Results 1 to 5 of 5

Thread: Integration

  1. #1
    Member
    Joined
    Nov 2008
    Posts
    103

    Integration

    It is given that . Find the values of the constants and .

    How would you solve this?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    o_O
    o_O is offline
    Primero Espada
    o_O's Avatar
    Joined
    Mar 2008
    From
    Canada
    Posts
    1,410
    Thanks
    1
    Let $\displaystyle I = \int_0^{\frac{\pi}{2}} e^{4y} \sin (3y) dy$

    Use integration by parts twice.

    For the first round, use: $\displaystyle u = e^{4y}$ and $\displaystyle dv =\sin (3y) dy$

    To get: $\displaystyle I = -\frac{1}{3}e^{4y}\cos (3y) \bigg|_0^{\frac{\pi}{2}} + \frac{4}{3} {\color{blue}\int_0^{\frac{\pi}{2}} e^{4y} \cos (3y) dy}$

    Use integrationg by parts again on the blue using: $\displaystyle u = e^{4y}$ and $\displaystyle dv = \cos (3y) dy$

    To get: $\displaystyle I = -\frac{1}{3}e^{4y}\cos (3y)\bigg|_0^{\frac{\pi}{2}} + \frac{4}{3} \left(\frac{1}{3} e^{4y} \sin (3y) \bigg|_0^{\frac{\pi}{2}} - \frac{4}{3} \int_0^{\frac{\pi}{2}} e^{4y} \sin (3y) dy \right)$

    You should notice that you get $\displaystyle I$ on the right hand side again. So juse use a little bit of algebra to solve for $\displaystyle I$:

    $\displaystyle {\color{red}I} = -\frac{1}{3}e^{4y}\cos (3y)\bigg|_0^{\frac{\pi}{2}} + \frac{4}{3} \left(\frac{1}{3} e^{4y} \sin (3y) \bigg|_0^{\frac{\pi}{2}} - \frac{4}{3} {\color{red}I} \right)$
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    12,028
    Thanks
    848
    Hello, Haris!

    It is given that: .$\displaystyle \int^{\frac{\pi}{2}}_0 e^{4x}\sin(3x)\,dx \;=\; pe^{2\pi} + q$

    Find the values of the constants $\displaystyle p$ and $\displaystyle q.$

    Obviously, we do the integration first . . .

    . . $\displaystyle \int^{\frac{\pi}{2}}_0e^{4x}\sin(3x)\,dx\;=\;\frac {e^{4x}}{25}\bigg[4\sin(3x) - 3\cos(3x)\bigg]^{\frac{\pi}{2}}_0$ ..
    This takes more work than I want to show here


    Evaluate: .$\displaystyle \frac{e^{4(\frac{\pi}{2})}}{25} \bigg[4\sin\tfrac{3\pi}{2} -3\cos\tfrac{3\pi}{2}\bigg] - \frac{e^0}{25}\bigg[4\sin0 - 3\cos0 \bigg]$

    . . . . . .$\displaystyle = \;\frac{e^{2\pi}}{25}\bigg[4(-1) - 3(0)\bigg] - \frac{1}{25}\bigg[4(0) - 3(1)\bigg] $

    . . . . . .$\displaystyle = \;\frac{e^{2\pi}}{25}(-4) - \frac{1}{25}(-3)$

    . . . . . .$\displaystyle =\;-\frac{4}{25}e^{2\pi} + \frac{3}{25} $


    Therefore: .$\displaystyle p \:=\:-\frac{4}{25},\;\;q \:=\:\frac{3}{25}$

    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    Nov 2008
    Posts
    103
    $\displaystyle \int^\frac{\pi}{2}_0 e^{4x}sin(3x) dx$

    but $\displaystyle sin(3x)=Im[e^{3ix}]$ so,

    $\displaystyle \int^\frac{\pi}{2}_0 e^{4x}e^{3ix} dx$

    $\displaystyle \int^\frac{\pi}{2}_0 e^{(4+3i)x} dx$

    $\displaystyle I=\bigg[\frac{1}{4+3i} e^{(4+3i)x}\bigg]^\frac{\pi}{2}_0$

    $\displaystyle I=\bigg[\frac{1}{25} (4-3i)e^{(4+3i)x}\bigg]^\frac{\pi}{2}_0$

    $\displaystyle I=\bigg[\frac{1}{25} (4-3i)e^{4x}e^{3ix}\bigg]^\frac{\pi}{2}_0$

    $\displaystyle I=\bigg[\frac{1}{25}e^{4x} (4-3i)(cos(3x)+isin(3x))\bigg]^\frac{\pi}{2}_0$

    $\displaystyle I=\bigg[\frac{1}{25}e^{4x}(4cos(3x)+4isin(3x)-3icos(3x)+3sin(3x))\bigg]^\frac{\pi}{2}_0$

    Now take the imaginary part of the result:

    $\displaystyle I=\bigg[\frac{1}{25}e^{4x}(4sin(3x)-3cos(3x))\bigg]^\frac{\pi}{2}_0$

    Is this method correct?
    Last edited by Haris; Jan 28th 2009 at 07:00 AM.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Math Engineering Student
    Krizalid's Avatar
    Joined
    Mar 2007
    From
    Santiago, Chile
    Posts
    3,656
    Thanks
    14
    Yes, that's another way of doing it.

    When (in particular) studying the improper integral $\displaystyle \int_0^\infty e^{-\alpha x}\sin(\beta x)\,dx$ its calculation is very easy by using that method.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 3
    Last Post: Nov 3rd 2010, 12:54 AM
  2. Replies: 2
    Last Post: Nov 2nd 2010, 04:57 AM
  3. Replies: 8
    Last Post: Sep 2nd 2010, 12:27 PM
  4. Replies: 2
    Last Post: Feb 19th 2010, 10:55 AM
  5. Replies: 6
    Last Post: May 25th 2009, 06:58 AM

Search Tags


/mathhelpforum @mathhelpforum