- anonymous

hey guys hope y"all are well so i have this advance level definite integral question which i have no clue how to slove help , appreciate it :)
https://fbcdn-sphotos-f-a.akamaihd.net/hphotos-ak-xfp1/v/t34.0-12/11909631_821736251278657_492814160_n.jpg?oh=3a6369ebefa89e9e8c6cabd375c6fabc&oe=55DC7BCA&__gda__=1440576288_0b59d00ed7b50754e40cf9942f4f8f3f

- katieb

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- freckles

\[\Delta x=\frac{b-a}{N}\]
this will be the base length of each rectangle

- freckles

you will need to find the height formula...
that is the distance from the curve to the x-axis is actually sqrt(x)
the height is sqrt(x)

- freckles

so starting at left endpoint you would do
x=0
x=0+(b-a)/N
x=0+2(b-a)/N
x=0+3(b-a)/N
...
xi=0+i(b-a)/N
...
where i=0...35
this will give you 36 rectangles

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## More answers

- freckles

right endpoint you would just use i=1..36 instead

- anonymous

what is 1..36?

- anonymous

oh wait from 1st rectangle to 36th rectangle ?

- freckles

\[A_\text{ \left endpoint rule }=\lim_{n \rightarrow \infty} \sum_{i=0}^{35} \Delta x \cdot f(a+i \frac{b-a}{N}) \\ A_{\text{ \right endpoint rule }}=\lim_{n \rightarrow \infty} \sum_{i=1}^{36} \Delta x f(a+i \cdot \frac{b-a}{N})\]

- anonymous

thats the answer for a ?

- freckles

no

- freckles

it can help you get to the answer though

- ganeshie8

|dw:1440392239536:dw|

- freckles

if you understand it

- freckles

lol i did 36 rectangles

- freckles

I forgot it said 18

- anonymous

See I do understand the definitions nd even can solve the question however I am struggling to understand what this question ultimately asking , I don't understand where U and L fit in the equation SRry about that

- anonymous

Yes that's why I was confused why were u doing 36

- freckles

i=0..17
i=1..18

- freckles

now that will give you 18 rectangles

- freckles

use i=0..17 for leftendpoint rule
this will start your rectangles from x=0
use i=1..18 for rightendpoint tule
this wil start your rectangles from x=0+(b-a)/N

- ganeshie8

basically, the problem boils down to finding the area under the curve \(\large f(x)=\sqrt{~x}\) :
|dw:1440392648846:dw|

- ganeshie8

suppose you don't know calculus or anything,
do you have any hope in finding the area of that blue region ?

- anonymous

No @ganeshie8

- anonymous

Wait unless u know the size of each square

- ganeshie8

sure you know the size of each square

- ganeshie8

what riemann did is very simple :
he divided that area into simple rectangles whose area can be found by simply multiplying width and height

- ganeshie8

we will see what U and L mean shortly,
but just so you get used to this rectangles business, lets do a quick example maybe ?

- anonymous

Sure @ganeshie8

- ganeshie8

can you find the area of below two big rectangles ?
|dw:1440393131543:dw|

- ganeshie8

notice that they both have same width : 18
you will need to find the height of each of them before using area of rectangle formula

- anonymous

120unit^2

- ganeshie8

how ?

- anonymous

90unit^2

- anonymous

When I saw the 18 I just did 18times 4

- ganeshie8

18 times 4 is 72, not 90

- ganeshie8

but from where did u get "4" ?

- anonymous

U said area together right ? So (4*9) +(6*9)

- anonymous

I just took approximation

- ganeshie8

Ahh nice, you have the right idea, but your answer is wrong because the width of each rectangle is 18
|dw:1440393575737:dw|

- ganeshie8

it seems you're assuming that the width of each rectangle is just 9, which is wrong..

- anonymous

Ok

- ganeshie8

try again

- anonymous

Can I take the height of each rectangle 4 and 6

- ganeshie8

you may, but the exact heights are \(\sqrt{18}\) and \(\sqrt{36}\)

- anonymous

(4*18) + (6*16) = 168 unit^2

- ganeshie8

looks good!
lets do one more exampl

- anonymous

Sure

- freckles

|dw:1440393845103:dw|