I was trying to evaluate the following integral $\int_0^{10} 2^x$ using Riemann sum.
What I tried:
$\triangle x_i= \frac{10-0}{n}=\frac{10}{n}$
$x_i=x_0 + i\triangle x_i = 0 + i\frac{10}{n}=i\frac{10}{n}$
Let $f(x)=2^x$.
Hence $f(x_i)=2^{x_i}=2^{i\frac{10}{n}}$
$S_n=\sum_{i=0}^{n-1} f(x_i).\triangle x_i$
$S_n=\sum_{i=0}^{n-1} \frac{10}{n}.2^{i\frac{10}{n}}$
$S_n=\frac{10}{n}\sum_{i=0}^{n-1} 2^{i\frac{10}{n}}$
$S_n=\frac{10}{n}({2^{\frac{10}{n}}}^0+{2^{\frac{10}{n}}}^1+...+{2^{\frac{10}{n}}}^{n-1})$
$S_n=\frac{10}{n}.\frac{1-{2^{\frac{10}{n}}}^n}{1-2^{\frac{10}{n}}}$
$S_n=\frac{10}{n}.\frac{1-2}{1-2^{\frac{10}{n}}}$
$S_n=\frac{10}{(2^\frac{10}{n}-1)(n)}$
$\lim \limits_{n \to \infty} S_n =\frac{1}{log2}$ Which doesn't evaluate to $\int_0^{10} 2^x$
Can someone tell me where I've gone wrong ?
Thanks in advance!
$$\lim_{n\to\infty}\dfrac{10}n\cdot\dfrac{1-(2^{10/n})^n}{1-2^{10/n}}=\dfrac{2^{10}-1}{\lim_{n\to\infty}\dfrac{2^{10/n}-1}{\dfrac{10}n}}$$
Now setting $10/n=h,$ $${\lim_{n\to\infty}\dfrac{2^{10/n}-1}{\dfrac{10}n}}=\ln2\cdot\lim_{h\to0}\dfrac{e^{h\ln2}-1}{h\ln2}=?$$