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 | 1 | initial version | posted 2014-07-17 11:12:06 -0600  |
Okay, I just realized that I didn't change by bounds of integration when I did my $u$ substitution. It should be: $$1/3 \int_1^2 \sqrt{u} du$$ $$1/3(2/3 \sqrt{u^3} ) \biggr|_1^2$$ $$=4/9 \sqrt{2}-2/9$$ simple mistake!
| | 2 | No.2 Revision |
Okay, I just realized that I didn't change by bounds of integration when I did my $u$ substitution. It should be: $$1/3 \int_1^2 \sqrt{u} \hspace{2 mm} du$$ $$1/3(2/3 \sqrt{u^3} ) \biggr|_1^2$$ $$=4/9 \sqrt{2}-2/9$$ simple mistake!
