# Thread: A tough definite integral

1. ## A tough definite integral

Is there any elementary reasoning to get $\displaystyle \displaystyle\int^1_0 \sqrt{x(1-x)}\ dx = \pi/8$? The actual antiderivative is too long, and I don't think computing it is the right approach for the question I am working on. Could there be a polar coordinate transformation? Since the answer has a $\displaystyle \pi$ in it. I just need a tiny hint, thanks.

2. Originally Posted by Crescent
Is there any elementary reasoning to get $\displaystyle \displaystyle\int^1_0 \sqrt{x(1-x)}\ dx = \pi/8$? The actual antiderivative is too long, and I don't think computing it is the right approach for the question I am working on. Could there be a polar coordinate transformation? Since the answer has a $\displaystyle \pi$ in it. I just need a tiny hint, thanks.

Note that $\displaystyle \displaystyle \int_0^1\sqrt{x(1-x)}dx=\frac{1}{2}\int_{-1}^{1}\sqrt{\left(x-\frac{1}{2}\right)^2-\frac{1}{2^2}}dx$, trying interpreting this geometrically as the area of a common shape.

3. Originally Posted by Drexel28
Note that $\displaystyle \displaystyle \int_0^1\sqrt{x(1-x)}dx=\frac{1}{2}\int_{-1}^{1}\sqrt{\left(x-\frac{1}{2}\right)^2-\frac{1}{2^2}}dx$, trying interpreting this geometrically as the area of a common shape.
Thanks for the hint! I used the result that $\displaystyle \sqrt{x(1-x)}=\sqrt{-(x-\frac{1}{2})^2+\frac{1}{4}}$ and the square root of parabola with negative leading coefficient and maximum point of positive coordinates is an ellipse (circle in this case) to get the result, that seems more intuitive to me.

4. Originally Posted by Crescent
Thanks for the hint! I used the result that $\displaystyle \sqrt{x(1-x)}=\sqrt{-(x-\frac{1}{2})^2+\frac{1}{4}}$ and the square root of parabola with negative leading coefficient and maximum point of positive coordinates is an ellipse (circle in this case) to get the result, that seems more intuitive to me.
Correct! So you can (sorry for forgetting the negative) realize this integral as the area of a semi-circle.

5. Does the fact that the integrand has the same value for both limits not help at all?

$\displaystyle \displaystyle \sqrt{x(1-x)}\bigg|_{x=0} = \sqrt{x(1-x)}\bigg|_{x=1}$ ... it's probably just a useless observation.

6. It tells you the graph is symmetric about x= 1/2. (It is, of course, a circle).

7. since i am newbie to this forum so it will take sometime to write $\displaystyle \displaystyle \sqrt{x(1-x)}\bigg|_{x=0} = \sqrt{x(1-x)}\bigg|_{x=1}$ in this style but it is very easy if u take polar coordinates........ just take $\displaystyle x=\sin^2 \theta$ and solve it u will get limit as lower limit=0 and upper limit = $\displaystyle \pi/2$.

8. Originally Posted by sushant1221
since i am newbie to this forum so it will take sometime to write $\displaystyle \displaystyle \sqrt{x(1-x)}\bigg|_{x=0} = \sqrt{x(1-x)}\bigg|_{x=1}$ in this style but it is very easy if u take polar coordinates........ just take $\displaystyle x=\sin^2 \theta$ and solve it u will get limit as lower limit=0 and upper limit = $\displaystyle \pi/2$.
You can definitely parameterize this integral with polar, but in essence what I said just skips the parameterization seeing what it will be.

9. That's correct, although I think Wallis's formula is a little overkill haha.