- 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.

- chestercat

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- anonymous

https://gyazo.com/d4d434542599aa197b53e5f947deb365

- anonymous

- zzr0ck3r

so where will out summation start?

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## More answers

- zzr0ck3r

remember we go to the right of the first interval

- zzr0ck3r

the intervals here are -1,0
0,1
1,2

- anonymous

Would we not need to divide into four intervals? That's what I got from you from the other posts.

- anonymous

Three rectangles = three intervals

- anonymous

I feel stupid....thanks @SithsAndGiggles

- anonymous

So can you also explain the right or left endpoints? My AP Calc teacher last year never taught us anything like this.

- 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

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

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

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

the value * the width

- 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

And the height of the rectangle would also be 1?

- 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

Ok, one last thing, the sigma starts with 1 because the width is 1 correct?

- anonymous

So if correct it would be C because of 3 intervals starting from 1?

- 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 \(x_i=i-1\).)