help with integral

• Oct 28th 2006, 10:22 PM
scorpion007
help with integral
http://img207.imageshack.us/img207/9109/math1yk3.png

In the solution, it is amazingly simple, but i don't understand why they do what they did.
• Oct 28th 2006, 11:01 PM
CaptainBlack
Quote:

Originally Posted by scorpion007
http://img207.imageshack.us/img207/9109/math1yk3.png

In the solution, it is amazingly simple, but i don't understand why they do what they did.

It is because of the symmetry of the cosine curve (actually the anti-symmetry
of cos about pi/2) guarantees that the part of the area above y=b/2 fits
exactly (when rotated through 90 degrees) above the part from a/2 to form
a rectangle of length a and width b/2.

RonL
• Oct 29th 2006, 07:34 AM
Soroban
Hello, scorpion007!

Quote:

http://img207.imageshack.us/img207/9109/math1yk3.png

In the solution, it is amazingly simple, but i don't understand why they do what they did.

We could have tried: . $\int^a_0\cos^{-1}\left(\frac{2x-a}{a}\right)\,dx$
. . but it is far easier to integrate with respect to $y.$

We have: . $y \;= \;\frac{b}{\pi}\cos^{-1}\left(\frac{2x-a}{a}\right)\quad\Rightarrow\quad\frac{\pi}{b}y\;= \;\cos^{-1}\left(\frac{2x - a}{a}\right)$ . $\quad\Rightarrow\quad\frac{2x-a}{a}\;=\;\cos\left(\frac{\pi}{b}y\right)
$

. . $2x - a \;=\;a\cos\left(\frac{\pi}{b}y\right)\quad\Rightar row\quad 2x \;=\;a\cos\left(\frac{\pi}{b}y\right) + a \quad\Rightarrow$ . $x\;=\;\frac{a}{2}\left[\cos\left(\frac{\pi}{b}y\right)+1\right]$

So we have: . $\frac{a}{2}\int^b_0\left[\cos\left(\frac{\pi}{b}y\right) + 1\right]\,dy \;= \;\frac{a}{2}\left[\frac{b}{\pi}\sin\left(\frac{\pi}{b}y\right) + y\right]^b_0$

. . $= \;\frac{a}{2}\left\{\left[\frac{b}{\pi}\sin\left(\frac{\pi}{b}\cdot b\right) + b\right] - \underbrace{\left[\frac{b}{\pi}\sin(0) + 0\right]}_{0}\right\}$

. . $= \;\frac{a}{2}\left[\frac{b}{\pi}\underbrace{\sin(\pi)}_0 + b\right] \;=\;\frac{ab}{2}$ . . . answer (c)

• Oct 29th 2006, 07:52 AM
CaptainBlack
Quote:

Originally Posted by Soroban
Hello, scorpion007!

We could have tried: . $\int^a_0\cos^{-1}\left(\frac{2x-a}{a}\right)\,dx$
. . but it is far easier to integrate with respect to $y.$

We have: . $y \;= \;\frac{b}{\pi}\cos^{-1}\left(\frac{2x-a}{a}\right)\quad\Rightarrow\quad\frac{\pi}{b}y\;= \;\cos^{-1}\left(\frac{2x - a}{a}\right)$ . $\quad\Rightarrow\quad\frac{2x-a}{a}\;=\;\cos\left(\frac{\pi}{b}y\right)$ $
$

. . $2x - a \;=\;a\cos\left(\frac{\pi}{b}y\right)\quad\Rightar row\quad 2x \;=\;a\cos\left(\frac{\pi}{b}y\right) + a \quad\Rightarrow$ . $x\;=\;\frac{a}{2}\left[\cos\left(\frac{\pi}{b}y\right)+1\right]$

So we have: . $\frac{a}{2}\int^b_0\left[\cos\left(\frac{\pi}{b}y\right) + 1\right]\,dy \;= \;\frac{a}{2}\left[\frac{b}{\pi}\sin\left(\frac{\pi}{b}y\right) + y\right]^b_0$

. . $= \;\frac{a}{2}\left\{\left[\frac{b}{\pi}\sin\left(\frac{\pi}{b}\cdot b\right) + b\right] - \underbrace{\left[\frac{b}{\pi}\sin(0) + 0\right]}_{0}\right\}$

. . $= \;\frac{a}{2}\left[\frac{b}{\pi}\underbrace{\sin(\pi)}_0 + b\right] \;=\;\frac{ab}{2}$ . . . answer (c)

But the whole point is that we don't need to do the integral, sure the LaTeX looks cool,
but all this manipulation is error prone. We should never minimise the value of knowlege
when used to avoid work.

RonL
• Oct 29th 2006, 03:53 PM
scorpion007
ah, thank you, in my original attempt at integration, i made a stupid algebraic mistake, which caused the answer to be wrong. In general, that's is how i attempted it the first time, before reading what CaptainBlack said.