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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- Feb 25th 2009, 03:23 PMfattydqEvaluating 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?

- Feb 25th 2009, 03:28 PMReckoner
- Feb 25th 2009, 03:35 PMProve It
$\displaystyle \int{(3x - 7)^{1.7}\,dx} = \frac{1}{3}\int{(3x - 7)^{1.7}\cdot 3\,dx}$

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

So the integral becomes

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

I trust you can go from here :) - Feb 25th 2009, 03:36 PMfattydq
- Feb 25th 2009, 04:01 PMReckoner