1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

New Calculus student needs help with integrals

  1. Aug 28, 2011 #1
    I'm a new Calculus student, and it's killing me! Please help! :(
    Thank you

    1. The problem statement, all variables and given/known data

    Find to 2-decimal place of the following integrals.


    2. Relevant equations

    1. (Upper limit= 33, lower limit= 18)
    1 / ([7x+8]^0.25) dx

    2. (Upper limit = 33, lower limit = 5)
    (x^1.6 + 26) / (x) dx


    I am so at a loss, please help me. Thank you
     
  2. jcsd
  3. Aug 28, 2011 #2

    gb7nash

    User Avatar
    Homework Helper

    What have you tried so far?
     
  4. Aug 28, 2011 #3
    For Question 1:
    : [ln|(7x+8)^0.25|] (upper=33, lower=18)
    Replace x with upper and lower, and subtract, I've got
    3.931872942 - 3.402328159 = 0.529544783

    For Question 2:
    : (33^1.6 + 26) / 33 - (5^1.6 + 26) / 5
    = 8.936954709 - 7.826527804 = 1.110426905

    Am I right? Thanks! =)
     
  5. Aug 28, 2011 #4

    gb7nash

    User Avatar
    Homework Helper

    Unfortunately, no.

    1) Do you know how to use u-substitution?

    2) Divide both terms out by x.
     
  6. Aug 28, 2011 #5
    Hello!

    For 1) I don't know u-substitution.

    For 2), what do you mean by dividing the terms by x?
    I replaced the equation once with 33 (upper limit for x), and another time with 5 (lower limit for x), then substract them together.
    What is the correct way?

    Thanks! =)
     
  7. Aug 28, 2011 #6

    gb7nash

    User Avatar
    Homework Helper

    You can't do 1) if you don't know u-substitution. Here's a link if you need a refresher:

    http://en.wikipedia.org/wiki/Integration_by_substitution

    I don't know what you're doing with 2). You plug in the numbers after you integrate. By dividing, I mean:

    [tex]\frac{x^{1.6}+26}{x} = x^{0.6}+\frac{26}{x}[/tex]

    Do you see how to integrate this now?
     
  8. Aug 28, 2011 #7
    Hello, please pardon me as I'm very weak in Math and Calculus.

    For Question 2:
    If it becomes x^0.6 + 26/x, according to my textbook "26/x" will become "26ln(x)"?

    x^0.6 is supposed to be f'(x) right?
    Am I supposed to convert it back to f(x)? How do I do that with the power being 0.6?

    Thank you!
    Jack
     
  9. Aug 28, 2011 #8

    gb7nash

    User Avatar
    Homework Helper

  10. Aug 28, 2011 #9
    Thank you gb7nash!

    I've worked out the following for question 2:
    "(x^1.6 + 26) / x" will become "(x^1.6 / 1.6) + 26ln(x)"

    Replacing x with the upper and lower limits:
    ((33^1.6 / 1.6) + 26ln(33)) - ((5^1.6 / 1.6) + 26ln(5))
    I get: 258.9838875 - 50.05328511 = 208.9306018

    Is this correct?

    Thank you!
    Jack
     
  11. Aug 28, 2011 #10

    gb7nash

    User Avatar
    Homework Helper

    That looks much better.
     
  12. Aug 28, 2011 #11
    Hello, for question 1, this is what I've worked out.

    Original Question: 1 / (7x + 8)^0.25 dx (upper=33, lower=18)

    This becomes:
    ==> (7x + 8) ^ -0.25 dx
    ==> ((7x + 8) ^ 0.75) / ( 7 ^ 0.75)
    ==> 4/21 [(7x + 8) ^ 0.75]

    Replacing x with upper limit of 33, and lower limit 18
    ==> 4/21 [(7(33) + 8) ^ 0.75 - (7(18) + 8) ^ 0.75]
    ==> 4/21 [60.78528058 - 39.38479586]
    ==> 4/21 [21.40048472]
    Final answer: 4.076282804

    Am I correct for question 1?

    Thank you =)
     
  13. Aug 28, 2011 #12

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Yes, you are. But your derivation isn't making me comfortable. The integral isn't ((7x + 8) ^ 0.75) / ( 7 ^ 0.75) and then you magically went to the correct 4/21 [(7x + 8) ^ 0.75], what happened??
     
  14. Aug 28, 2011 #13
    I have another question 3.

    Question: f(x) = x^4 + ax^2
    When "a" = 191, there is more than 1 critical path.
    Considering the x-coordinates of the critical path, find the greatest value of x to 2 decimal places.

    My working:
    ==> f(x) = x^4 + (191)x^2
    ==> f'(x) = 4x^3 - (382)x = 0
    ==> 2x(2x^2 - 191) = 0
    ==> Therefore: 2x^2 = 191, solving it, final answer: x = 9.772410143

    Is this right too?
    Thank you so much!
     
  15. Aug 28, 2011 #14
    Hello, I'm sorry, it should be
    (7x + 8) ^ 0.75 / (7(0.75))

    I wrote wrongly, I'm sorry!
     
  16. Aug 28, 2011 #15

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    That's much better. Thanks. But you still may need to show more steps on how you did that. Like using u-substitution.
     
  17. Aug 28, 2011 #16

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Partially. x=9.77.. is a critical point (not critical path) of f(x). But it's not the only solution of f'(x)=0. There are three of them. And I think you probably want to find the greatest value of f(x). Finding the 'greatest value of x' doesn't make much sense. And the greatest value of f(x) doesn't really work either, it has a minimum, it doesn't have a maximum. Are you sure that's the correct question?
     
  18. Aug 29, 2011 #17
    Hello Dr Dick, the question is: "When a = 191, there is more than one critical point. Considering the x-coordinates of the critical points, what is the value of the greatest x-coordinate, giving your answer to 2 decimal places?"

    Will my answer be correct now? Sorry I'm very weak in Calculus.
     
  19. Aug 29, 2011 #18
    Another question: Find the enclosed area between the curves, with the equations:
    y = x^2 - 14x
    y = 5.8x - x^2

    My working
    Using the rule: x = - (b/2a)
    f(x) = x^2 - 14x
    x = - (-14/2(1)) = 7
    f(x) = 7^2 - 14(7) = -49
    Minimum turning point = (7, -49)

    g(x) = 5.8x - x^2
    x = -(5.8/2(-1)) = 2.9
    g(x) = 5.8(2.9) - (2.9^2) = 8.41
    Maximum turning point = (2.9, 8.41)

    f(x) = g(x)
    ==> x^2 - 14x = 5.8x - x^2
    ==> 2x^2 -19.8x = 0
    ==> 2x(x = 9.9) = 0
    Therefore, x = 0 or x = 9.9

    Solving with upperlimit = 9.9, lowerlimit = 0, top - bottom:
    (5.8x - x^2) - (x^2 - 14x) dx
    ==> (19.8x - 2x^2) dx
    ==> [(19.8/2)x^2 - (2/3)x^3]
    Replacing x = 9.9 and 0
    ==> (970.299 - 646.866) - (0)
    Final Answer, area = 323.433

    Is this correct?

    Thank you! =)
     
  20. Aug 29, 2011 #19

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Yes, I think your answer is correct.
     
  21. Aug 29, 2011 #20

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    That's fine as well. But you didn't need your 'turning points' (I would call them critical points) to calculate the area, did you?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: New Calculus student needs help with integrals
  1. Need help with calculus (Replies: 10)

Loading...