1. ## Easy Integral

I have an integral that I think is probably super easy, but I'm just totally missing the point...

integrate: x(1-x)^10

I can't substitute anything here, can I? It looks like questions I've done a thousand times before, but this one just isn't clicking...

2. Sure you can: $\displaystyle {\color{red}u = 1 - x} \ \Rightarrow \ du = - dx \ \Leftrightarrow \ {\color{blue}-du = dx}$

Also note that if you rearrange the red, we have: $\displaystyle \color{magenta}x = 1 - u$

So: $\displaystyle \int {\color{magenta}x}({\color{red}1-x})^{10} \ {\color{blue}dx} = \int ({\color{magenta}1-u}) {\color{red}u}^{10} ({\color{blue}-du}) = \cdots$

3. Originally Posted by o_O
Sure you can: $\displaystyle {\color{red}u = 1 - x} \ \Rightarrow \ du = - dx \ \Leftrightarrow \ {\color{blue}-du = dx}$

Also note that if you rearrange the red, we have: $\displaystyle \color{magenta}x = 1 - u$

So: $\displaystyle \int {\color{magenta}x}({\color{red}1-x})^{10} \ {\color{blue}dx} = \int ({\color{magenta}1-u}) {\color{red}u}^{10} ({\color{blue}-du}) = \cdots$

Ok, that is tricky. So then I would multiply $\displaystyle ({\color{magenta}1-u}) {\color{red}u}^{10}$, and then integrate?

4. Yep: $\displaystyle (1-u)u^{10} = u^{10} - u^{11}$

And I'm sure you're familiar with integrating powers

5. Originally Posted by Hellreaver
Ok, that is tricky. So then I would multiply $\displaystyle ({\color{magenta}1-u}) {\color{red}u}^{10}$, and then integrate?
You can also do it by parts. Although, that's probably more time consuming.