anonymous
  • 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
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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freckles
  • freckles
\[\Delta x=\frac{b-a}{N}\] this will be the base length of each rectangle
freckles
  • 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
  • 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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freckles
  • freckles
right endpoint you would just use i=1..36 instead
anonymous
  • anonymous
what is 1..36?
anonymous
  • anonymous
oh wait from 1st rectangle to 36th rectangle ?
freckles
  • 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
  • anonymous
thats the answer for a ?
freckles
  • freckles
no
freckles
  • freckles
it can help you get to the answer though
ganeshie8
  • ganeshie8
|dw:1440392239536:dw|
freckles
  • freckles
if you understand it
freckles
  • freckles
lol i did 36 rectangles
freckles
  • freckles
I forgot it said 18
anonymous
  • 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
  • anonymous
Yes that's why I was confused why were u doing 36
freckles
  • freckles
i=0..17 i=1..18
freckles
  • freckles
now that will give you 18 rectangles
freckles
  • 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
  • ganeshie8
basically, the problem boils down to finding the area under the curve \(\large f(x)=\sqrt{~x}\) : |dw:1440392648846:dw|
ganeshie8
  • ganeshie8
suppose you don't know calculus or anything, do you have any hope in finding the area of that blue region ?
anonymous
  • anonymous
No @ganeshie8
anonymous
  • anonymous
Wait unless u know the size of each square
ganeshie8
  • ganeshie8
sure you know the size of each square
ganeshie8
  • 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
  • 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
  • anonymous
Sure @ganeshie8
ganeshie8
  • ganeshie8
can you find the area of below two big rectangles ? |dw:1440393131543:dw|
ganeshie8
  • 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
  • anonymous
120unit^2
ganeshie8
  • ganeshie8
how ?
anonymous
  • anonymous
90unit^2
anonymous
  • anonymous
When I saw the 18 I just did 18times 4
ganeshie8
  • ganeshie8
18 times 4 is 72, not 90
ganeshie8
  • ganeshie8
but from where did u get "4" ?
anonymous
  • anonymous
U said area together right ? So (4*9) +(6*9)
anonymous
  • anonymous
I just took approximation
ganeshie8
  • 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
  • ganeshie8
it seems you're assuming that the width of each rectangle is just 9, which is wrong..
anonymous
  • anonymous
Ok
ganeshie8
  • ganeshie8
try again
anonymous
  • anonymous
Can I take the height of each rectangle 4 and 6
ganeshie8
  • ganeshie8
you may, but the exact heights are \(\sqrt{18}\) and \(\sqrt{36}\)
anonymous
  • anonymous
(4*18) + (6*16) = 168 unit^2
ganeshie8
  • ganeshie8
looks good! lets do one more exampl
anonymous
  • anonymous
Sure
freckles
  • freckles
|dw:1440393845103:dw|