Question

Evaluate the limit (see below) over the region bounded by the graphs of the equations. (see below)

Answer 1

\[\lim_{n \rightarrow \infty} \sum_{i =1}^{n} f(c _{1}) \Delta x _{i}\]
for
\[f(x) = \sqrt{x}\]
y = 0
x = 0
x = 2

Answer 2

So, this is a Reimmann Sum Qestion essentially. And you know that
\[\sqrt{x} = x^\frac{ 1 }{ 2 }\]

Answer 3

mhm

Answer 4

I didn't use antiderivs :P
\[\int\limits_{a}^{b}f(x)dx = \lim_{n \rightarrow ∞}\sum_{i=1}^{n}f(x_i^*)\Delta x\]Reimann sum's is basiclaly the same thing as:

Answer 5

If c_1 is supposed to mean x_i, then you have:\[\lim_{n \rightarrow \infty}\sum_{i=1}^{n} \sqrt{x_i}\Delta x_i:=\int\limits_{0}^{2}\sqrt{x}dx\]It is the definition of the integral.

Answer 6

\[\Delta x = \frac{ b-a }{ n }\]

Answer 7

n is a constant while i changes.

Answer 8

i know it's the integral, but i think they want us to use the riemann sum with the xi stuff.
if it is like that, then i just choose a random n?
the attachment is an example.

Answer 9

You don't choose a random n, you simply keep n there. For instance, if you have 0<x<2, then ∆x = 2-0/n = 2/n

Answer 10

ohh that...

Answer 11

but what's the y = 0 for? is it just saying it's above the x axis?

Answer 12

They surely make your life difficult! The whole reasoning behind Riemann sums is: at the moment it is not meant as an approximation anymore, but the real thing (i.e. lim n to infty), the Fundamental Theorem of Calculus comes to help and we "just" have to calc the difference of two values of a primitive function.

Answer 13

Well of course, the graph of √x starts at 0 and goes up. You can't have a neg number under the radical. then u'd be dealing w/ imaginary numbers, which isn't pleasant.

Answer 14

oh haha ok.

Answer 15

Useful formulas to know for this problem:
\[\sum_{i=1}^{n}c = cn\] \[\sum_{i=1}^{n}i = \frac{ n(n+1) }{ 2 }\] \[\sum_{i=1}^{n}i^2 = \frac{ n(n+1)(2n+1) }{ 6 }\]
\[\sum_{i=1}^{n}i^3 = \left[ \frac{ n(n+1) }{ 2 } \right]^2\]

Answer 16

so then the xi = 2/n? and then u plug it into √x to get √(2i/n) and you just take the limit as n--> ∞ ?
oh i forgot. it also says (HInt: Let
\[ c_{i} = \frac{ 2i^2 }{ n }\]
with that, does that mean i just use that as xi?

Answer 17

As you cycle through the integers from 1 to n in the summation only "i" changes and so anything that isn’t an "i" will be a constant and can be factored out of the summation. So you will be using:
the 3rd formula I provided and taking the limit of that as n->∞

Answer 18

yeah i know the formulas, thanks.
okay thanks :)

Answer 19

Welcome. If you get stuck, let me know :)
But it sounds to me like you understand it. Good luck.

Answer 20

thanks. yeah i kinda forgot about this after learning integrals. :/