Riemann Sum of $f(x)=2^x$

Question

Using Riemann Sums, how can I compute the integral $$\int_{0}^{2} 2^x dx$$

I don't know how can I take the Partition and then compute the sums , someone can help to understand this method of Riemann Sums please.

Thanks for your time.

Answer 1

Since $\;f(x)=2^x\;$ is a continuous function it is Riemann integrable in any bounded interval and thus you can choose that interval's partition as you like and also the points in each subinterval where the function is going to be evaluated.

Thus, we can choose

$$P:=\left\{x_0=0<x_1=\frac2n<x_2=\frac4n<\ldots<x_k=\frac{2k}n<\ldots<x_n=\frac{2n}n=2\right\}$$

and the points $\;c_k=x_k\in[x_{k-1},x_k]\;$ , and form the Riemann sum

$$\sum_{k=0}^nf(c_k)(x_k-x_{k-1})=\frac2n\sum_{k=0}^n2^{\frac{2k}n}=\frac2n\sum_{k=0}^n\left(4^{1/n}\right)^k=$$

$$=\frac2n\cdot\frac{1-4^{\frac{n+1}n}}{1-4^{1/n}}=\frac2n\cdot\frac{1-e^{\frac{n+1}n\log4}}{1-e^{\frac1n\log4}}$$

Now, observe that

$$\lim_{n\to\infty}\frac1{n(1-e^{\frac1n\log4})}=\lim_{n\to\infty}\frac{\frac1n}{1-e^{\frac1n\log4}}\stackrel{\text{l'Hospital}}=\lim_{n\to\infty}\frac{-\frac1{n^2}}{\frac{\log4}{n^2}e^{\frac1n\log4}}=-\frac1{\log4}$$

and thus we get

$$\lim_{n\to\infty}\frac2n\cdot\frac{1-e^{\frac{n+1}n\log4}}{1-e^{\frac1n\log4}}=-\frac2{\log4}\left(1-4\right)=\frac3{\log2}$$

Checking:

$$\int_0^22^xdx=\left.\frac1{\log2}2^x\right|_0^2=\frac1{\log2}\left(4-1\right)=\frac3{\log2}$$