 or x = 4

∴ A =

=

= [32 – 64/3]

= 32/3 sq. units (Ans)