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!

Definite Integrals

  1. Feb 12, 2009 #1
    1. The problem statement, all variables and given/known data
    I'm not sure if I'm doing this right or not:

    If F(x)=[tex]\int_0^x{\sqrt{t^3+1}dt}[/tex], then find F'(2)

    3. The attempt at a solution
    [tex]\int_0^x[/tex]1/2(t3+1)-1/2*3t2

    1/2(x3+1)-1/2*3x2

    1/2(23+1)-1/2*3(2)2

    Answer=2
     
  2. jcsd
  3. Feb 12, 2009 #2
    Think back to the (first) fundamental theorem of calculus.

    By the way, your integration is wrong. What leads you to assume that

    [tex]
    \int_0^x{\sqrt{t^3+1}dt}
    [/tex]

    is equal to
    [tex]
    \int_0^x{1/2(t3+1)-1/2*3t2}
    [/tex] ??

    It looks like you were thinking of a u-substitution, but that doesn't work.
    Letting [tex]u = t^3+1[/tex], [tex]du=3t^2 dt[/tex], so [tex]dt = \frac{du}{3t^2}[/tex]
    You would then have to substitute this dt into your integral for a valid u-sub, but you notice that you would still have an expression in terms of t, so you cannot integrate with respect to u.
     
  4. Feb 12, 2009 #3
    I was thinking of integration by reverse chain rule :confused:
     
  5. Feb 12, 2009 #4

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Reverse chain rule is substitution. It doesn't work. As Knissp said, you can't integrate that function. You don't want the integral anyway, you want the derivative of the integral. Fundamental theorem of calculus.
     
  6. Feb 12, 2009 #5
    Still not sure if I'm doing it right

    [tex]\int_0^x{\sqrt{t^3+1}dt}[/tex]

    [tex]{\sqrt{x^3+1}}[/tex]

    [tex]{\sqrt{2^3+1}}[/tex]-[tex]{\sqrt{0^3+1}}[/tex]

    3-1=2
     
  7. Feb 12, 2009 #6
    that is correct
     
  8. Feb 12, 2009 #7

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Uh, not quite. Suppose F'(t)=sqrt(t^3+1), i.e. the antiderivative is F(t). Then the integral from 0 to x of sqrt(t^3-1)=F(x)-F(0), right? Find d/dx of that. The derivative of F(0) is zero, correct? It's a constant.
     
  9. Feb 12, 2009 #8

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    It is not.
     
  10. Feb 13, 2009 #9

    Mark44

    Staff: Mentor

    I concur with Dick.

    Also, it would be helpful if you identified the things you're working with, rather than write a bunch of disconnected expressions.

    [tex]F(x) = \int_0^x{\sqrt{t^3+1}dt}[/tex]
    [tex]So, F'(x) = \sqrt{x^3+1}[/tex]
    So far, so good. Now, what is F'(2)?
     
  11. Feb 13, 2009 #10
    Ok, thanks got the answer (3). Yea, I probably should write out the expressions more often
     
  12. Feb 13, 2009 #11

    statdad

    User Avatar
    Homework Helper

    Yea, I probably should write out the expressions more often

    should read

    Yea, I should write out the expressions more often.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Definite Integrals
  1. Definite Integral (Replies: 10)

  2. Definite integration (Replies: 4)

  3. Definite integral (Replies: 4)

  4. Definite integral (Replies: 6)

Loading...