Results 1 to 4 of 4

Math Help - Definite Integral - no clue.....

  1. #1
    Newbie
    Joined
    Sep 2012
    From
    Austria
    Posts
    15

    Definite Integral - no clue.....

    Hi,

    my math lector gave me the following "homework"

    calculate the definite integral:

    I tried using substitution where U=
    √(x^2+2x) and dx=(3*du)/2*(x^2+x^3/3)^3/2 but I don't really know where to go with this problem

    ∫[0,2] √(x^2+2x)*(2x+2) dx

    the intgegral is between 0 to 2..... ∫[0,2]

    Thank you in advance.....
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor MarkFL's Avatar
    Joined
    Dec 2011
    From
    St. Augustine, FL.
    Posts
    1,988
    Thanks
    734

    Re: Definite Integral - no clue.....

    Try the substitution u=x^2+2x instead, and don't forget the rewrite the limits of integration in terms of the new variable...
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,591
    Thanks
    1445

    Re: Definite Integral - no clue.....

    Quote Originally Posted by FireTheft View Post
    Hi,

    my math lector gave me the following "homework"

    calculate the definite integral:

    I tried using substitution where U=
    √(x^2+2x) and dx=(3*du)/2*(x^2+x^3/3)^3/2 but I don't really know where to go with this problem

    ∫[0,2] √(x^2+2x)*(2x+2) dx

    the intgegral is between 0 to 2..... ∫[0,2]

    Thank you in advance.....
    You are making life difficult for yourself with your particular substitution. When dealing with integration by substitution, in your integrand, you need an "inner" function, and you also need the derivative of this inner function as a factor. Here, it's pretty easy to see that an easy inner function is \displaystyle \begin{align*} x^2 + 2x \end{align*}, because its derivative, \displaystyle \begin{align*} 2x + 2 \end{align*} is a factor.

    Anyway, making the substitution \displaystyle \begin{align*} u = x^2 + 2x \implies du = (2x + 2)\,dx \end{align*}, and noting that \displaystyle \begin{align*} u(0) = 0 \end{align*} and \displaystyle \begin{align*} u(2) = 8 \end{align*} gives

    \displaystyle \begin{align*} \int_0^2{\sqrt{x^2 + 2x}\left(2x + 2\right)dx} &= \int_0^8{\sqrt{u}\,du} \\ &= \int_0^8{u^{\frac{1}{2}}\,du} \end{align*}

    which is an easy integral to solve.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Sep 2012
    From
    Austria
    Posts
    15

    Re: Definite Integral - no clue.....

    thx a lot guys.....

    the substitution u=x^2+2x makes it lot easier ;-)......
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 5
    Last Post: December 5th 2011, 05:21 PM
  2. Replies: 4
    Last Post: April 13th 2011, 02:08 AM
  3. definite integral
    Posted in the Calculus Forum
    Replies: 4
    Last Post: February 20th 2011, 10:24 AM
  4. definite integral/ limit of integral
    Posted in the Calculus Forum
    Replies: 1
    Last Post: March 22nd 2010, 04:00 AM
  5. Definite integral help please!
    Posted in the Calculus Forum
    Replies: 1
    Last Post: April 15th 2009, 09:12 AM

Search Tags


/mathhelpforum @mathhelpforum