Integration of a func raised to a fraction

[mojomike]

Homework Statement

I need to solve the following integral:

{ x * ((1-x)/3)^(1/7) dx

where { represents the integral symbol b/c I don't know how to type that.

The Attempt at a Solution

I've tried integration by parts and don't get anywhere. None of the integration techiques I know seem to work and nothing I've found online says what to do when you have this kind of a situation.

Any help would be greatly appreciated. I'm lost. Thanks everyone!

,Mike

 

[LeonhardEuler]

Can you think of a substitution that would make the (1/7) power part easy to integrate?

 

[mg0stisha]

I don't think he's coming back (as that was his only post), but I'm relatively new to calc and would like help solving this. Anyone willing to help?

 

[mojomike]

Thanks for the response,

Well, I tried to substitute that inside the parenthesis, but there's that blasted extra x outside of it to deal with. In integrating by parts, I made x the integration term(v) and the other part the differential term, but the result is just another integral only more complicated.

 

[VietDao29]

Homework Statement

I need to solve the following integral:

{ x * ((1-x)/3)^(1/7) dx

where { represents the integral symbol b/c I don't know how to type that.

The Attempt at a Solution

I've tried integration by parts and don't get anywhere. None of the integration techiques I know seem to work and nothing I've found online says what to do when you have this kind of a situation.

Any help would be greatly appreciated. I'm lost. Thanks everyone!

,Mike

Homework Statement

Homework Equations

The Attempt at a Solution

No, you don't use Integration by Parts in this problem.

To type maths beautifully, one should learn how to https://www.physicsforums.com/showthread.php?t=8997". And you can begin to dig in the awesome world of LaTeX today.

So, your integral is:

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

The first step is to take all constants out of it:

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

The product of (1 - x)^{1 / 7} and x is a little bit complicated to integrate, right? Because you just cannot expand the part (1 - x)^{1 / 7}.

If the problem was: \int x \left( 1 - x \right) ^ {2} dx, then it would be much easier, because we can expand (1 - x)² = 1 - 2x + x². And the whole integral would become: \int (x ^ 3 - 2 x ^ 2 + x) dx, which should be like, a piece of cake.

So, back to the problem, we then think about a simple u-substitution, which eliminates the needs to expand (1 - x)^{1 / 7}. Let's see if you can figure out what u should be. :)

 

[mg0stisha]

Would it be u=1/7?

 

[Elucidus]

Would it be u=1/7?

Letting u equal a constant implies du = 0 and the integral is trivial. The difficulty with the integrand was the (1 - x) being raised to a fractional power. If the base were simpler...


--Elucidus

 

[mg0stisha]

So right now we have <br /> \frac{1}{\sqrt[7]{3}} \int x \left( 1 - x \right) ^ {\frac{1}{7}} dx<br /> correct?
i guess we could distribute the x in front of the parentheses so we'll have<br /> \frac{1}{\sqrt[7]{3}} \int \left( x - x^2 \right) ^ {\frac{1}{7}} dx<br />? then simplify that and pull the negative sign out front? but then there's no substitution...

the only other idea would be to make a substitution so that u=(1-x)?

 

[darkmagic]

Bring all constants outside then integrate. Use integration by parts. u=x, du=dx,
dv=(1-x)^1/7 dx, v=-7/8(1-x)^8/7. Any corrections if I integrated it wrong.

 

[Cyosis]

So right now we have <br /> \frac{1}{\sqrt[7]{3}} \int x \left( 1 - x \right) ^ {\frac{1}{7}} dx<br /> correct?
i guess we could distribute the x in front of the parentheses so we'll have<br /> \frac{1}{\sqrt[7]{3}} \int \left( x - x^2 \right) ^ {\frac{1}{7}} dx<br />? then simplify that and pull the negative sign out front? but then there's no substitution...

This is not correct, you cannot distribute unequal powers. Note that x=(x^7)^{\frac{1}{7}}. Now that they are equal powers you can distribute, which would yield x(1-x)^{\frac{1}{7}}=(x^7-x^8)^{\frac{1}{7}}. This doesn't make things any easier.
As hinted before use u=1-x.

 

[Elucidus]

Bring all constants outside then integrate. Use integration by parts. u=x, du=dx,
dv=(1-x)^1/7 dx, v=-7/8(1-x)^8/7. Any corrections if I integrated it wrong.

To go from dv=(1-x)^{1/7}dx \text{ to }v=-\frac{7}{8}(1-x)^{8/7} requires integration by substitution (in fact, using the same substitution that would solve it directly). The integral can be done more simply than by integration by parts - although "by parts" is valid.

--Elucidus

 

[mg0stisha]

I just can't figure it out. I've never formally taken a calculus class, just taught myself out of a cheap self teach book from barnes and noble. I substituted so that u = (1-x) and now have \frac{1}{\sqrt[7]{3}} \int x \left( u\right) ^ {\frac{1}{7}} dx. And, once again I'm stuck.

 

[LeonhardEuler]

If u= 1- x, then what is 'x'? You need to make the substitution everywhere there is an 'x'. This means you must also integrate du instead of dx.

 

[Cyosis]

As pointed out you need to express every x in terms of u, including the differential. If you substitute u=f(x), then du=f'(x)dx. In your case f(x)=1-x. Can you continue from here?

 

[mg0stisha]

u= 1-x
du= -1
correct?

 

[Cyosis]

No, if f(x)=1-x and du=f'(x)dx, then f'(x)=? and du=?

 

[mg0stisha]

The derivative of 1-x isn't -1?

 

[Cyosis]

Yes that part is correct, however incomplete. You want to express the differential, du in terms of the differential dx.

 

[mg0stisha]

du/dx = -1? i have no clue why I'm so lost right now.

 

[Cyosis]

If you substitute u=f(x), then du=f'(x)dx

It's not much more than plug and play. Two differentials are related as du=\frac{du}{dx}dx. This is the exact same as the one in my qu0te, just with a different notation. Use these properties to find du in terms of x.

 

[mg0stisha]

I'm having huge problems, I'm going to just work it out visually.
So if du=f'(x)dx and du=\frac{du}{dx}dx, then f'(x)=du/dx, correct?

EDIT: That post was completely pointless. Let me work on this. Sorry.

 

[mg0stisha]

du = -1dx

 

[Cyosis]

Yep that's correct. Your previous post although pointless perhaps was correct as well. So now that you have found the new differential du in terms of dx can you write down the the new, u dependent, integral?

 

[mg0stisha]

I'll do my best. This piece of crap work computer won't let me copy the LaTeX code, so i'll try my best to re-write that on my own.

\frac{1}{\sqrt[7]{3}} \int x(u)^\frac{1}{7}du

where u=1-x and du=-1dx

 

[Cyosis]

No that's not right. You want to replace all old x variables with u variables. You now have a mix of u and x variables which is impossible to integrate.

The orginal integral:
\frac{1}{\sqrt[7]{3}} \int x \left( 1 - x \right) ^ {\frac{1}{7}} dx<br />

You want to write all these x-variables in terms of u.
If u=1-x then x=?
If du=-dx then dx=?

 

[mg0stisha]

Oh, okay. So then x=1-u and dx=-du.

\frac{1}{\sqrt[7]{3}} \int 1-u(1-u)^\frac{1}{7}(-du)

That doesn't look right..

 

[Cyosis]

How did you get the (1-u)^{\frac{1}{7}} term?

 

[mg0stisha]

btw, if you click on the LaTeX code reference, that is what i want the integral to be, why does it show up different on the actual image?

 

[mg0stisha]

Yes that was wrong as well. it should be (1-(1-u))^1/7 shouldn't it?

 

[Cyosis]

Looking at the LaTeX code the actual image is correct.

 

[Cyosis]

Are you just guessing now?

You have
u=1-x
x=1-u
dx=-du.

Put these values into the original integral.

 

[mg0stisha]

No, I wasn't guessing. The original integral is \frac{1}{\sqrt[7]{3}} \int x \left( 1 - x \right) ^ {\frac{1}{7}} dx. So plugging in x=1-u and dx=-du, I see \frac{1}{\sqrt[7]{3}} \int 1-u \left( 1 - (1-u) \right) ^ {\frac{1}{7}} -du

 

[Cyosis]

You're getting very close now, although you're taking a detour. You need some more brackets though, especially the first term seems awkward.

 

[mg0stisha]

Would you plug -x=-1+u into the parentheses so that everything cancels to simply u?

And I feel like I'm wasting, if this isn't the case that's awesome, i'll work with this as long as you want to. If so, feel free to tell me at anytime and i'll stop/seek help from someone else. Thanks!

\frac{1}{\sqrt[7]{3}} \int (1-u) \left( 1 -1+u) \right) ^ {\frac{1}{7}} -du = \frac{1}{\sqrt[7]{3}} \int (1-u) \left(u) \right) ^ {\frac{1}{7}} -du

 

[Cyosis]

We're not going to stop.

Lets do it in steps.
We started out with the substitution u=1-x, using this substitution the integral becomes (ignoring the constant in front of it):

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

But we need all x-variables in terms of u-variables. We also know that dx=-du so let's put that into the integral as well. The integral now becomes:

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

Now we only have one x left and we know that x=1-u. Can you do the final step now by getting rid of the last x?

 

[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:
<br /> -\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