How to Integrate a Square Root Expression with Limits?

[Math Monster]

Homework Statement

Integrate (x-x^2)^1/2 from 1/2 to 1.

I tried to use substitution with x=sinu, dx=cos du to get:
sinu(1-sinu)^1/2 cosu du
but no idea where to go from there!

Also tried integration by parts, but it didn't work!

Help!

 

[Dickfore]

Perform the Euler substitution:
[tex]
\left( x (1 - x) \right)^{\frac{1}{2}} = x \, t
[/tex]
[tex]
x (1 - x) = x^2 \, t^2
[/tex]
[tex]
1 - x = x \, t^2
[/tex]
[tex]
x = \frac{1}{1 + t^2} \Rightarrow \left(x (1 - x) \right)^{\frac{1}{2}} = x \, t = \frac{t}{1+t^2}, \ dx = -\frac{2 \, t \, dt}{(1 + t^2)^2}
[/tex]
[tex]
x = \frac{1}{2} \Rightarrow t = 1, \ x = 1 \Rightarrow t = 0
[/tex]
So, your integral becomes:
[tex]
I = \int_{1}^{0}{\frac{t}{1 + t^2} \, \frac{-2 t}{(1 + t^2)^2} \, dt} = 2 \, \int_{0}^{1}{\frac{t^2}{(1 + t^2)^3} \, dt}
[/tex]

Try using integration by parts on this integral. What should your u and v be?

 

[Math Monster]

That is a great help I think!

I have worked through what you did, and got:

u=t^2
du = 2t dt

dv = (1+t^2)^-3
v = (1+t^2)^-2 / -4t

so integral of udv =
((t^2)(1+t^2)^-2) / (-4t) evaluated 0-1
+ integral_0 ^1 (1+t^2)^(-2) / -2 dt (the -4t from v has canceled with the 2t from du)

If this is right, I'm able to finish it off from here!

Thanks very much. I've never seen the Euler method before, if you can provide a bit more information how to use it in other examples I would be grateful.

Kind regards

Harriet

 

[Dickfore]

Your v does not follow from your dv. It seems you don't know how integration works.

 

[Math Monster]

To work out v i:

1. Increased the power of the bracket from -3 to -2.
2. Divide by the new power (-2)
3. Divide by the differential of the bracket (2t)

Could you explain where I went wrong? Thanks Harriet

 

[Dickfore]

In step 3. Notice that "the differential of the bracket" depends on t, your dummy variable.

 

[Mentallic]

dv = (1+t^2)^-3
v = (1+t^2)^-2 / -4t

If you went backwards and tried to differentiate v with respect to t, you'd have to use the quotient rule.

 

[Math Monster]

Oh okay, I'm a bit lost. I thought you could introduce the dummy variable and then differentiate/integrate as normal as you changed limits etc. Could you explain how it would be different?

Thanks

 

[Dickfore]

An alternative substitution is obtained by completing the square under the square root:
[tex]
x - x^2 = -\left(x^2 - 2 \frac{1}{2} x + \frac{1}{4} - \frac{1}{4} \right) = \left(\frac{1}{2}\right)^2 - \left( x - \frac{1}{2} \right)^2
[/tex]
Then, try the substitution:
[tex]
x - \frac{1}{2} = \frac{1}{2} \sin t
[/tex]

What are the new limits of integration? What does the integral transform to?

 

[Math Monster]

I rearranged to find the integral becomes 1/2cost
with t= 0, t=pi/2 for the new limits?

Any good?

 

[Dickfore]

The limits are correct. Where did 1/2 cos t come from?

 

[Math Monster]

Once you complete the square, substitute x-1/2 = 1/2sint to get 1/4 - 1/4sin^2 t
Factorise to get 1/4 ( 1-sin^2 t) = 1/4 cos^2 t
As we are looking at sqrt (x-x^2), take sqrt of this to get 1/2 cost

Does that make sense?

 

[Dickfore]

Once you complete the square, substitute x-1/2 = 1/2sint to get 1/4 - 1/4sin^2 t
Factorise to get 1/4 ( 1-sin^2 t) = 1/4 cos^2 t
As we are looking at sqrt (x-x^2), take sqrt of this to get 1/2 cost

Does that make sense?

Yes. So, it came from simplifying the square root. How did you transform dx then?

 

[Math Monster]

dx = 1/2 cos t ?

Completely forgot about that!

So integral should be 1/4 cos^2 t?

 

[Dickfore]

Yes. Now, when faced with a square of a sine or cosine, use the double-angle formula:
[tex]
\cos^2 t = \frac{1 + \cos 2 t}{2}
[/tex]
Then, your integral should become elementary.

 

[quackdesk]

I agree with Dickfore :) I had an example like that on my last exam :)