Results 1 to 4 of 4

Math Help - Substituting definate integrals

  1. #1
    Member
    Joined
    Jun 2007
    Posts
    79

    Substituting definate integrals

    I have been having problems substituting and one instance of basic integration. 1 and 3 are subs and 2 is the basic integration (I believe)
    1. /int (sqrt pi)_0 x(cos x)^2 the answer is 0
    2. /int (x^2)(2x^3)^1/2 the answer is (2/9)(sqrt 2)(x^(9/2))+C
    3. /int ((x^2)(2x^3-5)^1/2) the answer is (1/9)(2x^3-5)^3/2+C

    For 2. i laid it out as x^2*(2x^3)^1/2. mult. for it to be, just 2x^3 which i know is wrong. plz help.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Eater of Worlds
    galactus's Avatar
    Joined
    Jul 2006
    From
    Chaneysville, PA
    Posts
    3,001
    Thanks
    1
    For #1, you can use parts. \int[xcos^{2}(x)]dx

    Let u=x, \;\ dv=cos^{2}(x)dx, \;\ du=dx , \;\ v=\int[cos^{2}(x)]dx=\frac{xsin(x)cos(x)}{2}+\frac{x^{2}}{2}=\frac{x  sin(2x)+2x^{2}}{4}

    For #2: \int[x^{2}\sqrt{2x^{3}}]dx

    Let u=2x^{3}, \;\ du=6x^{2}dx, \;\ \frac{1}{6}du=x^{2}dx

    For #3: \int[x^{2}\sqrt{2x^{3}-5}]dx

    Same substitutions as above, except use u=2x^{3}-5
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Jun 2007
    Posts
    79
    Quote Originally Posted by galactus View Post
    For #1, you can use parts. \int[xcos^{2}(x)]dx

    Let u=x, \;\ dv=cos^{2}(x)dx, \;\ du=dx , \;\ v=\int[cos^{2}(x)]dx=\frac{xsin(x)cos(x)}{2}+\frac{x^{2}}{2}=\frac{x  sin(2x)+2x^{2}}{4}

    For #2: \int[x^{2}\sqrt{2x^{3}}]dx

    Let u=2x^{3}, \;\ du=6x^{2}dx, \;\ \frac{1}{6}du=x^{2}dx

    For #3: \int[x^{2}\sqrt{2x^{3}-5}]dx

    Same substitutions as above, except use u=2x^{3}-5
    I get #1 and #3, for #2, can you tell me the steps to get that particular answer?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,738
    Thanks
    644
    Hello, driver327!

    2)\;\; \int x^2(2x^3)^{\frac{1}{2}}dx

    The answer is: . \frac{2\sqrt{2}}{9}x^{\frac{9}{2}} + C

    We have: . x^2(2x^3)^{\frac{1}{2}} \;=\;x^2\cdot2^{\frac{1}{2}}\cdot x^{\frac{3}{2}} \;=\;\sqrt{2}\,x^{\frac{7}{2}}

    Now integrate: . \sqrt{2}\int x^{\frac{7}{2}}\,dx

    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. solving definate integrals
    Posted in the Calculus Forum
    Replies: 1
    Last Post: March 29th 2009, 10:41 AM
  2. areas and definate integrals
    Posted in the Calculus Forum
    Replies: 7
    Last Post: December 13th 2008, 01:04 PM
  3. Definate Integrals
    Posted in the Calculus Forum
    Replies: 1
    Last Post: March 4th 2008, 09:27 PM
  4. Tricky? Definate Integrals
    Posted in the Calculus Forum
    Replies: 6
    Last Post: March 4th 2008, 03:28 PM
  5. definate integrals
    Posted in the Calculus Forum
    Replies: 4
    Last Post: August 14th 2007, 08:43 PM

Search Tags


/mathhelpforum @mathhelpforum