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!

Homework Help: Definite integration 2 & 3

  1. Aug 10, 2007 #1
    1. The problem statement, all variables and given/known data
    2. /int 1_0 (5u^7+pi^2) dx the answer is (5/8)+pi^2
    3./int 4_0 (x^(1/2))(x+1) the answer is 272/15.

    2. Relevant equations

    3. The attempt at a solution
    For 2. I already have the 5/8, my question do I integrate the pi^2? I tried integrating that with no success.

    For 3. (x^(3/2)/3/2)((x^2/2)+x)
    then (2x^(3/2)/3)((1/2)x^2+x)
    plug 4: (16/3)(8+4=12)=192/3=64
  2. jcsd
  3. Aug 10, 2007 #2
    for 2: yes you integrate .Pi/2 is a constant.How do you integrate a constant.
    for 3: your method is not acceptable.

    If you take derivative of (2x^(3/2)/3)((1/2)x^2+x) you don't obtain
    (x^(1/2))(x+1) But you should have done so.

    So Try to expand (x^(1/2))(x+1) ( get rid of the paranthesis ) Then apply the rule you know about this type of integration.

    you can't integrate two parts seperately. Remember for example how do you take derivative of product of functions. it is NOT just the product of derivatives of functions themselves.
  4. Aug 10, 2007 #3


    User Avatar
    Science Advisor
    Homework Helper

    How did you try?

    /int 1_0 (5u^7+pi^2)dx ; where is the x? The variable of integration should be x..
    Anyway, I assume that was a type-error.

    pi^2 is just a constant...

    for ./int 4_0 (x^(1/2))(x+1)

    I assume you mean:

    /int 4_0 (x^(1/2))(x+1)dx

    Have you tried to multiply (x^(1/2)) into (x+1) ?
  5. Aug 10, 2007 #4
    I found what i did wrong now. thanx
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook