Integration of a func raised to a fraction

[mg0stisha]

$$-\int 1-u u^\frac{1}{7}du$$ or $$-\int 1-u^\frac{2}{7}du$$?

Still stuck.

 

[Cyosis]

No, x is getting multiplied by u^(1/7) so with whatever we replace x that something needs to be multiplied by u^(1/7) as a whole. What you have done here is multiplying only a part of x instead of all of it.

To break it down in a number example this is what you've done. We have two numbers 5=1+4 and 6. We want to multiply them together.
5*6=30
Now we do it your way by replacing 5 with 1+4, which should be perfectly valid.
5*6=1+4*6=25?

Do you see what's going wrong now?

 

[mg0stisha]

I understand it completely in the example you gave, not 100% sure about in the integral. In this case, we have x and u which we want to multiply together. and x=1-u.

so for the above example, x*u=xu
and I'm doing it x*u= (1-u)(u) = (u-u^2)=-u?

i basically just tried pluging our variables into your example, didnt quite work out i don't think.

 

[Elucidus]

$$-\int 1-u u^\frac{1}{7}du$$ or $$-\int 1-u^\frac{2}{7}du$$?

Still stuck.

The first integral is so close to correct. You are correct that u = 1 - x implies x = 1 - u. When you substitute, it is always a good idea to encapsulate the substituted expression in parentheses in order to avoid sign and distribution errors (which is what happened here).

Once parentheses are put in the proper place, this integral should sail to the answer.

--Elucidus

 

[mg0stisha]

$$-\int (1-u) u^\frac{1}{7}du$$

Can you simply just integrate and simplify from here?

 

[Elucidus]

$$-\int (1-u) u^\frac{1}{7}du$$

Can you simply just integrate and simplify from here?

Well, simplify (distribute), integrate (power rule), and then revert to the original context of x, yes.

--Elucidus

 

[mg0stisha]

And by distributing all I'm coming up with is u^(1/7) - u^(2/7) du...

 

[Elucidus]

And by distributing all I'm coming up with is u^(1/7) - u^(2/7) du...

Be careful, $u=u^1=u^{(7/7)}$

--Elucidus

 

[VietDao29]

And by distributing all I'm coming up with is u^(1/7) - u^(2/7) du...

1 + 1/7 is not 2/7 man.. Be careful.. After distributing, hopefully, you can go from there, right?

 

[mg0stisha]

Yes, I can integrate this easily after i distribute, I'm not sure why I'm having such a hard time

$$\int u^\frac{7}{7}-u^\frac{8}{7}du = \int -u^\frac{1}{7}du = -\int u^\frac{1}{7}du$$

Integrating gives $$\frac{u^\frac{8}{7}}{7}u$$ or $$\frac{u^\frac{15}{7}}{7}$$

yes or no?

 

[Cyosis]

I don't want to put you down, but your issues are not with calculus. The problem is that your mathematical basis seems very weak. You should really review symbolic/fractional subtraction,multiplication,division and addition.

Lets provide you with an example again. Take x^4 and x^2. According to you x^4-x^2=x^2. Let's write it out:

x*x*x*x-x*x=x*x

That certainly doesn't look right, let's put in a number, 2.
2*2*2*2-2*2=12!=4
You cannot simplify that expression.

If they had been multiplied then you could simplify the expression.
Example:
x^4*x^2=(x*x*x*x)*(x*x)=x*x*x*x*x*x=x^6

This is the integral you want to evaluate:
$$-\int u^\frac{1}{7}-u^\frac{8}{7}du=\int -u^\frac{1}{7}du+\int u^\frac{8}{7}du$$
There is no more simplification to be done from this point, you can integrate the fractional terms in the same way you integrate an expression like u^2.

 

[mg0stisha]

Know of anyway I can do that before I start my calc class Monday?

 

[Cyosis]

Maybe this page is useful to you http://online.math.uh.edu/Math1330/index.html .

On another note how's the integral going?

 

[mg0stisha]

Thanks! Well the first integral would be (u^(-2))/2 I believe, and I think the secnd one would be (u^(15/7))/7 ? I wouldve typed it in LaTeX but I'm on my phone, sorry.

 

[Elucidus]

Small snag in your plans, methinks.

$$(1-u)u^{1/7} \neq u^{7/7}-u^{8/7}.$$

In slow motion, the distribution goes:

$$(1-u)u^{1/7} = (1)(u^{1/7})-(u)(u^{1/7}).$$

Can you get the rest of the way?

--Elucidus

 

[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. :)