Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I am having a bit of a hard time figuring this one out. Would be very nice if anyone would like to help me (no more forehead banging against wall :))

while: $ 0 \leq x \leq L$

The function is: $\xi(x) = k\sqrt{x(L-x)} $ & $\xi(x)dx = k\sqrt{x(L-x)}dx $

$m = \int_{0}^{L}k\sqrt{x(L-x)}dx = k\int_{0}^{L}\sqrt{Lx-x^2}dx = k\int_{0}^{L}\sqrt{\frac{L^2}{4}-(x-\frac{L}{2})^2}dx$

I cannot understand how they develop that last step. Any help is much appreciated!

Thank you!

share|improve this question

2 Answers 2

up vote 1 down vote accepted

Note that $$\left(x-\frac L2\right)^2=x^2-xL+\frac{L^2}4 \hspace{5pt}\Rightarrow\hspace{5pt}\frac{L^2}4-\left(x-\frac L2\right)^2=xL-x^2$$

share|improve this answer
Such a simple mistake! Thank you very much for your answer! –  Lukas Arvidsson Nov 15 '12 at 10:21

First recall that $-(x-a)^2=-x^2+2ax -a^2$ for any $a$. You are given $Lx-x^2=Lx-x^2+\frac{L^2}{4} - \frac{L^2}{4} =\frac{L^2}{4}-x^2+Lx - \frac{L^2}{4}=\frac{L^2}{4}-x^2+2\frac{L}{2}x - (\frac{L}{2})^2=\frac{L^2}{4}-(x-\frac{L}{2})$. This is done by just using the first formula I listed.

share|improve this answer
Thank you for your answer! –  Lukas Arvidsson Nov 15 '12 at 10:32

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.