Results 1 to 9 of 9

Math Help - Anti-derivatives

  1. #1
    Newbie
    Joined
    Feb 2008
    Posts
    10

    Anti-derivatives

    Simple anti-derivative problems. I missed the lecture on this and need to be walked through this:

    Indefinite integral for: (4t-9)^-3

    and

    (x^2+2x-3)/(x^4)

    thanks a lot
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor Mathstud28's Avatar
    Joined
    Mar 2008
    From
    Pennsylvania
    Posts
    3,641

    ok

    Quote Originally Posted by coach2uf View Post
    Simple anti-derivative problems. I missed the lecture on this and need to be walked through this:

    Indefinite integral for: (4t-9)^-3

    and

    (x^2+2x-3)/(x^4)

    thanks a lot
    you have \int{(4t-9)^3}dt...now we can rewrite this as \frac{1}{4}\int{(4t-9)^3}dt....and now using the rules of \int{f(g(x))\cdot{g'(x)} we can see the answer easily


    and you also have \int{x^2+2x-3}{x^4}...split the fraction apart to get \frac{x^2}{x^4}+\frac{2x}{x^4}-\frac{3}{x^4} and the answer should be readily apparent
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Feb 2008
    Posts
    10
    That's just it; I can't see the answer.

    Quote Originally Posted by Mathstud28 View Post
    you have \int{(4t-9)^3}dt...now we can rewrite this as \frac{1}{4}\int{(4t-9)^3}dt....and now using the rules of \int{f(g(x))\cdot{g'(x)} we can see the answer easily


    and you also have \int{x^2+2x-3}{x^4}...split the fraction apart to get \frac{x^2}{x^4}+\frac{2x}{x^4}-\frac{3}{x^4} and the answer should be readily apparent
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor Mathstud28's Avatar
    Joined
    Mar 2008
    From
    Pennsylvania
    Posts
    3,641
    Quote Originally Posted by coach2uf View Post
    That's just it; I can't see the answer.
    the first answer is \frac{(4t-9)^4}{16}+C...the second answer is \frac{-1}{x}-\frac{1}{x^2}+\frac{1}{x^3}+C
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Feb 2008
    Posts
    10
    How do you get there though?

    Quote Originally Posted by Mathstud28 View Post
    the first answer is \frac{(4t-9)^4}{16}+C...the second answer is \frac{-1}{x}-\frac{1}{x^2}+\frac{1}{x^3}+C
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor Mathstud28's Avatar
    Joined
    Mar 2008
    From
    Pennsylvania
    Posts
    3,641

    ok

    Quote Originally Posted by coach2uf View Post
    How do you get there though?
    this is something you will have to learn but I will try to help \int{ax^n}dx=anx^{n-1}...and if you have \int{f(g(x))\cdot{g'(x)}} you integrate it as though it was just f(x) and then replace the x with g(x)
    Follow Math Help Forum on Facebook and Google+

  7. #7
    o_O
    o_O is offline
    Primero Espada
    o_O's Avatar
    Joined
    Mar 2008
    From
    Canada
    Posts
    1,408
    \int(4t-9)^{-3}dt

    Let u = 4t - 9. So: du = 4 dt \Rightarrow dt = \frac{du}{4}

    So performing your substitution:
    \int(4t - 9)^{-3}dt = \int u^{-3} \cdot \frac{du}{4} = \frac{1}{4} \int u^{-3} du

    And this should be a standard one.

    For the second one, what MathStud was saying was to split the fraction up:

    \int \frac{x^{2} - 2x + 3}{x^{4}} dx = \int \left( \frac{x^{2}}{x^{4}} - \frac{2x}{x^{4}} + \frac{3}{x^{4}}\right) dx

    See how each term cancels into something you can integrate rather easily?
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Newbie
    Joined
    Feb 2008
    Posts
    10
    Quote Originally Posted by o_O View Post
    So performing your substitution:
    \int(4t - 9)^{-3}dt = \int u^{-3} \cdot \frac{du}{4} = \frac{1}{4} \int u^{-3} du
    Wouldn't it be u^{-4} ?
    Follow Math Help Forum on Facebook and Google+

  9. #9
    o_O
    o_O is offline
    Primero Espada
    o_O's Avatar
    Joined
    Mar 2008
    From
    Canada
    Posts
    1,408
    What do you mean? I just replaced 4t + 9 with u. The exponent -3 remains the same.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. anti-derivatives
    Posted in the Calculus Forum
    Replies: 3
    Last Post: May 5th 2010, 12:24 AM
  2. Anti-derivatives
    Posted in the Calculus Forum
    Replies: 1
    Last Post: March 14th 2009, 10:49 AM
  3. Trig derivatives/anti-derivatives
    Posted in the Calculus Forum
    Replies: 1
    Last Post: February 10th 2009, 02:34 PM
  4. ...Anti-Derivatives?
    Posted in the Calculus Forum
    Replies: 5
    Last Post: October 20th 2007, 07:39 PM
  5. Anti-Derivatives
    Posted in the Calculus Forum
    Replies: 7
    Last Post: April 15th 2007, 12:18 AM

Search Tags


/mathhelpforum @mathhelpforum