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Thread: 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.

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

    $\displaystyle \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'.

    $\displaystyle \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.

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

    $\displaystyle \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'.

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

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

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

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

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

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

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

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

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

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

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

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

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

    Thank you,

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

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

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

    so

    $\displaystyle \int cos3x \times 3dx = sin3x + c $

    so shouldn't it just be $\displaystyle \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 $\displaystyle \int 15(3x-2)^{4} dx $

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

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

    so

    $\displaystyle \int cos3x \times 3dx = sin3x + c $

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

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

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