Answer:
The area enclosed by given curves is [tex]\frac{4}{3}[/tex] square units.
Step-by-step explanation:
The given equations are
[tex]y=6x[/tex]
[tex]y=x^2+8[/tex]
Draw the graph of given equations.
From the below graph it is clear that both curves intersect each other at (2,12) and (4,24).
The area of region enclosed by curves is
[tex]A=\int_{2}^{4}6xdx-\int_{2}^{4}(x^2+8)dx[/tex]
[tex]A=6(\frac{1}{2}[x^2]_{2}^{4})-\frac{1}{3}[x^3]_{2}^{4}-8[x]_{2}^{4}[/tex]
[tex]A=3\left(4^2-2^2\right)-\frac{1}{3}\left(4^3-2^3\right)-8\left(4-2\right)[/tex]
[tex]A=3[12]-\frac{1}{3}[56]-8[2][/tex]
[tex]A=36-\frac{56}{3}-16[/tex]
[tex]A=\frac{4}{3}[/tex]
Therefore the area enclosed by given curves is [tex]\frac{4}{3}[/tex] square units.
