Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

How did they do this?

  1. Oct 20, 2013 #1
    How did they did they do the second line? Any help appreciated, thanks in advance.
     

    Attached Files:

  2. jcsd
  3. Oct 20, 2013 #2
    You know the rules of integration right? Imagine differentiating [itex]x^3[/itex]. You get [itex]3x^2[/itex].
    Now reverse the differentiation.

    So integrating [itex]3x^2[/itex] you first increase the power and then divide by the new power. So...

    1) Raise the power: [itex]3x^3[/itex]
    2) Divide by the new power: [itex]\frac{3}{3}x^3[/itex] = [itex]x^3[/itex]

    That's all that's happening on the second line.
     
  4. Oct 20, 2013 #3
    ##\int 3x^\frac{3}{4}+7x^{-5}+\frac{1}{6}x^{-\frac{1}{2}}\ dx=\int 3x^\frac{3}{4}\ dx+\int 7x^{-5}\ dx+\int \frac{1}{6}x^{-\frac{1}{2}}\ dx## using the sum rule for integrals


    ##\int 3x^\frac{3}{4}\ dx+\int 7x^{-5}\ dx+\int \frac{1}{6}x^{-\frac{1}{2}}\ dx=3\int x^\frac{3}{4}\ dx+7\int x^{-5}\ dx+\frac{1}{6}\int x^{-\frac{1}{2}}\ dx## using the constant multiple rule for integrals


    ##3\int x^\frac{3}{4}\ dx+7\int x^{-5}\ dx+\frac{1}{6}\int x^{-\frac{1}{2}}\ dx=3\cdot\frac{1}{\frac{3}{4}+1} x^{\frac{3}{4}+1}+7\cdot\frac{1}{-5+1}x^{-5+1}+\frac{1}{6}\cdot\frac{1}{-\frac{1}{2}+1} x^{-\frac{1}{2}+1}+c## using the integral formula ##\int x^\alpha\ dx=\frac{1}{a+1}x^{\alpha+1}+c## when ##\alpha\neq-1##
     
  5. Oct 20, 2013 #4
    Profuse thanks dear sir.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook