# Evaluating this indefinite integral

• Feb 25th 2009, 04:23 PM
fattydq
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?
• Feb 25th 2009, 04:28 PM
Reckoner
Quote:

Originally Posted by fattydq
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.
• Feb 25th 2009, 04:35 PM
Prove It
Quote:

Originally Posted by fattydq
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 :)
• Feb 25th 2009, 04:36 PM
fattydq
Quote:

Originally Posted by Reckoner
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
$\int{(3x - 7)^{1.7}\,dx} = \frac{1}{3}\int{(3x - 7)^{1.7}\cdot 3\,dx}$

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?
• Feb 25th 2009, 05:01 PM
Reckoner
Quote:

Originally Posted by fattydq
So my final answer would just be 1/3 times (3x-7)^2.7/2.7?

Yes, plus an arbitrary constant.