# A tough definite integral

• Nov 13th 2010, 08:22 PM
Crescent
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.
• Nov 13th 2010, 08:27 PM
Drexel28
Quote:

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.
• Nov 13th 2010, 09:11 PM
Crescent
Quote:

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.
• Nov 13th 2010, 09:17 PM
Drexel28
Quote:

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.
• Nov 13th 2010, 10:00 PM
Hardwork
Does the fact that the integrand has the same value for both limits not help at all? (Thinking)

$\displaystyle \displaystyle \sqrt{x(1-x)}\bigg|_{x=0} = \sqrt{x(1-x)}\bigg|_{x=1}$ ... it's probably just a useless observation.
• Nov 14th 2010, 05:50 AM
HallsofIvy
It tells you the graph is symmetric about x= 1/2. (It is, of course, a circle).
• Nov 14th 2010, 01:23 PM
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 = $\displaystyle \pi/2$.
• Nov 14th 2010, 01:43 PM
Drexel28
Quote:

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.
• Nov 14th 2010, 02:14 PM
sushant1221
• Nov 14th 2010, 02:15 PM
Drexel28
That's correct, although I think Wallis's formula is a little overkill haha.