Answer:
The area of the region between the curves y=6x^2 and y=4x is 8/27
Step-by-step explanation:
Use the diagram to visualize the problem, the area colored of blue is the one that needs to be found, let's do it in 3 parts:
Part 1: Find the intersection points of the curves
To do this we put both equations in one and solve it for x:
[tex]6x^2=4x[/tex]
[tex]6x^2-4x=0\\2x(3x-2)=0[/tex]
This equation has 2 possible solutions:
x=0 and x=2/3, so the interval for integration is 0 <= x <= 2/3
Part 2: Find the area below each curve
[tex]A_{blue}=\int\limits^0_{2/3} {6x^2} \, dx \\A_{blue}=2x^3[/tex], evaluate in 0 and 2/3
[tex]A_{blue}=\frac{16}{27}[/tex]
[tex]A_{red}=\int\limits^0_{2/3} {4x} \, dx \\A_{red}=2x^2[/tex], evaluate in 0 and 2/3
[tex]A_{red}=\frac{8}{9}[/tex]
Part 3: Substract the area of the blue curve (y=6x^2) to the area of the red curve (y=4x)
[tex]Area=\frac{8}{9}-\frac{16}{27}\\Area=\frac{8}{27}[/tex]
