J Jason76 Senior Member Joined Oct 19, 2012 Messages 1,180 Aug 10, 2013 #1 \(\displaystyle \int(x - 2)\sqrt{x} \. dx\) Should we go with "integration by parts" on this one? Also how can i make a blank space before dx ? Last edited: Aug 10, 2013
\(\displaystyle \int(x - 2)\sqrt{x} \. dx\) Should we go with "integration by parts" on this one? Also how can i make a blank space before dx ?
MarkFL Super Moderator Staff member Joined Nov 24, 2012 Messages 3,021 Aug 10, 2013 #2 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. Last edited: Aug 10, 2013
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.
J Jason76 Senior Member Joined Oct 19, 2012 Messages 1,180 Aug 10, 2013 #3 MarkFL said: 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. Click to expand... This is a trick question. A lot of people would be thinking of some more difficult way to solve.
MarkFL said: 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. Click to expand... This is a trick question. A lot of people would be thinking of some more difficult way to solve.
