A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

anonymous

  • one year ago

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

  • This Question is Closed
  1. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  2. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  3. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    |dw:1439170371150:dw|

  4. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    not bad!

  5. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  6. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  7. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  8. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  9. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    i don't get the negative 3 part

  10. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  11. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    so pull the 3 outside the summation

  12. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  13. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    not what you wrote

  14. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  15. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    nope

  16. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    then there is no \(-3\)

  17. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    which type of way are you solving for this problem?

  18. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    you have that part above right?

  20. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I did it like this

  21. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  22. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  23. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  24. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  25. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    is that totally clear or no?

  27. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    ah, I see it

  28. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  29. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  30. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    ok lets go real slow here

  31. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    lol thanks

  32. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  33. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    mhm

  34. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  35. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    n!

  36. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  37. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    i.e \(n\)

  38. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    similarly \[\sum_{i=1}^n c=n\times c\] for any constant \(c\)

  39. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    okay well that makes everything more clear

  40. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  41. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  43. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    is the answer -5/2?

  44. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    i really doubt it

  45. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    yea, me too

  46. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  47. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  48. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  49. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    yea

  50. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  51. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    yep

  53. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  54. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    yep, have that already

  55. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I got the answer!

  57. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    is it clear that the limit is 54?

  58. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    yay

  59. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I just made some silly mistake and yea

  60. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    lol thanks! Appreciate it the help

  61. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    and i guess \[6-18+54=40\] or sommat

  62. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    it equals 42 buddy

  63. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  64. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    oops

  65. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    lol you're fine

  66. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  67. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    ....no lol, scroll up

  68. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    u need some rest

  69. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    who don't know that?

  70. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.