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Math Help - Evaluating this indefinite integral

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    Evaluating this indefinite integral

    It's the indefinite integral of (3x-7)^1.7 dx I don't even know where I'd begin finding that antiderivative, maybe substitution of some sort? Could someone walk me through this?
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    Quote Originally Posted by fattydq View Post
    It's the indefinite integral of (3x-7)^1.7 dx I don't even know where I'd begin finding that antiderivative, maybe substitution of some sort? Could someone walk me through this?
    Let u=3x-7 and use the power rule. This one is pretty easy.
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    Quote Originally Posted by fattydq View Post
    It's the indefinite integral of (3x-7)^1.7 dx I don't even know where I'd begin finding that antiderivative, maybe substitution of some sort? Could someone walk me through this?
    \int{(3x - 7)^{1.7}\,dx} = \frac{1}{3}\int{(3x - 7)^{1.7}\cdot 3\,dx}

    Let u = 3x - 7 so that \frac{du}{dx} = 3.

    So the integral becomes

    \frac{1}{3}\int{u^{1.7}\frac{du}{dx}\,dx} = \frac{1}{3}\int{u^{1.7}\,du}.

    I trust you can go from here
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    Quote Originally Posted by Reckoner View Post
    Let u=3x-7 and use the power rule. This one is pretty easy.
    So it would end up being 1/3 the integral of u^1.7 which would end up being (u^2.7)/2.7 times 1/3, with u resub'd back in? so 1/3 times (3x-7)^2.7 over 2.7?

    Quote Originally Posted by Prove It View Post
    \int{(3x - 7)^{1.7}\,dx} = \frac{1}{3}\int{(3x - 7)^{1.7}\cdot 3\,dx}

    Let u = 3x - 7 so that \frac{du}{dx} = 3.

    So the integral becomes

    \frac{1}{3}\int{u^{1.7}\frac{du}{dx}\,dx} = \frac{1}{3}\int{u^{1.7}\,du}.

    I trust you can go from here
    So my final answer would just be 1/3 times (3x-7)^2.7/2.7?
    Last edited by mr fantastic; February 26th 2009 at 02:52 AM. Reason: Merged posts
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    Quote Originally Posted by fattydq View Post
    So my final answer would just be 1/3 times (3x-7)^2.7/2.7?
    Yes, plus an arbitrary constant.
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