Results 1 to 3 of 3

Thread: Definite Integral Int by Parts

  1. #1
    Member
    Joined
    Nov 2007
    Posts
    232

    Definite Integral Int by Parts

    Please check. I have problems solving for the second part of:

    $\displaystyle \int_a^b u \, dv = uv \mid_a^b - \int_a^b v \, du$

    Use integration by parts to solve the given integral.

    $\displaystyle \int_1^9 (7t - 42)e^{7 - t} \, dt$

    Let
    $\displaystyle u = 7t - 42$

    $\displaystyle du = 7 \, dt$

    $\displaystyle dv = e^{7 - t} \, dt$

    $\displaystyle v = -e^{7 - t}$

    $\displaystyle I = [(7t - 42)(-e^{7-t}) \mid_1^9] - \int_1^9 (-7e^{7-t}) \, dt$

    $\displaystyle I = [(-7(9)e^{7 - 9}) - (-7(1)e^{7-1})] + 7 \int_1^9 e^{7-t} \, dt$

    $\displaystyle I = -56e^4 + [7(-e^{7-1})(-e^{7-1}) \mid_1^9]$

    $\displaystyle I = -56e^4$
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member

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

    I would factor out the 7


    $\displaystyle \int_1^9 (7t - 42)e^{7 - t} \, dt$

    We have: .$\displaystyle 7\!\int^9_1(t-6)\,e^{7-t}\,dt$

    . . $\displaystyle \begin{array}{ccccccc}
    u &=& t-6 & & dv &=& e^{7-t}dt \\ du &=& dt & & v &=& \text{-}e^{7-t} \end{array}$

    We have: . $\displaystyle 7\left[\text{-}(t-6)e^{7-t} + \int e^{7-t}\,dx\right]_1^9 \;=\;7\bigg[\text{-}(t-6)e^{7-t} - e^{7-t}\bigg]^9_1 $

    . . . . . . $\displaystyle = \;\text{-}7e^{7-t}\bigg[(t-6) + 1\bigg]^9_1 \;=\;\text{-}7e^{7-t}(t - 5)\,\bigg]^9_1 $


    Then: . $\displaystyle -7\bigg[e^{-2}(9-5) - e^6(1-5)\bigg] \;=\;-7\left[4e^{-2} + 4e^6\right]$

    . . . . . $\displaystyle = \;-28\left(e^{-2} + e^6\right) \;=\;\frac{\text{-}28(1 + e^8)}{e^2} $

    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Nov 2007
    Posts
    232
    Quote Originally Posted by Soroban View Post
    Hello, Macleef!

    I would factor out the 7



    We have: .$\displaystyle 7\!\int^9_1(t-6)\,e^{7-t}\,dt$

    . . $\displaystyle \begin{array}{ccccccc}
    u &=& t-6 & & dv &=& e^{7-t}dt \\ du &=& dt & & v &=& \text{-}e^{7-t} \end{array}$

    We have: . $\displaystyle 7\left[\text{-}(t-6)e^{7-t} + \int e^{7-t}\,dx\right]_1^9 \;=\;7\bigg[\text{-}(t-6)e^{7-t} - e^{7-t}\bigg]^9_1 $

    . . . . . . $\displaystyle = \;\text{-}7e^{7-t}\bigg[(t-6) + 1\bigg]^9_1 \;=\;\text{-}7e^{7-t}(t - 5)\,\bigg]^9_1 $


    Then: . $\displaystyle -7\bigg[e^{-2}(9-5) - e^6(1-5)\bigg] \;=\;-7\left[4e^{-2} + 4e^6\right]$

    . . . . . $\displaystyle = \;-28\left(e^{-2} + e^6\right) \;=\;\frac{\text{-}28(1 + e^8)}{e^2} $

    why is the last step $\displaystyle \frac{\text{-}28(1 + e^8)}{e^2}$ ? I thought $\displaystyle (e^{-2} + e^6)$ would be... $\displaystyle e^4$...regardless...

    . $\displaystyle = \;-28\left(e^{-2} + e^6\right) \;=\;\frac{\text{-}28(1 + e^8)}{e^2} $
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 5
    Last Post: Dec 5th 2011, 05:21 PM
  2. [SOLVED] Definite integral involving parts
    Posted in the Calculus Forum
    Replies: 1
    Last Post: Apr 10th 2010, 07:38 PM
  3. Replies: 5
    Last Post: Feb 23rd 2010, 12:33 PM
  4. Replies: 3
    Last Post: Sep 9th 2008, 02:13 AM
  5. Integration by Parts - definite integral
    Posted in the Calculus Forum
    Replies: 2
    Last Post: Sep 2nd 2008, 07:17 PM

Search Tags


/mathhelpforum @mathhelpforum