Find the area bounded by y = 3x^2 + 1, x = 0, x = 2, y = 0
The Attempt at a Solution
Not sure how to do this? Is this like finding the upper and lower sums?
?? What happened to " BuBbLeS01". I don't know what you mean by "without the y= 0"! Without some lower boundary you don't have a region to find the area of. The upper boundary is given y= 3x2+ 1 and the lower boundary by y= 0 so a "thin rectangle" would have length 3x2+ 1- 0= x2+ 1, and width dx. That is what you integrate (from 0 to 2 not from 2 to 0).I think the Area is 10 Square Units with or without the y=0.
I don't know how to integrate 0.
1 bounded area +ve
A= 2[int]0 ydx
A=2[int]0 3x^2 +1 dx
A= [x^3 + x]2/0
Problem with this is it covers the -negative area aswell when the y=0 should stop it.
Can't find anything in my notes.Can't integrate zero.