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In high school calculus, I was first taught that the area under a curve $f(x)$ between $x=a$ and $x=b$ is given by:

$$ A = \lim_{\delta x \rightarrow 0} \sum \limits_{a}^{b} f(x) \delta x $$

Then this limit was defined as being the definite integral of $f(x)$ from $a$ to $b$, and I was told to take it on trust that this was equal to $F(b) - F(a)$, where $F(x)$ is the indefinite integral of $f(x)$.

Of course, I now know that this is the FTC, but back then I didn't, so I came up with my own "proof" via some hand-waving. What I would like to know is, to what extent was this a proof of the FTC? What's missing?

Divide the interval $(a,b)$ into $N$ regions of width $\delta x = \frac{b-a}{N}$. Then:

$$ A = \lim_{\delta x \rightarrow 0} \delta x \sum_{i=1}^{N} f(x_i) $$

Let $F(x) = \frac{d}{dx} f(x)$, then $f(x_i) \approx \frac{F(x_{i+1}) - F(x_i)}{\delta x}$, so that:

$$ A = \lim_{\delta x \rightarrow 0} \delta x \left[ \frac{F(x_2)-F(x_1)}{\delta x} + \frac{F(x_3)-F(x_2)}{\delta x} + \dots + \frac{F(x_{N+1})-F(x_N)}{\delta x}\right]\\=F(x_{N+1})-F(x_1) $$

MGA
  • 9,869

2 Answers2

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What is missing is the remainder estimate. You replace $f(x_i)$ by $(F(x_i+\delta x)-F(x_i))/\delta x$ but do not write out an estimate for the difference between these. Will the sum of all $N$ these differences multiplied by $\delta x$ tend to zero?

Vladimir
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Your argument can be turned into a correct proof when we are allowed to assume that the convergence $$\lim_{h\to0}{F(x+h)-F(x)\over h}=f(x)$$ is uniform in $x$. This means that, given any $\epsilon>0$, there is a $\delta>0$ such that $$\left|{F(x+h)-F(x)\over h}-f(x)\right|<\epsilon\qquad(a\leq x\leq b, \ 0<h<\delta)\ .$$ If this is the case we have $$f(x) h-\epsilon h\leq F(x+h)-F(x)=f(x)h + \epsilon h\qquad(a\leq x\leq b, \ 0<h<\delta)\ .\tag{1}$$ We now choose $N$ so large that $h:={b-a\over N}<\delta$ and that at the same time $\sum_{i=1}^N f(x_i) h$ is within $\epsilon$ from $\int_a^b f(x)\ dx$. Then $(1)$ implies (note the telescoping sum!) $$\sum_{i=1}^N f(x_i) h-\epsilon(b-a)\leq F(b)-F(a)\leq \sum_{i=1}^N f(x_i) h+\epsilon(b-a)\ ,$$ so that $$\left|\int_a^b f(x)\ dx-\bigl(F(b)-F(a)\bigr)\right|\leq\bigl(1+b-a\bigr)\epsilon\ .$$ As $\epsilon>0$ was arbitrary the desired formula follows.

Christian Blatter
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