Integration using summation How do you integrate $\sqrt{x}$ from an arbitrary constant $a$ to another $b$ by summation ?
 A: Let us divide the region $a\le x\le b$ into $n$ sub-region as $a,ar,ar^2,\cdots ar^n=n$ such that $\delta_i=ar^i-ar^{i-1}$
If real $m\ne-1,$
$$\int_a^b x^mdx=\lim_{n\to\infty}\left(\sum_{1\le i\le n}(ar^i)^m\delta_m\right)$$
$$=\lim_{n\to\infty}\left(\sum_{1\le i\le n} (ar^i)^m(ar^i-ar^{i-1})\right)$$
$$=a^{m+1}\lim_{r\to1}(r-1)\sum_{1\le i\le n} (r^{i-1})^{m+1} $$
$$=a^{m+1}\lim_{r\to1}(r-1)\sum_{1\le i\le n} (r^{m+1})^{i-1} $$
$$=a^{m+1}\lim_{r\to1}(r-1)\cdot1\cdot\frac{(r^{m+1})^n-1}{r^{m+1}-1} $$
$$=a^{m+1}\lim_{r\to1}\frac{\left(\frac ba\right)^{m+1}-1}{\frac{r^{m+1}-1}{r-1}} $$ as $\frac ba= r^n\implies (r^{m+1})^n=(r^n)^{m+1}=\left(\frac ba\right)^{m+1}$
$$=(b^{m+1}-a^{m+1})\cdot\frac1{\lim_{r\to1}\left(\frac{r^{m+1}-1}{r-1}\right)}$$
$$=(b^{m+1}-a^{m+1})\cdot \frac1{(m+1)\cdot1^m}$$
$$\text{  as }\lim_{x\to a}\frac{x^n-a^n}{x-a}=nx^{n-1} \text{ using L'Hospital Rule }$$
A: Do you mean something like this:
$f(x)= \sqrt{x}$, 
$$\int_a^b \sqrt{x}=\lim_{n\to \infty} \dfrac{b-a}{n} \sum_1^n\sqrt{x_i}$$
A: Since the function $x\mapsto\sqrt{x}$ is nondecreasing, every Riemann sum associated to the subdivision $s=(x_i)_{0\leqslant i\leqslant n}$ is between $R^+(s)$ and $R^-(s)$ with
$$
R^-(s)=\sum_{i=1}^n(x_i-x_{i-1})\sqrt{x_{i-1}},\qquad R^+(s)=\sum_{i=1}^n(x_i-x_{i-1})\sqrt{x_{i}}.
$$
If the mesh of $s$ is $\delta(s)$,
$$
R^+(s)-R^-(s)=\sum_{i=1}^n(x_i-x_{i-1})(\sqrt{x_{i}}-\sqrt{x_{i-1}})\leqslant\delta(s)\sum_{i=1}^n\sqrt{x_{i}}-\sqrt{x_{i-1}}.
$$
Thus, $R^+(s)-R^-(s)\leqslant\delta(s)(\sqrt{b}-\sqrt{a})\to0$ when $\delta(s)\to0$, that is, the function is Riemann integrable. Furthermore, for every $i$,
$$
(x_i-x_{i-1})\sqrt{x_{i-1}}\leqslant\tfrac23(x_i\sqrt{x_i}-x_{i-1}\sqrt{x_{i-1}})\leqslant(x_i-x_{i-1})\sqrt{x_{i}},\qquad (\ast)
$$
hence
$$
R^-(s)\leqslant\sum_{i=1}^n\tfrac23(x_i\sqrt{x_i}-x_{i-1}\sqrt{x_{i-1}})\leqslant R^+(s),
$$
and the sum in the middle is telescoping hence all this shows that
$$
\int_a^b\sqrt{x}\,\mathrm dx=\tfrac23(b\sqrt{b}-a\sqrt{a}).
$$
Finally, the result hinges on the algebraic inequalities $(\ast)$, which we now prove. 
Dividing $(\ast)$ by $\frac13(\sqrt{x_i}-\sqrt{x_{i-1}})\gt0$ and renaming $\sqrt{x_{i-1}}=x$ and $\sqrt{x_i}=y$, one must show that, for all $0\leqslant x\lt y$,
$$
3x(y+x)\leqslant2(y^2+yx+x^2)\leqslant3y(y+x),
$$
or,
$$
-(y-x)(2y+x)\leqslant0\leqslant(y-x)(y+2x).
$$
Since $y-x\gt0$, $2y+x\gt0$ and $y+2x\gt0$, this proves $(\ast)$.
