Integration of a func raised to a fraction

[mg0stisha]

(1)(u^\frac{1}{7})=u^\frac{1}{7}
(u)(u^\frac{1}{7})=u^\frac{8}{7}
therefore we have
u^\frac{1}{7}-u^\frac{8}{7}

 

[VietDao29]

(1)(u^\frac{1}{7})=u^\frac{1}{7}
(u)(u^\frac{1}{7})=u^\frac{8}{7}
therefore we have
u^\frac{1}{7}-u^\frac{8}{7}

And if you integrate this, what will you get? Note that:

\int x ^ {\alpha} dx = \frac{x ^ {\alpha + 1}}{\alpha + 1} + C, \mbox{ where } \alpha \in \mathbb{R} \backslash \{ - 1\}.

 

[mg0stisha]

So we'll integrate \int u^\frac{1}{7}du - \int u^\frac{8}{7}du

and we get \frac{7u^\frac{8}{7}}{8} - \frac{7u^\frac{15}{7}}{15} correct?

Hopefully.

 

[VietDao29]

So we'll integrate \int u^\frac{1}{7}du - \int u^\frac{8}{7}du

and we get \frac{7u^\frac{8}{7}}{8} - \frac{7u^\frac{15}{7}}{15} correct?

Hopefully.

Yup, correct, except that you're missing the "Constant of Integration" + C.

But the problem does not ask you to integrate ∫(1 - u)u^1/7du. The original problem is:

\int x \left( \frac{1 - x}{3} \right) ^ {\frac{1}{7}} dx

So, what you have to do next is to change u back to x, add the "Constant of Integration", and also, don't forget a constant that I took out from the beginning so that the integral looks easier.

 

[mg0stisha]

So literally right now we have \frac{1}{\sqrt[7]{3}}(\frac{7(1-x)^\frac{8}{7}}{8} - \frac{7(1-x)^\frac{15}{7}}{15}) +C?

 

[VietDao29]

So literally right now we have \frac{1}{\sqrt[7]{3}}(\frac{7(1-x)^\frac{8}{7}}{8} - \frac{7(1-x)^\frac{15}{7}}{15}) +C?

Congratulations, now that you've finally solved it. :)

Oh, and btw, you can make your answer looks a little bit nicer by pulling the 7 out. Like this:

{\color{red}-} \frac{7}{\sqrt[7]{3}}\left( \frac{(1-x)^\frac{8}{7}}{8} - \frac{(1-x)^\frac{15}{7}}{15} \right) +C

----------------

EDIT: Wait, you are forgetting a minus sign, when finding du = -dx.

 

[mg0stisha]

Nothing further!? thank god :) Thanks a ton guys (Cyosis, VietDao29, Euclidus). I might have been in a little over my head with this one, but thanks for not giving up on me!

P.S. I bought an algebra refresher book. Much needed. Thanks for being honest Cyosis :)

EDIT: thanks, i remember the negative sign being there now!

 

[Elucidus]

You are very welcome.

--Elucidus

 

[VietDao29]

Nothing further!? thank god :) Thanks a ton guys (Cyosis, VietDao29, Euclidus). I might have been in a little over my head with this one, but thanks for not giving up on me!

P.S. I bought an algebra refresher book. Much needed. Thanks for being honest Cyosis :)

One minor mistake left is that you have forgotten a minus sign, when finding du = -dx. There should be a minus sign also. I'm sorry for not spotting this out earlier. =.="

Just take a break now. When you fell like it, try to do this problem again. You've already had the result. Just try to see if you can solve it from scratch. :)