## anonymous one year ago 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

1. freckles

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

2. 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)

3. 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

4. freckles

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

5. anonymous

what is 1..36?

6. anonymous

oh wait from 1st rectangle to 36th rectangle ?

7. freckles

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

8. anonymous

thats the answer for a ?

9. freckles

no

10. freckles

11. ganeshie8

|dw:1440392239536:dw|

12. freckles

if you understand it

13. freckles

lol i did 36 rectangles

14. freckles

I forgot it said 18

15. 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

16. anonymous

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

17. freckles

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

18. freckles

now that will give you 18 rectangles

19. 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

20. ganeshie8

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

21. ganeshie8

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

22. anonymous

No @ganeshie8

23. anonymous

Wait unless u know the size of each square

24. ganeshie8

sure you know the size of each square

25. 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

26. 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 ?

27. anonymous

Sure @ganeshie8

28. ganeshie8

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

29. 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

30. anonymous

120unit^2

31. ganeshie8

how ?

32. anonymous

90unit^2

33. anonymous

When I saw the 18 I just did 18times 4

34. ganeshie8

18 times 4 is 72, not 90

35. ganeshie8

but from where did u get "4" ?

36. anonymous

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

37. anonymous

I just took approximation

38. ganeshie8

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

39. ganeshie8

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

40. anonymous

Ok

41. ganeshie8

try again

42. anonymous

Can I take the height of each rectangle 4 and 6

43. ganeshie8

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

44. anonymous

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

45. ganeshie8

looks good! lets do one more exampl

46. anonymous

Sure

47. freckles

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