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Integration by parts - Does this make sense?

  1. Feb 11, 2012 #1
    I'm confused. I was making up some of my own problems involving higher powers of x to integrate. For example:

    [tex]\displaystyle\int x^5 e^{5x}dx[/tex]

    I set about going about finding [tex]\frac{dy}{dx}[/tex] up to [tex]\frac{d^6y}{dx^6}[/tex].

    [tex]u=x^5[/tex]
    [tex]\frac{du}{dx}=5x^4[/tex]
    [tex]\frac{d^2u}{dx^2}=20x^3[/tex]
    [tex]\frac{d^3u}{dx^3}=60x^2[/tex]
    [tex]\frac{d^4u}{dx^4}=120x[/tex]
    [tex]\frac{d^5u}{dx^5}=120[/tex]
    [tex]\frac{d^6u}{dx^6}=0[/tex]

    Conversely I found [tex]v=\displaystyle\int \frac{dv}{dx}=\displaystyle\int e^{5x}dx=\frac{1}{5}e^{5x} + C_1[/tex] up to


    [tex]\displaystyle\int \displaystyle\int \displaystyle\int \displaystyle\int \displaystyle\int e^{5x}dx=\frac{1}{3125}e^{5x} + C_5[/tex]

    I'm not even sure if you can write 5 integrals next to eachother. If not, what's the correct notation?

    Anyway, I then substituted the above values into the IBP formula:

    [tex]\displaystyle\int u\frac{dv}{dx}dx=uv - \displaystyle\intv\frac{du}{dx}dx[/tex]

    [tex]\displaystyle\int x^5e^{5x}dx=\frac{1}{5}x^5e^5x-[\displaystyle\int x^4e^{5x}dx][/tex]

    [tex]\frac{1}{5}x^5e^{5x}=[\displaystyle\int \frac{5}{4}x^4e^{5x}dx]=\frac{5}{25}x^4e^{5x} - \displaystyle\int \frac{20}{25}x^3e^{5x}dx[/tex]

    [tex]\frac{1}{5}x^5e^{5x}-\frac{5}{25}x^4e^{5x}= [\displaystyle\int \frac{20}{25}x^3e^{5x}dx]=\frac{20}{125}x^3e^{5x}- [\displaystyle\int \frac{60}{125}x^2e^{5x}dx][/tex]... You get the idea.

    It's kind of like reduction, reducing the integral to obtain an answer if that makes sense?

    Anyway, if there's an easier way to integrate higher powers of x using IBP I'd like to know.

    There's a pattern there but I can't uncover it. Sort of like the Binomial expansion, same principle it seems.

    Thanks! :)

    My working will help too.

    JvcH7.jpg
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Feb 11, 2012 #2

    LCKurtz

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    Look at http://math.ucsd.edu/~wgarner/math20b/int_by_parts.htm [Broken]

    Scroll down a little to see an example similar to yours.
     
    Last edited by a moderator: May 5, 2017
  4. Feb 11, 2012 #3

    HallsofIvy

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    I'm not clear why you were doing all that. Differentiating [itex]x^5[/itex] and integrating [itex]e^{5x}[/itex] separately doesn't help you a whole lot in integrating their product. Instead, use "integration by parts" as suggested in your title (what you did was NOT integration by parts). And do it 5 times.

    1) Let [itex]u= x^5[/itex] and [itex]dv= e^{5x}dx[/itex]. Then [itex]du= 5x^4dx[/itex] and [itex]v= (1/5)e^{5x}dx[/itex] so
    [tex]\int udv= uv- \int vdu[/tex]
    [tex]\int x^5e^{5x}dx= (1/5)x^5e^{5x}- \int x^4e^{5x}dx[/tex]
    and you have reduced the power of x to 4.

    2) Let [itex]u= x^4[/itex] and [itex]dv= e^{5x}dx[/itex]. Then [itex]du= 4x^3dx[/itex] and [itex]v= (1/5)e^{5x}[/itex]. Now the integral is
    [tex]\int x^5e^{5x}dx= (1/5)x^5e^{5x}- ((1/5)x^4e^{5x}- (4/5)\int x^3e^{5x}dx[/tex]
    [tex]= (1/5)x^4e^{5x}- (1/5)x^4e^{5x}+ (4/5)\int x^3e^{5x}dx[/tex]
    where you have reduce the power of x to 3.

    Continue until you reduce the power of x to 0 and then integrate the exponential.
     
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