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I think the following example explains the fundamental theorem of calculus quite intuitively. Or more precisely, that's what I thought; now I'm starting to have some doubts.

Suppose $v(t)$ is the velocity of a car driving along the highway. The units for $t$ are in hours and the units for $v(t)$ are in miles per hour. Assume $v(t)$ is continuous and nonnegative. What is the displacement of the car over one hour (ie., $t \in [0,1]$)?

Well, if we subdivide $[0,1]$ into $n$ subintervals of equal length, in each subinterval $\left[\frac{k}{n}, \frac{k+1}{n}\right]$ the velocity doesn't change too much for large $n$ and hence can be approximated by $v(\frac{k}{n})$. Therefore, the displacement in $\left[\frac{k}{n}, \frac{k+1}{n}\right]$ is equal to $\frac{1}{n} v(\frac{k}{n}) + \epsilon(k, n)$ where $\epsilon(k, n)$ is a small error dependent on $k$ and $n$.

Hence $$ \text{Displacement} = x(1) - x(0) = \sum_{k=0}^{n-1} \frac{1}{n} v\left(\frac{k}{n}\right) + \sum_{k=0}^{n-1} \epsilon(k, n) $$

Note that the above equality holds for all $n$, since we have accounted for the error. If we assume that $ \sum_{k=0}^{n-1} \epsilon(k, n) \to 0$ as $n \to \infty$, then it's easy to see that $$x(1) - x(0) = \lim \sum_{k=0}^{n-1} \frac{1}{n} v\left(\frac{k}{n}\right) = \int_0^1 v(t) \ dt$$

However, it's not obviously clear to me why $\sum_{k=0}^{n-1} \epsilon(k, n) \to 0$ as $n \to \infty$. Why should this hold intuitively?

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  • $\begingroup$ I believe that comes through the mean value theorem ;) $\endgroup$ – Simply Beautiful Art Mar 11 '17 at 1:01
  • $\begingroup$ @SimplyBeautifulArt Fair enough, but that's more of a formal explanation. I am looking for intuition. $\endgroup$ – MathematicsStudent1122 Mar 11 '17 at 1:02
  • $\begingroup$ Well, better than $\epsilon (k,n)$ you can write $\frac{b-a}{n}\epsilon (k,n)$ for the error term in each sub-interval. Now, if we put a uniform bound of $\epsilon$ on all $\epsilon (k,n)$'s, which will exist for nice functions, then the sum of all these error terms will be bounded by $\epsilon.$ $\endgroup$ – Behnam Esmayli Mar 11 '17 at 1:20
  • $\begingroup$ I would hope the units of $v(t)$ are miles/hour, not miles. $\endgroup$ – eyeballfrog Mar 11 '17 at 1:45
  • $\begingroup$ @eyeballfrog Oops, that's right $\endgroup$ – MathematicsStudent1122 Mar 11 '17 at 2:05
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The function $v$ is uniformly continuous on $[0,1]$. Given any $\epsilon > 0$, there is an $N$ such that whenever $n \geq N$, we have, for all $k$ and all $x \in [k/n, (k+1)/n]$, the inequality $|v(x) - v(k/n)| \leq \epsilon$. Thus your error $\epsilon(k,n)$ is bounded in absolute value by $\epsilon/n$. Summing over $k$, the total error in the displacement is bounded in absolute value by $\epsilon$, so long as $n \geq N$.

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In broad strokes:

Suppose $F(x)$ is the cumulative area under the curve of $f(x)$ from $0$ to $x$ i.e. the green region below?

enter image description here

What is the derivative of $F(x)$? How fast is $F(x)$ changing as $x$ changes?

$F(x+h)-F(x)$ is the red region.

$F'(x) = f(x)$

Update:

The Riemann Sum.

We can partition the domain

$a=x_0<x_1<x_2<\cdots<x_i<\cdots<x_n =b$

$\int_a^b f(x) dx = \sum_{i=1}^{n} f(x_i^*)(x_i - x_{i-1})$

Where $x_i^*\in [x_{i-1},x_i]$

I say, that if the partition is fine enough, it does matter if we take the biggest possible value of $f(x_i^*)$ the smallest or something in between.

enter image description here

Suppose we choose $x_i^*$ to always produce the smallest $f(x_i)$. We would call this the lower sum, and we would get the green area.

And if we choose $x_i^*$ to produce the largest $f(x_i)$ we get the upper sum. The green + red.

And the difference is just the red boxes.

As the partition gets to be increasingly fine, the errors (the red boxes) only get smaller, and eventually their net area goes to 0.

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    $\begingroup$ Doesn't answer the question. $\endgroup$ – MathematicsStudent1122 Mar 11 '17 at 2:06
  • $\begingroup$ This is the intuitive explanation of the fundamental theorem of calculus. I suppose after re-reading your question, what you are really looking for is an explanation why the Riemann Integral converges. $\endgroup$ – Doug M Mar 11 '17 at 2:31

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