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Does anyone know how to prove that $f(x)=e^x$ is Riemann integrable using right or left hand Riemann sums?

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    $\begingroup$ if you want to do it directly, split your interval into pieces of equal length $\epsilon$, and the upper and lower Riemann sums are both geometrical series, with the upper one $e^\epsilon$-times bigger than the lower one $\endgroup$
    – user8268
    Commented Mar 24, 2011 at 22:07

4 Answers 4

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If the goal is simply to show $e^x$ is integrable on some interval $[a,b]$ (as opposed to actually computing the value of the integral), it's not even necessary to sum the geometric series. Since $e^x$ is increasing, the left-hand sum is a lower sum ($L$) and the right-hand sum is an upper sum ($U$) and so the difference $U-L$ (if we use a uniform partition) is given by $$ U-L = \frac{b-a}{n} (e^b - e^a). $$ This is less than any $\epsilon > 0$ for suitable choice of $n$.

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$e^x$ uniformly continuous in every interval $[a,b]$. Explicitly, $|e^x-e^y|\le e^b |x-y|$ for $x,y\in [a,b]$.

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WLOG take the range of integration $[0,b]$. Then then the left-hand Riemann sum is given by $$\text{LH} = \frac{1}{N} \sum_{j=0}^{N-1} e^{j b/N}=\frac{e^b -1}{e^{b/N}-1}.$$ The right-hand sum can also be obtained $$\text{RH} = \frac{1}{N} \sum_{j=1}^{N} e^{j b/N}=e^{b/N}\, \text{LH}.$$ For $b$ constant and $N\to\infty$, $$ \text{RH} = \text{LH} = e^b-1$$ which proves that $e^x$ is Riemann integrable.

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Your task is easily handled if you use one of the Riemann integrability theorems for continuous or for monotone functions.

Otherwise, if you really have to check integrability through the definition, you can do as follows.

Take an interval $[a,b]$ and a decomposition $D=\{ a=x_0<x_1\ldots <x_n<x_{n+1}=b\} \subseteq [a,b]$ of the interval; write explicitly the values of:

$s_D(e^x)= \sum_{k=0}^n \inf_{x\in [x_k,x_{k+1}]} e^x\ (x_{k+1}-x_k)$ and

$S_D(e^x)= \sum_{k=0}^n \sup_{x\in [x_k,x_{k+1}]} e^x\ (x_{k+1}-x_k)$;

then consider $S_D(e^x)-s_D(e^x)$ and prove that this difference can be bounded from above by small $\varepsilon >0$ uniformly w.r.t. $D$ (i.e. uniformly w.r.t. the number of point of $D$) as long as $D$ has small diameter $\delta :=\max \{ x_{k+1} -x_k| k=0,\ldots ,n\}$, in the sense that $\delta<C \varepsilon$. In order to properly choose $\delta$, you have to use the hint of lhf.

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