For this problem, we need to use our integral knowledge to set up the problem:
The bounds will be from 2 to 6, cutting tiny horizontal bars into the graph, to approximate the area
Now for the function inside the integral:
⇒ it is the top function minus the bottom function
⇒ function within integral = x² - 0
Let's put it all together and solve:
[tex]Area=\int\limits^6_2 {x^2-0} \, dx =\int\limits^6_2 {x^2} \, dx=\frac{6^3}{3} -\frac{2^3}{3} \\Area=\frac{6^3-2^3}{3} =\frac{216-8}{3} =\frac{208}{3}[/tex]
Answer: 208/3 ≈ 69.333
Hope that helps!
