anonymous
  • anonymous
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.
Calculus1
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
So the interval would be from [0,2] [2,4] [4,6]?
anonymous
  • anonymous
yes
anonymous
  • anonymous
do you know which endpoints you're using?

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anonymous
  • anonymous
Would the endpoints not be 1 to 6?
anonymous
  • anonymous
you only have three intervals. I meant endpoints in terms of right or left of the interval
anonymous
  • anonymous
can you draw the rectangles? |dw:1441160832388:dw|
anonymous
  • anonymous
|dw:1441160946061:dw| Something like this?
anonymous
  • 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?
anonymous
  • anonymous
Ok, so how would this help me setup my integral?
anonymous
  • 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
anonymous
  • anonymous
So (4+8+12)=24?
anonymous
  • 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|
anonymous
  • anonymous
Why 8?
anonymous
  • anonymous
sorry, typo. it's supposed to be 6
anonymous
  • anonymous
118, final answer?
anonymous
  • anonymous
yes
anonymous
  • anonymous
Appreciate it.
anonymous
  • anonymous
no problem

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