Integration Square Root Problem

[Jason76]

\(\displaystyle \int(x - 2)\sqrt{x} \. dx\)

Should we go with "integration by parts" on this one?

Also how can i make a blank space before dx ?

 

[MarkFL]

You could simply distribute, and then integrate term by term:

\(\displaystyle \displaystyle \int(x-2)\sqrt{x}\,dx=\int x^{\frac{3}{2}}-2x^{\frac{1}{2}}\,dx\)

To insert the space, use \, instead.

 

[Jason76]

You could simply distribute, and then integrate term by term:

\(\displaystyle \displaystyle \int(x-2)\sqrt{x}\,dx=\int x^{\frac{3}{2}}-2x^{\frac{1}{2}}\,dx\)

To insert the space, use \, instead.

This is a trick question. A lot of people would be thinking of some more difficult way to solve.