Question

Find the area of the region included between the parabola y = ${\displaystyle \frac{3x^2}{4}}$ and the line 3x – 2y + 12 = 0.

Answer 1

Solving the equations of the given curves y = ${\displaystyle \frac{3x^2}{4}}$ and 3x – 2y + 12 = 0

We get 3x² – 6x – 24 = 0

⇒ (x – 4)(x + 2) = 0

⇒ x = 4, x = –2

Which give y = 12, y = 3

From Fig.8.6, the required area = area of ABC

= ${\displaystyle {\int }_{-2}^4\left(\frac{12+3x}{2}\right)\text{d}x-{\int }_{-2}^4\frac{3x^2}{4} \text{d}x}$

= ${\displaystyle {(6x+ \frac{3x^2}{4})}_{-2}^4-{\left|\frac{3x^3}{12}\right|}_{-2}^4}$

= 27 sq.units