Question

Find the area enclosed between 3y = x², X-axis and x = 2 to x = 3.

Answer 1

3y = x²

∴ y = ${\displaystyle \frac{x^2}{3}}$

Required area = ${\displaystyle {\int }_2^{3}y dx}$

= ${\displaystyle {\int }_2^3\frac{x^2}{3}dx}$

= ${\displaystyle {\left[\frac{x^3}{9}\right]}_2^3}$

= ${\displaystyle 3-\frac{8}{9}}$

= ${\displaystyle \frac{27-8}{9}}$

= ${\displaystyle \frac{19}{9}}$ sq.units