anonymous
  • anonymous
Write the summation to estimate the area under the curve y = 1 + x^2 from x = –1 to x = 2 using 3 rectangles and right endpoints.
Calculus1
chestercat
  • chestercat
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anonymous
  • anonymous
https://gyazo.com/d4d434542599aa197b53e5f947deb365
anonymous
  • anonymous
zzr0ck3r
  • zzr0ck3r
so where will out summation start?

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zzr0ck3r
  • zzr0ck3r
remember we go to the right of the first interval
zzr0ck3r
  • zzr0ck3r
the intervals here are -1,0 0,1 1,2
anonymous
  • anonymous
Would we not need to divide into four intervals? That's what I got from you from the other posts.
anonymous
  • anonymous
Three rectangles = three intervals
anonymous
  • anonymous
I feel stupid....thanks @SithsAndGiggles
anonymous
  • anonymous
So can you also explain the right or left endpoints? My AP Calc teacher last year never taught us anything like this.
anonymous
  • anonymous
Sure. I found I learned it best with a visual representation (which I'll be doing in parts for color-coding purposes) |dw:1441158115140:dw|
anonymous
  • anonymous
So the point here is to find the area under the curve, but suppose we don't know the various rules for antiderivatives. Instead, we can approximate the area using rectangles (among other shapes), like so: |dw:1441158222375:dw|
anonymous
  • anonymous
The area under the curve is then *approximately* equal to the total area under the curve. The "left/right endpoint" approximations refer to those that use rectangles that are determined by their left/right endpoints. For your question, you're looking at the right endpoints: |dw:1441158349598:dw|
anonymous
  • anonymous
Consequently, the area of each rectangle is equivalent to the value of the function \(x^2+1\) at these endpoints, i.e. \(x=0,1,2\). The width of each rectangle is fixed at \(1\). So, \[\int_{-1}^2(x^2+1)\,dx\approx \sum (\text{areas of rectangles})\]
zzr0ck3r
  • zzr0ck3r
the value * the width
anonymous
  • anonymous
Specifically, \[\int_{-1}^2(x^2+1)\,dx\approx \sum_{i=1}^n (\text{height of rectangle }i)(\text{width of rectangle }i)\] Like I mentioned before, the width of each rectangle is fixed at \(1\), so \[\int_{-1}^2(x^2+1)\,dx\approx \sum_{i=1}^n (\text{height of rectangle }i)\]
anonymous
  • anonymous
And the height of the rectangle would also be 1?
anonymous
  • anonymous
That depends on the rectangle you're referring to! For the first rectangle, the right endpoint occurs at \(x=0\). The height is given by the function value at that point, and \(0^2+1=1\), so yes, but only for this rectangle.
anonymous
  • anonymous
Ok, one last thing, the sigma starts with 1 because the width is 1 correct?
anonymous
  • anonymous
So if correct it would be C because of 3 intervals starting from 1?
anonymous
  • anonymous
Not quite. The numbers denoted by \(i\) in sigma notation are called the indices. The way I wrote it, \(i=1\) means that we're talking about the first object that belongs to some set. In this case, \(i=1\) means that we're talking about rectangle \(1\), \(i=2\) refers to the next one, and so on: |dw:1441159080551:dw| This is the info we're using: \[\begin{array}{c|c} i&\text{right endpoint}\\ \hline 1&x_1=0\\ 2&x_2=1\\ 3&x_3=2 \end{array}\] So in fact, the area is approximately \[\begin{align*} \sum_{i=1}^3 ({x_i}^2+1)&=({x_1}^2+1)+({x_2}^2+1)+({x_3}^2+1)\\ &=(0^2+1)+(1^2+1)+(2^2+1) \end{align*}\] But this is not one of the answer choices (not exactly, anyway). Why? It's because \(x_i\neq i\). However, the above sum IS equal to \[\begin{align*} \sum_{i=0}^2 (i^2+1)&=(0^2+1)+(1^2+1)+(2^2+1) \end{align*}\] (As it turns out, the way I wrote it, I ended up using \(x_i=i-1\).)