Question

Which is the area between the x-axis and y=x^2 from x=2 to x=6

Answer 1

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

• top function: y= x²
• bottom function: x-axis ⇒ y =0

⇒ function within integral = x² - 0

Let's put it all together and solve:

`Area=\int\limits^6_2 {x^2-0} \, dx =\int\limits^6_2 {x^2} \, dx=(6^3)/(3) -(2^3)/(3) \\Area=(6^3-2^3)/(3) =(216-8)/(3) =(208)/(3)`

Answer: 208/3 ≈ 69.333

Hope that helps!