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Thread: integration power rule

  1. #1
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    integration power rule

    the integral of $\displaystyle \int\frac{x}{\sqrt{4-x^2}} dx$ is $\displaystyle -\sqrt{4-x^2} + C$.

    this book shows a middle step that i don't understand: $\displaystyle -\frac{1}{2} \int (4-x)^{-\frac{1}{2}}(-2x) dx$

    i'm trying to use power rule, but where did that $\displaystyle -\frac{1}{2}$ come from that is before the $\displaystyle \int$?
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by dataspot View Post
    the integral of $\displaystyle \int\frac{x}{\sqrt{4-x^2}} dx$ is $\displaystyle -\sqrt{4-x^2} + C$.

    this book shows a middle step that i don't understand: $\displaystyle -\frac{1}{2} \int (4-x)^{-\frac{1}{2}}(-2x) dx$

    i'm trying to use power rule, but where did that $\displaystyle -\frac{1}{2}$ come from that is before the $\displaystyle \int$?
    Consider the derivative of the inside of the radical?

    Or more conservatively

    Let $\displaystyle \psi=4-x^2$
    so then $\displaystyle d\psi=-2xdx$

    now looking at our integral we have

    $\displaystyle \frac{-1}{2}\int\frac{d\psi}{\sqrt{\psi}}=\frac{-1}{2}\cdot{2\sqrt{\psi}}$

    Now making our back sub we get

    $\displaystyle -\sqrt{4-x^2}$
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  3. #3
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by dataspot View Post
    the integral of $\displaystyle \int\frac{x}{\sqrt{4-x^2}} dx$ is $\displaystyle -\sqrt{4-x^2} + C$.

    this book shows a middle step that i don't understand: $\displaystyle -\frac{1}{2} \int (4-x)^{-\frac{1}{2}}(-2x) dx$

    i'm trying to use power rule, but where did that $\displaystyle -\frac{1}{2}$ come from that is before the $\displaystyle \int$?
    It's not that obvious at first. But its easy to understand.

    $\displaystyle \int\frac{x}{\sqrt{4-x^2}}\,dx$

    Let $\displaystyle u=4-x^2\implies\,du=-2x\,dx$

    Note that in the integral, we have $\displaystyle \int\frac{{\color{red}x}}{\sqrt{4-x^2}}\color{red}\,dx$, not $\displaystyle -2x\,dx$. To make the $\displaystyle -2x\,dx$ appear, multiply the integral by $\displaystyle \frac{-2}{-2}$ (equivalent to 1)

    Thus, we get the following:

    $\displaystyle \frac{-2}{-2}\int\frac{x}{\sqrt{4-x^2}}\,dx\implies {\color{red}\frac{-1}{2}}\int\frac{-2x}{\sqrt{4-x^2}}\,dx$.

    Now make the substitutions:

    $\displaystyle \frac{-1}{2}\int u^{-\frac{1}{2}}\,du$.

    Hope this makes sense!
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by dataspot View Post
    the integral of $\displaystyle \int\frac{x}{\sqrt{4-x^2}} dx$ is $\displaystyle -\sqrt{4-x^2} + C$.

    this book shows a middle step that i don't understand: $\displaystyle -\frac{1}{2} \int (4-x)^{-\frac{1}{2}}(-2x) dx$

    i'm trying to use power rule, but where did that $\displaystyle -\frac{1}{2}$ come from that is before the $\displaystyle \int$?
    the -1/2 is to account for the -2 that was put in front of the x (noticed the -2x?). what they did was multiply the x by -2 and then divide by -2 (hence the -1/2) so as to not change anything. why? because -2x is the derivative of 4 - x^2. why do we care? well, in this set up, it is "easy" to see that we have an expression that is the result of using the chain rule to differentiate a function. basically what happened here is doing substitution without actually saying "let u = 4 - x^2." this is a more clever and concise way of doing it


    EDIT: Beaten again...twice...and with better explanations too good job guys
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    Quote Originally Posted by Chris L T521 View Post
    It's not that obvious at first. But its easy to understand.

    $\displaystyle \int\frac{x}{\sqrt{4-x^2}}\,dx$

    Let $\displaystyle u=4-x^2\implies\,du=-2x\,dx$

    Note that in the integral, we have $\displaystyle \int\frac{{\color{red}x}}{\sqrt{4-x^2}}\color{red}\,dx$, not $\displaystyle -2x\,dx$. To make the $\displaystyle -2x\,dx$ appear, multiply the integral by $\displaystyle \frac{-2}{-2}$ (equivalent to 1)

    Thus, we get the following:

    $\displaystyle \frac{-2}{-2}\int\frac{x}{\sqrt{4-x^2}}\,dx\implies {\color{red}\frac{-1}{2}}\int\frac{-2x}{\sqrt{4-x^2}}\,dx$.

    Now make the substitutions:

    $\displaystyle \frac{-1}{2}\int u^{-\frac{1}{2}}\,du$.

    Hope this makes sense!
    ok...

    if the resulting $\displaystyle du$ had equaled $\displaystyle x dx$, then you wouldn't have had to multiply by 1 (rather, $\displaystyle \frac{-2}{-2}$), but since it didn't, you had to even things out.

    yeah?
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  6. #6
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    Quote Originally Posted by Chris L T521 View Post
    Now make the substitutions:

    $\displaystyle \frac{-1}{2}\int u^{-\frac{1}{2}}\,du$.
    i'm not sure i could finish the problem from here either. been a while.
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by dataspot View Post
    i'm not sure i could finish the problem from here either. been a while.
    from here, we use the power rule: $\displaystyle \int x^n~dx = \frac {x^{n + 1}}{n + 1} + C$ for $\displaystyle n \ne -1$

    so, $\displaystyle - \frac 12 \int u^{-1/2}~du = - \frac 12 \cdot \frac {u^{1/2}}{1/2} + C = -u^{1/2}$
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    Quote Originally Posted by dataspot View Post
    i'm not sure i could finish the problem from here either. been a while.
    Use the power rule for integration:

    $\displaystyle \int x^n\,dx = \frac1{n + 1}x^{n + 1} + C,\text{ for }n\neq-1$
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  9. #9
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    thanks for the help!!!
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