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!

Basic Calculus Questions - polar integration and roots

  1. Aug 25, 2009 #1
    It's been a while since I studied calculus and basically I have a review sheet for a course I'm taking, but not a graded assignment. So, I was hoping if anyone knew a resource to point me in the right direction with a couple of problems:

    [tex]
    \int_0^\theta x^a dx
    [/tex]

    Where a is not an element of the set {0, -1} and theta > 0. I'm going over my old calculus text but still at a loss as to how to set this problem up, but maybe I'm making it more difficult than it is.

    Also, I had a question involving finding real values of x such that a given function is 0, and justifying my answer. I just did this algebraically, but was wondering if there was another way that I've forgotten using calculus:

    [tex]
    f(x) = x^2\ +\ 2*x\ +\ 2
    [/tex]
    [tex]
    (x+1)*(x+1)+1 = 0
    [/tex]
    [tex]
    (x+1)^2 = -1
    [/tex]
    [tex]
    x = -1 - i, -1 + i
    [/tex]

    Thanks.
     
  2. jcsd
  3. Aug 25, 2009 #2

    Mark44

    Staff: Mentor

    The antiderivative of xa is 1/(a + 1) xa + 1.
     
  4. Aug 25, 2009 #3
    Thanks, but I was feeling that maybe I was overlooking something. At this point I just stopped.

    [tex]

    \int_0^\theta x^a dx\ =\ (\theta^(a+1))/(a\ +\ 1)

    [/tex]
     
  5. Aug 25, 2009 #4

    Mark44

    Staff: Mentor

    I think you meant this as [itex]\theta^{a + 1}/(a + 1)[/itex]
    but it didn't quite come out that way.
     
  6. Aug 25, 2009 #5

    Mark44

    Staff: Mentor

    For your other problem, find the zeroes of f(x) = x2 + 2x + 2, the algebraic (as opposed to calculus) way is one way to go. In fact, calculus is not really appropriate for this type of problem.

    f(x) = x2 + 2x + 2 = x2 + 2x + 1 + 1 = (x + 1)2 + 1

    The squared term is always >= 0, so adding 1 gives a value that is always >= 1, hence there are no real values for which f(x) = 0.

    Looking at this graphically, the function's graph is a parabola that opens up, and whose vertex is at (-1, 1). Since the vertex is the lowest point on the graph, and it is above the x-axis, there are no values of x for which f(x) is less than or equal to zero.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Basic Calculus Questions - polar integration and roots
  1. Root integral question (Replies: 1)

Loading...