Question

Find the area enclosed by the parabolas y = 4x – x² and y = x² – x.

Answer 1

To area enclosed by

Y = 4x – x² (1)

4x – x² = x² – x

2x² – 5x = 0

x = 0 or x = \(\frac{5}{2}\)

y = 0 or y = \(\frac{15}{4}\)

And y = x² – x

\(\left(y+\frac{1}{4} \right)+\left(x-\frac{1}{2} \right)^2 ...(2)\)

Equation (1) represents a parabola downward with vertex at (2, 4) and meets axes at (4, 0), (0, 0).

Equation (2) represents a parabola upward whose vertex is \(\left(\frac{1}{2},\frac{1}{4} \right)\) and meets axes at Q(1, 0), (0, 0).Points of intersection of parabolas are O (0, 0) and A \(\left(\frac{5}{2},\frac{15}{4} \right)\).

These are shown in the graph below :

Required area = Region OQAP

The area enclosed by the parabolas y = 4x – x² and y = x² – x is \(\frac{125}{24}\) sq. units.