im sure my questions are getting frustrating and i'm sorry, but i don't see where the 2/3 comes from after 1/3 derivative sqrt U du.
sqrt u was converted to the form u^3/2 and i understand how that happened. but i dont get where the 2/3 is from. everything else i get. i'm sure its a simple rule im forgetting:sad:
Don't worry about being frustrating, ask until you understand!
Recall: \(\displaystyle f(x) = \int \! f'(x)\, \mathrm{d} x \)
So this means we can test any anti-derivative and see if it is correct by differentiating it again. If you get the integrand, great ! Your answer is correct.
So applied to your problem, what happens when we take the derivative of: \(\displaystyle u^{\frac{3}{2}}\)
Well we get: \(\displaystyle \frac{3}{2}\sqrt{u}\) but we want what is in the integrand! Which is \(\displaystyle \sqrt{u}\). So, that's easy, lets just multiply the anti-derivative by \(\displaystyle \frac{2}{3}\) to correct for this.
Now we get \(\displaystyle \frac{d}{du}(\frac{2}{3}u^{\frac{3}{2}})=\) \(\displaystyle \frac{2}{3}\frac{3}{2}\sqrt{u}\) = \(\displaystyle \sqrt{u}\) Which is our original integrand, meaning our anti-derivative is correct !
Formally:
\(\displaystyle \int \! x^n\, \mathrm{d} x =\frac{x^{n+1}}{n+1}+C\)
