## anonymous one year ago Need Help! Use the definition of an integral to evaluate the integral: 5 to -1, (1+3x)*dx

1. anonymous

$\int_{-1}^5(1+3x)dx$?

2. anonymous

and you have to do it the hard way right? not the easy way ?

3. anonymous

|dw:1439170371150:dw|

4. anonymous

5. anonymous

Yea, thats it! and the hard way.... and sorry about the drawing above

6. anonymous

the first one is evidently a constant since $\sum_{i=1}^n 1=n$

7. anonymous

Yep! I dont know what to do about the negative 3 though

8. anonymous

so the first part $\int_{-1}^51dx=6$ which is not surprising since $$1\times 6=6$$

9. anonymous

i don't get the negative 3 part

10. anonymous

$\int_{-1}^53xdx=3\int_{-1}^5xdx$

11. anonymous

so pull the 3 outside the summation

12. anonymous

It's supposed to be a negative 3 though...

13. anonymous

not what you wrote

14. anonymous

i mean not $\int_{-1}^5(1+3x)dx$ that is a plus 3, is there a typo there?

15. anonymous

nope

16. anonymous

then there is no $$-3$$

17. anonymous

which type of way are you solving for this problem?

18. anonymous

looks like same as you break the integral in to two pieces, first one is $\int_{-1}^51dx=\frac{6}{n}\sum_{1}^n1=\frac{6}{n}\times n=6$

19. anonymous

you have that part above right?

20. anonymous

I did it like this

21. anonymous

ok that might work too, although it is certainly more work than necessary

22. anonymous

Yea, but that's the way we gotta do it on our exam

23. anonymous

really? why can't you just pull the 3 out of the summation like a normal person would?

24. anonymous

You mean make the 6/n that's already on the outside and multiply that with negative 3?

25. anonymous

no matter lets see if this works, it should you have \$\frac{6}{n}\sum_{i=1}^n(-3)=\frac{6}{n}\times -3n=-18$

26. anonymous

is that totally clear or no?

27. anonymous

ah, I see it

28. anonymous

eh...actually how did you get the n onto the negative 3?

29. anonymous

so far we are at $$6-18=-12$$ now for the rest (i hope this wrods)

30. anonymous

ok lets go real slow here

31. anonymous

lol thanks

32. anonymous

the first sum is what you wrote above $\frac{6}{n}\sum_{i=1}^n1$right ?

33. anonymous

mhm

34. anonymous

now what does $\sum_{i=1}^n1$ mean in reality?

35. anonymous

n!

36. anonymous

right it is $\overbrace{1+1+...+1}^{\text{ n times}}$

37. anonymous

i.e $$n$$

38. anonymous

similarly $\sum_{i=1}^n c=n\times c$ for any constant $$c$$

39. anonymous

okay well that makes everything more clear

40. anonymous

so $\sum_{i=1}^n(-3)=-3\sum_{i=1}^n1=-3n$

41. anonymous

it is just the distributive law in action, which is why you should really start with $\int 1dx+3\int xdx$ to make your summations simpler i don't get the "do it this way on a test" part, but i believe you

42. anonymous

so now we have taken care of the first two sums correct?

43. anonymous

is the answer -5/2?

44. anonymous

i really doubt it

45. anonymous

yea, me too

46. anonymous

$y=3x+1$ is a line with slope 3 and y intercept 1 it would be easiest to compute this integral with geometry lets continue

47. anonymous

I think the answer was 42 in the back of the book

48. anonymous

all we have left is $\frac{6}{n}\frac{18}{n}\sum_{i=1}^ni$ right?

49. anonymous

yea

50. anonymous

the last summation after you distribute the $$\frac{6}{n}$$

51. anonymous

you made some sort of goofy algebra mistake in the last line should have 3 sums, not a product with -3 in it but you wrote the sum correctly $\sum_{i=1}^ni=\frac{n(n+1)}{2}$

52. anonymous

yep

53. anonymous

so now we are at $\frac{6\times 18}{n^2}\times\frac{n(n+1)}{2}$

54. anonymous

yep, have that already

55. anonymous

or more precisely $\lim_{n\to \infty}\frac{6\times 18}{n^2}\times\frac{n(n+1)}{2}$ a limit you compute with your eyeballs

56. anonymous

I got the answer!

57. anonymous

is it clear that the limit is 54?

58. anonymous

yay

59. anonymous

I just made some silly mistake and yea

60. anonymous

lol thanks! Appreciate it the help

61. anonymous

and i guess $6-18+54=40$ or sommat

62. anonymous

it equals 42 buddy

63. anonymous

btw it really really is easier if you pull the 3 out of the summation

64. anonymous

oops

65. anonymous

lol you're fine

66. anonymous

didn't you say it was 40 in the back of the book?

67. anonymous

....no lol, scroll up

68. anonymous

u need some rest

69. anonymous

who don't know that?