Solution:
Area enclosed between the parabola, 
Points of intersection of the given curves are 
![\begin{array}{l} \therefore \text { Area } O B A O=\text { Area } C D B A-(\text { Area } O D B O+\text { Area } O A C O) \\ =\int_{-2}^{1} \frac{1}{2}(3 x+12) d x-\int_{-2}^{1} \frac{3 x^{2}}{4} d x \\ =\frac{1}{2}\left[\frac{3 x^{2}}{2}+12 x\right]_{-2}^{4}-\frac{3}{4}\left[\frac{x^{3}}{3}\right]_{-2}^{4} \\ =\frac{1}{2}[24+48-6+24]-\frac{1}{4}[64+8] \\ =\frac{1}{2}[90]-\frac{1}{4}[72] \\ =45-18 \\ =27 \text { sq. units } \end{array}](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-94f9549343bf26c4e929babd262515ea_l3.png "Rendered by QuickLaTeX.com")
Find the area enclosed by the parabola
Solution:
Area enclosed between the parabola,
Points of intersection of the given curves are
