1. Not finding help here? Sign up for a free 30min 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!

Can someone explain to me why this is with finding an integral

  1. Nov 2, 2014 #1
    • Homework forum guidelines require the use of the homework template
    Alright, I know how to find the integral of a regular function, But an inverse, is throwing me off

    Can someone explain why 1/Sqrt(7t)dt is 2sqrt(t/7)

    I'm probably missing something, but why does the variable end up as a numerator?

    I know the inverse is 1/(r+1)x^(r+1)

    So I work it out as
    1/(-.5+1) which goes out to 2

    then the rest
    (7t)^-1/2+1
    (7t)^1/2
    (what am I missing here)
    =2sqrt(t/7)+C
     
  2. jcsd
  3. Nov 2, 2014 #2

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    So, you are saying that
    [tex] \int \frac{1}{\sqrt{7t}} \, dt = 2 \sqrt{t/7} + C[/tex]
    You can check by taking the derivative of your answer with respect to t and see if it gives you back the integrand. That is something you should always do, and when you do it you will have answered your own question.
     
  4. Nov 2, 2014 #3

    RUber

    User Avatar
    Homework Helper

    It always helps me to think of
    ##\sqrt{7t}=(7t)^{1/2}##. Or in your case,
    ##\int (7t)^{-1/2} dt##
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Can someone explain to me why this is with finding an integral
Loading...