I need to calculate $\int_0^1 e^x \; dx$ using Riemann sum.

Problem set gives a hint:"The sum is a geometric progression. You will need the limit $ \lim _{n\to \infty }n\left(e^{\frac{1}{n}}-1\right)$ . This can be evaluated putting h = 1/n and relating the limit to the derivative of $e^x$ at x = 0".

For future googling: problem comes from problem set for MIT Open Courses, Single Variable Calculus, Unit 3 - Integration, section 3B-Definite Integrals, problem 3B-6.

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    $\begingroup$ If you write down the Riemann sum, what do you get? $\endgroup$ – Kusma Sep 8 '18 at 21:25
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    $\begingroup$ Hint: The problem set gives you a hint. Have you proven that the things that the hint states? Have you tried using the hint to help you solve the problem? $\endgroup$ – Arthur Sep 8 '18 at 21:25
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    $\begingroup$ One Riemann sum approximation to $\int_a^b f(x) dx $ is $\sum_{k=0}^{n-1} f(a+{k\over n}(b-a)) {1 \over n}$. Note that the integrand is increasing. $\endgroup$ – copper.hat Sep 8 '18 at 21:37
  • $\begingroup$ Arthur, I did the following reasoning: $\lim _{n\to \infty }n\left(e^{\frac{1}{n}}-1\right) = \frac{e^{\left(0+\Delta x\right)}-e^0}{\Delta \:x}=\frac{d}{dx}e^x\:at\:0\:=\:e^0\:=\:1$ $\endgroup$ – Ivan Koshelev Sep 8 '18 at 21:40


We have that

$$\int_0^1 e^x \; dx=\lim_{n\to \infty}\frac1n \sum_{k=0}^n e^{\frac k n}=\lim_{n\to \infty}\frac1n \sum_{k=0}^n \left(e^{\frac 1 n}\right)^k$$

For the limit we have that

$$\lim_{n\to \infty }n\left(e^{\frac{1}{n}}-1\right)=\lim_{n\to \infty }\frac{\left(e^{\frac{1}{n}}-1\right)}{\frac1n}=\lim_{h\to 0 }\frac{e^{h}-1}{h}=1$$

and we can deduce that by definition of derivative or by l'Hopital.

  • $\begingroup$ I got to both of those conclusions, but not sure, how to tie them together. $\endgroup$ – Ivan Koshelev Sep 8 '18 at 23:10
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    $\begingroup$ Recall that by geometric series $$\sum_0^n r^k=\frac{r^{n+1}-1}{r-1}$$ then let $r=e^{1/n}$ and finally take the limit. $\endgroup$ – gimusi Sep 8 '18 at 23:12
  • $\begingroup$ Ok, applying the formula for series $\sum_0^n r^k=\frac{r^{n+1}-1}{r-1}$ , and provided $\lim_{n\to \infty }$ I got $ \frac{1}{n}\left(\frac{1-\left(1+\frac{1}{n}\right)^{n+1}}{1-\left(1+\frac{1}{n}\right)}\right)=\frac{\frac{1}{n}\left(1-\left(1+\frac{1}{n}\right)^{n+1}\right)}{-\frac{1}{n}}=-\left(1-\left(1+\frac{1}{n}\right)^{n+1}\right)=\left(1+\frac{1}{n}\right)^{n+1}-1=e-1=e^1-e^0 $ . This agrees with FTC $\int _0^1\:e^xdx\:=\:e^1-e^0$ . Correct? $\endgroup$ – Ivan Koshelev Sep 9 '18 at 15:47
  • $\begingroup$ The result is correct but you need to use $r=e^{1/n}$ in the geometric series. $\endgroup$ – gimusi Sep 9 '18 at 15:52

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