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I'm trying to calculate the area under the curve of $ y=x^2 $ between $ x=1 $ and $x = 3$ and above $y=0$ using the sum of infinitely many rectangles.

So far I've tackled it by first defining the width of every rectangle to be $\Delta x = \frac{3-1}{n} = \frac{2}{n}$.

After that I proceeded by stating that the area of every rectangle under the curve combined is

$\sum_{i=1}^n (\frac{2i}{n})^2*(\frac{2}{n})$

I'm told that $\sum_{i=1}^n i^2 = \frac{n(n+1)(2n+1)}{6}$ which I then used in my calculations:

$\sum_{i=1}^n (\frac{2}{n})^2*i^2*(\frac{2}{n}) = \frac{8}{n^3}*(\frac{n(n+1)(2n+1)}{6})$

Simplifying to:

$8 * \frac{(n+1)(2n+1)}{6n^2} = \frac{8n^2+12n+4}{3n^2}$

Proceeding by calculating the limit as $x$ approaches infinity:

$\lim _{x\to \infty }\left(\frac{\left(8n^2+12n+4\right)}{3n^2}\right) = \frac{8}{3}$.

However, the answer should be $\frac{26}{3}$ which is indeed very close to what I have, but I have no idea where the problem lies. Any ideas?

SmhConfused
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    You forgot to use $x=1$. We should have $f(x_i)= (1+\frac{2i}{n})^2$ – Ahmed S. Attaalla Feb 22 '17 at 18:42
  • @AhmedS.Attaalla Why do we add $1$ ? Isn't the function just $f(x) = x^2$ ? What am I missing? – SmhConfused Feb 22 '17 at 18:47
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    To be precise, you are not adding the area of infinitely many rectangles. You are adding the area of a finite number of rectangles and letting the number of rectangles grow arbitrarily large. – Matthew Leingang Feb 22 '17 at 18:49
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    The points ${x_0,x_1,\dots,x_n}$ need to be a partition of $[1,3]$. So $x_0 = 1$, $x_1 = 1 + \Delta x$, $x_2 = 1 + 2 \Delta x$, et cetera, up to $x_n = 1 + n \Delta x = 3$. You apply $f$ to those values. – Matthew Leingang Feb 22 '17 at 18:50
  • @MatthewLeingang Thank you! – SmhConfused Feb 22 '17 at 18:56
  • Why not just integrate? The proof that reversing differentiation is equivalent to finding the area (i.e. FTC) uses the rectangles you talk about. –  Feb 22 '17 at 19:25
  • @selfawareuser I know. That is what I would normally do but it was specified in this task that I was to use this method. – SmhConfused Feb 22 '17 at 19:41

2 Answers2

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The x values you are using in your first sum are incorrect. Let i equal 1 for example to see this.

PM.
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Following up on my comment, you need $$ R_n = \sum_{i=1}^n \left(1 + \frac{2i}{n}\right)^2 \cdot \frac{2}{n} $$ Expand the square and collect the sums of powers: \begin{align*} R_n &= \sum_{i=1}^n \left(1 + \frac{4i}{n} + \frac{4i^2}{n^2}\right)\cdot \frac{2}{n} \\ &= \frac{2}{n} \sum_{i=1}^n 1 + \frac{8}{n^2} \sum_{i=1}^n i + \frac{8}{n^3} \sum_{i=1}^n i^2 \\ &= \frac{2}{n} \cdot n + \frac{8}{n^2} \cdot \frac{n(n+1)}{2} + \frac{8}{n^3} \frac{n(n+1)(2n+1)}{6} \end{align*} As $n\to \infty$, this tends to $2 + 4 + \frac{8}{3} = \frac{26}{3}$.

Matthew Leingang
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