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

Question

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.

Answer 1

HINT

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.