Answer:
[tex]A=4.5[/tex]
Step-by-step explanation:
The area between two curves over the interval [tex][a,b][/tex] is [tex]A=\int\limits^b_a {f(x)-g(x)} \, dx[/tex] where [tex]f(x)[/tex] is the upper function and [tex]g(x)[/tex] is the lower function.
Therefore:
[tex]A=\int\limits^3_0 {3x-x^2} \, dx[/tex]
[tex]A=\frac{3}{2}x^2-\frac{x^3}{3}\Big|_0^3[/tex]
[tex]A=[\frac{3}{2}(3)^2-\frac{(3)^3}{3}]-[\frac{3}{2}(0)^2-\frac{(0)^3}{3}][/tex]
[tex]A=\frac{3}{2}(9)-\frac{27}{3}[/tex]
[tex]A=\frac{27}{2}-\frac{27}{3}[/tex]
[tex]A=\frac{9}{2}[/tex]
[tex]A=4.5[/tex]
So, the area between [tex]f(x)=3x[/tex] and [tex]g(x)=x^2[/tex] on the interval [tex](0,3][/tex] is [tex]4.5[/tex] square units.
