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Math Help - Idea for general integral of a function of a linear

  1. #1
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    Idea for general integral of a function of a linear

    Hey, first post.
    I was looking through the textbook and their rules for the integral of a function done to a linear function, eg
    \int_{}^{} \sin{(ax+b)}dx
    or
    \int_{}^{} {(ax+b)}^{3}dx
    or
    \int_{}^{} \frac{1}{(ax+b)}dx
    all seemed to fit the general case
    \int_{}^{} f{(ax+b)}dx\:=\: \frac{1}{a}\times \int_{}^{} f{(ax+b)}d(ax+b)
    Simply, am I right?
    If so, is this a general rule that just isn't explained in high school?
    If not, could you please help me as to where I've gone wrong, or are there any amendments that would put me on the right track?
    I'm really enthusiastic about this sort of stuff, so any help/advice is great!
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  2. #2
    ynj
    ynj is offline
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    Quote Originally Posted by chug1 View Post
    Hey, first post.
    I was looking through the textbook and their rules for the integral of a function done to a linear function, eg
    \int_{}^{} \sin{(ax+b)}dx
    or
    \int_{}^{} {(ax+b)}^{3}dx
    or
    \int_{}^{} \frac{1}{(ax+b)}dx
    all seemed to fit the general case
    \int_{}^{} f{(ax+b)}dx\:=\: \frac{1}{a}\times \int_{}^{} f{(ax+b)}d(ax+b)
    Simply, am I right?
    If so, is this a general rule that just isn't explained in high school?
    If not, could you please help me as to where I've gone wrong, or are there any amendments that would put me on the right track?
    I'm really enthusiastic about this sort of stuff, so any help/advice is great!
    yes.It is right.
    In the most general case, \int f(x) \ dx=\int f\circ g(t)g'(t) \ dt.
    if you let g(t)=ax+b, it would become what you have said.
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  3. #3
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    Cheers mate
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  4. #4
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    It is explained in high school (at least in beginning Calculus courses). It a simple application of "substitution". In each of those, let u= ax+ b. Then du= a dx so dx= \frac{1}{a} du.
    The integral \int f(ax+ b)dx becomes \frac{1}{a}\int f(u) du which is exactly what you have.
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  5. #5
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    Ah I it recognise now, substitution certainly is in the course, but I guess because what I did was a generalisation of their specific cases it didn't occur to me that it could be found using their reasoning (substitution), lest they would have already done so.
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  6. #6
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    You are right, however, that substituting for a linear term is especially easy and worth noting for itself. That probably is not emphasized enough. It also helps to make the point that "substitution" is just the reverse of the "chain rule" for differentiation. Just as, when differentiating f(ax+ b), you multiply by the derivative of ax+b, a, so when integrating, you divide by it.

    The problem with non-linear terms is that their derivatives involve the variable and you can't just "move" the variable in and out of the integral- it has to already be in the integral.-
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  7. #7
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    Aye, that reversal does make sense.
    A teacher and I were also discussing whether you could or couldn't take out the front 1/ the derivative of the inner function for higher order functions, as, like you said, it would still contain the variable. Thanks for clearing that one up also!
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