Where have I gone wrong in my integral evaluation using summation?

Question

$$\int_a^b \dfrac{dx}{\sqrt x}$$ where $a,b>0$

I know how to do it using the formula of course but I want to know the error in my evaluation through first principles.

Attempt:

$\int_a^b\dfrac{dx}{\sqrt x }= \lim_{n\to \infty, h \to 0}\sum _{k=1}^{n}h(f(a+kh))$ (here: $f(x)= \frac 1 {\sqrt x}$ and $h = \dfrac{b-a}{n}$)

$= \lim_{n\to \infty} (h(\frac{1}{\sqrt{a+h}}+\frac{1}{\sqrt{a+2h}}...+\frac{1}{\sqrt{a+nh}}))$

Note that

$S= \frac{1}{\sqrt{a+h}}+ \frac{1}{\sqrt{a+2h}}+...+\frac{1}{\sqrt{a+nh}}$

$\sum(\frac{1}{\sqrt{a+nh}})< S <\sum(\frac{1}{\sqrt{nh}}) $

From squueze theorem $\lim _{h \to 0}S = \dfrac{n}{\sqrt a}$

So the initial riemann sum simplifies to:

$\lim_{n\to \infty} \dfrac{nh}{\sqrt a}$

Note that $nh = b-a$

$\implies $ the integral is equal to $\dfrac{b-a}{\sqrt a}$

Where have I gone wrong?

Answer 1

• $h$ is $n$-dependence, i.e. $$h=\dfrac{b-a}{n}$$

• $\displaystyle \int_a^b f(x)\, dx =\lim_{n\to \infty} \sum_{k=1}^{n} hf(a+kh)$

• $x_{k}=a+kh$ for $k=0,1,\ldots, n$

• As $f(x)$ is decreasing in this case, the sandwich should be

$$\frac{\frac{b-a}{n}}{\sqrt{a+\frac{(b-a)(k+1)}{n}}} < \int_{x_k}^{x_{k+1}} \frac{dx}{\sqrt{x}} < \frac{\frac{b-a}{n}}{\sqrt{a+\frac{(b-a)k}{n}}}$$

$$\sum_{k=0}^{n-1} \frac{\frac{b-a}{n}}{\sqrt{a+\frac{(b-a)(k+1)}{n}}} < \int_{a}^{b} \frac{dx}{\sqrt{x}} < \sum_{k=0}^{n-1} \frac{\frac{b-a}{n}}{\sqrt{a+\frac{(b-a)k}{n}}}$$

• $b>a>0$

• Your summands are independent of the running index $k$ making your sandwich too loose.

• The sandwich should be tight enough so that both the lower and upper bounds converge to the same limit.

• There is no mathematical advantage of evaluating the integral by Riemann sum. However, it'll be easy to find the sum by evaluating the integral. Usually, the function itself should behave nice enough so that its series bounds can be evaluated by standard series (e.g. AP).

• See an example of evaluation of Riemann sum here and also application of sandwich theorem in here and also this.

Answer 2

$${S_L} = \sum\limits_{k = 1}^n {\frac{1}{{\sqrt {a + nh} }}} = \sum\limits_{k = 1}^n {\frac{1}{{\sqrt {a + \left( {b - a} \right)} }}} = \frac{n}{{\sqrt b }} = \frac{{b - a}}{{\sqrt b h}}$$ $${S_R} = \sum\limits_{k = 1}^n {\frac{1}{{\sqrt {nh} }}} = \sum\limits_{k = 1}^n {\frac{1}{{\sqrt {b - a} }}} = \frac{n}{{\sqrt {b - a} }} = \frac{{\sqrt {b - a} }}{h}$$ Neither sum converges in the limit as $h \to 0$ (or, equivalently, as $n \to \infty$).