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Math Help - Integration by recognition, explanation needed

  1. #1
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    Integration by recognition, explanation needed

    I am trying to understand the example in my book, however can't seem to makc sense of it. If someone could just explain it to me.

     \frac{d}{dx}(3x-2)^{5} = 15(3x-2)^{4}

     \int 15(3x-2)^{4} dx = (3x-2)^{5} + C

    the next line I don't get, why do they divide both sides by 15? What's the need?

    the book says 'an adjustment factor of 1/15 is needed'.

     \int (3x-2)^{4} dx = \frac{1}{15}(3x-2)^{5} + c
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  2. #2
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    Quote Originally Posted by Tweety View Post
    I am trying to understand the example in my book, however can't seem to makc sense of it. If someone could just explain it to me.

     \frac{d}{dx}(3x-2)^{5} = 15(3x-2)^{4}

     \int 15(3x-2)^{4} dx = (3x-2)^{5} + C

    the next line I don't get, why do they divide both sides by 15? What's the need?

    the book says 'an adjustment factor of 1/15 is needed'.

     \int (3x-2)^{4} dx = \frac{1}{15}(3x-2)^{5} + c
    If evaluating the above line only, Tweety...

    \int{(3x-2)^4}dx=\frac{1}{15}(3x-2)^5+C

    because of \frac{d}{dx}\left[(3x-2)^5+C\right]=15(3x-2)^4

    hence \frac{1}{15}\ \frac{d}{dx}\left[(3x-2)^5+C\right]=(3x-2)^4

    therefore \int{(3x-2)^4}dx=\frac{1}{15}\left[(3x-2)^5+C\right]

    You can try it using substitution of course... u=3x-2,\ du=3dx,\ \frac{du}{3}=dx

    \int{u^4}\frac{du}{3}=\frac{1}{3}\int{u^4}du=\frac  {1}{3}\ \frac{1}{5}\ u^5+c=\frac{1}{15}(3x-2)^5+C
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  3. #3
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    Quote Originally Posted by Archie Meade View Post
    If evaluating the above line only, Tweety...

    \int{(3x-2)^4}dx=\frac{1}{15}(3x-2)^5+C

    because of \frac{d}{dx}\left[(3x-2)^5+C\right]=15(3x-2)^4

    hence \frac{1}{15}\ \frac{d}{dx}\left[(3x-2)^5+C\right]=(3x-2)^4

    therefore \int{(3x-2)^4}dx=\frac{1}{15}\left[(3x-2)^5+C\right]

    You can try it using substitution of course... u=3x-2,\ du=3dx,\ \frac{du}{3}=dx

    \int{u^4}\frac{du}{3}=\frac{1}{3}\int{u^4}du=\frac  {1}{3}\ \frac{1}{5}\ u^5+c=\frac{1}{15}(3x-2)^5+C

    Thank you,

    However I am still unclear about this, because we are trying to work out the  \int   15(3x-2)^{4} dx

    not   \int (3x-2)^{4} dx

    for example,  \frac{d}{dx} sin3x = cos3x \times 3

    so

     \int cos3x \times 3dx  = sin3x + c

    so shouldn't it just be  \int 15(3x-2)^{4} dx = (3x-2)^{5} + C ?
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  4. #4
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    Quote Originally Posted by Tweety View Post
    Thank you,

    However I am still unclear about this, because we are trying to work out the  \int   15(3x-2)^{4} dx

    not   \int (3x-2)^{4} dx

    for example,  \frac{d}{dx} sin3x = cos3x \times 3

    so

     \int cos3x \times 3dx  = sin3x + c

    so shouldn't it just be  \int 15(3x-2)^{4} dx = (3x-2)^{5} + C ?
    Hi Tweety,

    I think the book example is calculating \int{(3x-2)^4}dx using the result from \int{15(3x-2)^4}dx

    since \int{15(3x-2)^4}dx=15\int{(3x-2)^4}dx
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