Homework Help: Integration question

1. Feb 23, 2005

ToxicBug

Could anyone evaluate this integral for me? I got it in an exam and didn't know how to do it.

$$\int\sqrt{x-x^2}$$

Last edited: Feb 23, 2005
2. Feb 23, 2005

vincentchan

let u = x-1/2 and do the substitude,...

3. Feb 23, 2005

ToxicBug

What do you mean? Can you evaluate it completely showing the steps?

4. Feb 23, 2005

Jameson

Ok, Mathematica was definitely wrong. There's going to be a trig substitution in this integral somewhere.

Last edited: Feb 23, 2005
5. Feb 23, 2005

ToxicBug

Mathematica is **** and I'm sure thats not a right answer.

6. Feb 23, 2005

vincentchan

what do I mean? I thought I make it very clear already!
$$\int\sqrt{x-x^2}dx=\int\sqrt{x(1-x)}dx$$

let $$u=x-\frac{1}{2}$$

It become:

$$\int\sqrt{(\frac{1}{2}-u)(\frac{1}{2}+u)}du$$

$$=\int\sqrt{\frac{1}{4}-u^2 }du$$

$$=\frac{u}{4}\sqrt{1-4u^2}+\frac{1}{8}sin^{-1}(2u)+C$$

$$=\frac{x-1/2}{4}\sqrt{1-4(x-1/2)^2}+\frac{1}{8}sin^{-1}(2x-1)+C$$

7. Feb 24, 2005

ToxicBug

Sick, thanks.

8. Feb 25, 2005

dextercioby

And one more thing:"Mathematica" is ALWAYS RIGHT...:grumpy:

Daniel.

9. Feb 25, 2005

HallsofIvy

Cute factoring.

What I would have done is "complete the square": x- x2= 1/4-1/4+ x-x2= 1/4-(x-1/2)2 which would have led me to the u= x- 1/2 substitution. Once I had it in the form $\sqrt{\frac{1}{4}- u^2}$ I would use a trig substitution.