# Definite integral as Riemann sums

#### terryds

1. Homework Statement

Determine $\int_{0}^{2}\sqrt{x}dx$ using left riemann sums

2. Homework Equations

$\int_{a}^{b}f(x)dx = \lim_{n\rightarrow \infty}\sum_{i=0}^{n-1}(\frac{b-a}{n})f(x_i)$

3. The Attempt at a Solution

$\frac{b-a}{n}=\frac{2-0}{n}=\frac{2}{n}$
$\int_{0}^{2}\sqrt{x}dx = \lim_{n\rightarrow \infty}\frac{2}{n}(0+ \frac{\sqrt{2}}{n}+\frac{\sqrt{4}}{n}+...) =\lim_{n\rightarrow \infty}\frac{2}{n} \frac{\sqrt{2}}{n}(\sqrt{0} + \sqrt{1}+\sqrt{2}+\sqrt{3}+...)$

I'm stuck now. Please help. I don't know how to get the sums of the infinite radical series

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#### LCKurtz

Homework Helper
Gold Member
1. Homework Statement

Determine $\int_{0}^{2}\sqrt{x}dx$ using left riemann sums

2. Homework Equations

$\int_{a}^{b}f(x)dx = \lim_{n\rightarrow \infty}\sum_{i=0}^{n-1}(\frac{b-a}{n})f(x_i)$

3. The Attempt at a Solution

$\frac{b-a}{n}=\frac{2-0}{n}=\frac{2}{n}$
$\int_{0}^{2}\sqrt{x}dx = \lim_{n\rightarrow \infty}\frac{2}{n}(0+ \frac{\sqrt{2}}{n}+\frac{\sqrt{4}}{n}+...) =\lim_{n\rightarrow \infty}\frac{2}{n} \frac{\sqrt{2}}{\color{red}n}(\sqrt{0} + \sqrt{1}+\sqrt{2}+\sqrt{3}+...)$

I'm stuck now. Please help. I don't know how to get the sums of the infinite radical series
I think you may want a square root on that $n$ I colored red. But to answer your question, that is not a trivial sum to calculate, and I wouldn't expect to see this problem in a typical calculus exercise set. See, for example,
http://math.stackexchange.com/questions/1241864/sum-of-square-roots-formula

#### Mark44

Mentor
1. Homework Statement

Determine $\int_{0}^{2}\sqrt{x}dx$ using left riemann sums

2. Homework Equations

$\int_{a}^{b}f(x)dx = \lim_{n\rightarrow \infty}\sum_{i=0}^{n-1}(\frac{b-a}{n})f(x_i)$

3. The Attempt at a Solution

$\frac{b-a}{n}=\frac{2-0}{n}=\frac{2}{n}$
$\int_{0}^{2}\sqrt{x}dx = \lim_{n\rightarrow \infty}\frac{2}{n}(0+ \frac{\sqrt{2}}{n}+\frac{\sqrt{4}}{n}+...) =\lim_{n\rightarrow \infty}\frac{2}{n} \frac{\sqrt{2}}{n}(\sqrt{0} + \sqrt{1}+\sqrt{2}+\sqrt{3}+...)$
The sums in the last two expressions above do not have an infinite number of terms. What's the last term in each sum?
terryds said:
I'm stuck now. Please help. I don't know how to get the sums of the infinite radical series

#### terryds

The sums in the last two expressions above do not have an infinite number of terms. What's the last term in each sum?
I think you may want a square root on that $n$ I colored red. But to answer your question, that is not a trivial sum to calculate, and I wouldn't expect to see this problem in a typical calculus exercise set. See, for example,
http://math.stackexchange.com/questions/1241864/sum-of-square-roots-formula
$\lim_{n\rightarrow \infty }\frac{2}{n}\sqrt{\frac{2}{n}}(\sqrt{0} + \sqrt{1} + \sqrt{2} + ... + \sqrt{\frac{2}{n}}(n-1)))$

Alright, I've fixed my mistakes. And, I'm still confused how to get the sum of that series

#### Ray Vickson

Homework Helper
Dearly Missed
$\lim_{n\rightarrow \infty }\frac{2}{n}\sqrt{\frac{2}{n}}(\sqrt{0} + \sqrt{1} + \sqrt{2} + ... + \sqrt{\frac{2}{n}}(n-1)))$

Alright, I've fixed my mistakes. And, I'm still confused how to get the sum of that series
You should go back and read post #2, which explains why you probably cannot sum the series, no matter how hard you try. Of course, you may be able to find good approximations to the sum, but that is not the same as finding an exact value.

#### lurflurf

Homework Helper
Doing the sum is overkill for finding the limit.
I think you have an error I think you should have
$$\int_0^2 \! \sqrt{n} \, \mathrm{d}x=\lim_{n\rightarrow\infty} {\left( \frac{n}{2} \right)}^{-3/2}\sum_{i=1}^n \sqrt{n}$$
Instead of finding the sum we will make a few observations that suffice
we can replace the sum in the limit by a simple expression
$$\sum_{i=1}^n \sqrt{n}\sim {C} {\sqrt{n}}^3$$
The exponent is the only one that allows the limit to converge to a nonzero value
we find C by computing the limit
$$C=\lim_{n\rightarrow\infty}\frac{{\sqrt{n}}}{{\sqrt{n+1}}^3-{\sqrt{n}}^3}$$
because we must have
$$\sum_{i=1}^{n+1} \sqrt{n}-\sum_{i=1}^n \sqrt{n}\sim {\sqrt{n}}$$

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