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Why does :

$$\frac{\displaystyle\int_t^{t+h}f(s) \, ds}{h} = f(t)\text{ as }h \rightarrow 0\text{ ?}$$

Intuitively this makes sense, because the value of the integral is infinitesimally close to $f(t)$, you have $h$ of them, and you divide the result by $h$. However, I'd like a more formal explanation if possible.

Michael Hardy
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tmakino
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    You need $f$ to be continuous here. Then for $x$ in interval $[t,t+h]$, $f(x)\approx f(t)$. The integral then is approximately $h\cdot f(t)$. (Not very formal ...) – David Mitra May 31 '13 at 17:04

2 Answers2

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We will assume that $h$ approaches $0$ from the right. A similar argument deals with the general case. We will need to make some assumptions about $f$, since the result does not hold in general. We will show that if $f$ is continuous in some interval $[t,t+b]$, then $$\lim_{h\to 0^+} \frac{1}{h}\int_t^{t+h} f(s)\,ds=f(t).$$

We need to assume (or prove) certain properties of the definite integral. Let $m_h$ be the minimum of $f(s)$ in the interval $[t,t+h]$, and let $M_h$ be the maximum of $f(s)$ in this interval. These both exist by continuity.

From any of the standard definitions of the definite integral, we can show that $$h\, m_h \le \int_t^{t+h} f(s)\,ds\le h\, M_h.$$ Divide through by $h$. We get $$m_h \le \frac{1}{h} \int_t^{t+h} f(s)\,ds\le M_h.$$ Now let $h\to 0^+$. Then $m_h\to f(t)$ and $M_h\to f(t)$, and now Squeezing gives us the desired result.

Another way: Or else we can use more machinery, like the Fundamental Theorem of Calculus. Consider the function $g(h)$ given by $$g(h)=\frac{\int_t^{t+h}f(s)\,ds}{h}. $$ Using L'Hospital's Rule and the Fundamental Theorem, we find that $$\lim_{h\to 0}g(h)=\lim_{h\to 0}\frac{f(t+h)}{1}.$$ The limit on the right is $f(t)$.

Remark: We stayed away from using the term "infinitesimal," since in the reals there is no positive element with is $\lt \frac{1}{n}$ for every positive integer $n$. One could reframe the first argument above using the language of non-standard analysis, but there seems to be no great advantage to doing so.

André Nicolas
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  • The FTC is closely related to the above result. However, if a proof of the FTC has been given, then FTC can be used. The fact that during the proof of FTC we came within a whisker of proving the above result does not change that. – André Nicolas May 31 '13 at 18:04
  • Fair enough. =) – Pedro May 31 '13 at 18:07
  • @AndreNicolas: I'm having difficulty following how you reached the right hand side: $\lim_{h\to 0}g(h)=\lim_{h\to 0}\frac{f(t+h)}{1}$. Using L'Hopital's to differentiate, then the Fundamental Theorem to express in terms of antiderivatives, wouldn't you end up with $F(t+h) - F(h)$? – tmakino May 31 '13 at 18:21
  • If you want to think of FTC in terms of antiderivatives, we reach $\frac{F(t+h)-F(t)}{h}$ and then differentiate. Note that $F(t)$ is a constant, when we differentiate the top with respect to $h$ we get $F'(t+h)$ (times, if you wish, the derivative of $t+h$ with respect to $h$, which is $1$.) – André Nicolas May 31 '13 at 18:27
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If $f$ is continuous, then for every $\varepsilon>0$ there exists $\delta>0$ such that if $s$ differs from $t$ by less than $\delta$ then $f(s)$ differs from $f(x)$ by less than $\varepsilon$. Consequently if $|h|<\delta$ then the integral $\displaystyle\int_t^{t+h} f(s)\,ds$ is an integral of a function whose values are between $f(t)\pm\varepsilon$, over an interval of length $h$. Therefore the integral is between $h(f(t)\pm\varepsilon)$. Divide by $h$ and the quotient is between $f(t)\pm\varepsilon$.

So no matter how small $\varepsilon$ is, we can find $\delta$ small enough that if $|h|<\delta$ then the expression whose limit is taken differs from $f(t)$ by less than $\varepsilon$.

Michael Hardy
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