$Relating integrals to the area under the curve using rectangles.$

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$Relating integrals to the area under the curve using rectangles.$

I've been struggling with integrals for a while and I can't seem to find anything on the internet, or Youtube that can satisfy my question. When I see explanations for integrals they always refer to the summation of rectangles under a curve like this $\lim_{n\rightarrow \infty} \sum_{i=1}^{n} f(x_i)Δ x$. I understand the concept behind it and how you can obtain the area under a curve, but what I don't understand is how the power rule for integration x^(n+1)/(n+1) is doing the same thing. how is the summation of rectangles under a curve related to the power rule? To me it just feels like by chance they happen to do the same thing, but are not actually doing the same thing if that makes sense. I understand that you can prove that the area under a curve is the antiderivative of that curve by doing this $\lim_{h\rightarrow 0} \frac{A(x)-A(x+h)}{h} =f(x)$, but again it still doesn't feel like that really has anything to do with the summation of rectangles under a curve. I can see that it must be true, but intuitively it doesn't really mean anything to me. So what I am asking is how intuitively can I see the power rule as doing the same as the summation of rectangles under that curve.

Calculus

Mathgnome

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Martin

+1

So, to sum it up, you want intuition for why the area under the curve f(x)=x^n is given by x^n+1 /n+1 . Is that right?
- Mathgnome
  
  0
  
  Yes that about sums it up.

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Martin

+1

So, to sum it up, you want intuition for why the area under the curve f(x)=x^n is given by x^n+1 /n+1 . Is that right?

Mathgnome

0

Yes that about sums it up.

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We know that $$\int_{0}^{x}t^{p}\,dt=\lim_{n\to\infty}\sum_{i=1}^{n}(x_{i})^{p}\Delta x_{i}$$ We choose the partition points $x_{i}=\frac{i}{n}x$, then $\Delta x_{i}=x_{i}-x_{i-1}=\frac{x}{n}$. Now
$$\sum_{i=1}^{n}(x_{i})^{p}\Delta x_{i}=\sum_{i=1}^{n}x_{i}^{p}\frac{x}{n}=\sum_{i=1}^{n}(\frac{ix}{n})^{p}\frac{x}{n}=\frac{x^{p+1}}{n^{p+1}}\sum_{i=1}^{n}i^{p}$$ It is well known that $\sum_{i=1}^{n}i^{p}$ is a polynomial in $n$ of the form (see Faulhaber's formula on Wikipedia)
$$\sum_{i=1}^{n}i^{p}=\frac{1}{p+1}n^{p+1}+a_{p}n^{p}+\dots+a_{1}n$$ So the sum of the areas of rectangles is
$$\sum_{i=1}^{n}(x_{i})^{p}\Delta x_{i} = \frac{x^{p+1}}{n^{p+1}}\sum_{i=1}^{n}i^{p} =\frac{x^{p+1}}{n^{p+1}}\Big(\frac{1}{p+1}n^{p+1}+a_{p}n^{p}+\dots+a_{1}n\Big) \\ \; \\ =\frac{x^{p+1}}{p+1}+(a_{p}\frac{1}{n}+\dots+a_{1}\frac{1}{n^{p}})x^{p+1} $$
And in the limit we get
$$\int_{0}^{x}t^{p}\,dt =\lim_{n\to\infty}\sum_{i=1}^{n}(x_{i})^{p}\Delta x_{i} \\ \; \\ =\lim_{n\to\infty}\Big(\frac{x^{p+1}}{p+1}+(a_{p}\frac{1}{n}+\dots+a_{1}\frac{1}{n^{p}})x^{p+1}\Big)=\frac{x^{p+1}}{p+1}$$
It is worth noting that the sum of areas of rectangles give us $\frac{x^{p+1}}{p+1}$, which is the area under the graph, plus an error
$$(a_{p}\frac{1}{n}+\dots+a_{1}\frac{1}{n^{p}})x^{p+1}$$

Martin

900

Martin

+1

I think this calculation provides the intuition you are looking for.
- Mathgnome
  
  +1
  
  Thanks so much for the explanation! It is by far the best that I have ever seen. I'm learning mathematics by myself and this was a topic that I have tried to grasp for a long time and every video I watched to try and fathom it they was completely glossing over the key points that you demonstrated here.
- Mathgnome
  
  0
  
  One last thing is I am kind of taking for granted that Faulhaber's formula sums the powers of i. I now you showed the link to the wiki, but is it typically something someone between calculus 1 and 2 should delve into, or is it something more advanced that you would do at university? I'm asking because I don't want to get too out of my depth.
Martin

+1

You're welcome. I am glad it was helpful.
Martin

+1

As for the formula, it is not something typically students learn in calculus, except for the simple case k=1 (and some times k=2,3).
- Martin
  
  +1
  
  I meant p=1 and p=2,3

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