## anonymous one year ago Estimate the area under the curve f(x) = x^2 + 1 from x = 0 to x = 6 by using three circumscribed (over the curve) rectangles. Answer to the nearest integer.

1. anonymous

So the interval would be from [0,2] [2,4] [4,6]?

2. anonymous

yes

3. anonymous

do you know which endpoints you're using?

4. anonymous

Would the endpoints not be 1 to 6?

5. anonymous

you only have three intervals. I meant endpoints in terms of right or left of the interval

6. anonymous

can you draw the rectangles? |dw:1441160832388:dw|

7. anonymous

|dw:1441160946061:dw| Something like this?

8. anonymous

yes. so the width of all the rectangles will be 2. The heights will the y-values at 2, 4, and 6. Make sense?

9. anonymous

Ok, so how would this help me setup my integral?

10. anonymous

This is the integral$\int\limits_{0}^{6}(x^2+1) dx$ But that's not what you're being asked to find/evaluate. You're being asked for an estimate of the area, and you use the rectangle you drew to find it. For example, the area of the first rectangle is 2f(2) because 2 is its width and f(2) is its height. So you really just need to find the areas of the 3 rectangles and add them up

11. anonymous

So (4+8+12)=24?

12. anonymous

$A=2f(2)+2f(4)+2f(6)$ $A=2[f(2)+f(4)+f(6)]$ You have to plug in 2, 4, and 8 into $$f(x)=x^2+1$$ to get f(2), f(4), and f(6) |dw:1441161661208:dw|

13. anonymous

Why 8?

14. anonymous

sorry, typo. it's supposed to be 6

15. anonymous

16. anonymous

yes

17. anonymous

Appreciate it.

18. anonymous

no problem