1. ## Functions

A function
$\displaystyle y = f(x)$
under what geometrical conditions would
$\displaystyle f(x) = f^{-1} (x) ?$
and whats an example of this function for which
$\displaystyle f(x) = f^{-1} (x) ?$

&

does
$\displaystyle \int^{1}_{-1} \sqrt$$4-x^{2}$$$

$\displaystyle = 2{\pi}$

2. Originally Posted by nath_quam
A function
$\displaystyle y = f(x)$
under what geometrical conditions would
$\displaystyle f(x) = f^{-1} (x) ?$
For example, the identity function,
$\displaystyle f(x)=x$

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Quick I am proud that yiu deleted your post
(I am moderator I can see everything you do, in case you are wondering).

3. Originally Posted by nath_quam
$\displaystyle \int^{1}_{-1} \sqrt$$4-x^{2}$$$

$\displaystyle = 2{\pi}$
Yes, because,
$\displaystyle y=\sqrt{4-x^2}$ is a semi-circle.
From $\displaystyle -2\leq x\leq 2$ the area below is,
$\displaystyle \frac{1}{2}\pi (2)^2=2\pi$

4. Hello, Nath!

Given a function: $\displaystyle y = f(x)$
(a) under what geometrical conditions would $\displaystyle f(x) = f^{-1}(x)\;?$

(b) What's an example of this function for which: $\displaystyle f(x) = f^{-1} (x)\:?$

(a) The graph must be symmetric to the line $\displaystyle y = x$.

(b) A straight line like: .$\displaystyle y \;= \; 3 - x$
. . .A hyperbola like: .$\displaystyle y \:= \:\frac{1}{x}$

5. Would someone be able to solve this

$\displaystyle \int^{1}_{-1}\sqrt$$4-x^{2}$$$
using
$\displaystyle x = 2sin\theta\$

6. Hello, Nath!

$\displaystyle \int^1_{-1}\!\sqrt{4-x^2}\,dx$ . using $\displaystyle x = 2\sin\theta$

We can do it, but it takes a lot of work
and you're expected to be very familiar with Trigonometry.

I'll go through the whole thing in baby-steps . . .

Let $\displaystyle \x = 2\sin\theta\quad\Rightarrow\quadd x = 2\cos\theta\,d\theta$

Note that: .$\displaystyle \sqrt{4 - x^2} \:= \:\sqrt{4 - 2\sin^2\theta} \:=$ $\displaystyle \sqrt{4(1 - \sin^2\theta)} \:=\:\sqrt{4\cos^2\theta} \:=\:2\cos\theta$

Substitute: .$\displaystyle \int(2\cos\theta)(2\cos\theta\,d\theta) \:=\:4\int\cos^2\theta\,d\theta$

Double-angle identity: .$\displaystyle 4\int\frac{1 + \cos2\theta}{2}\,d\theta \;= \;2\int(1 + \cos2\theta)\,d\theta$

Integrate: .$\displaystyle 2\left[\theta + \frac{1}{2}\sin2\theta\right]$

Half-angle identity: .$\displaystyle 2\left[\theta + \frac{1}{2}(2\sin\theta\cos\theta)\right] \:= \:2\left[\theta + \sin\theta\cos\theta\right]$

Now we must back-substitute.
We have: .$\displaystyle x = 2\sin\theta\quad\Rightarrow\quad\sin\theta = \frac{x}{2}\quad\Rightarrow\quad\theta = \arcsin\frac{x}{2}$
. . and: .$\displaystyle \cos\theta = \frac{\sqrt{4-x^2}}{2}$

So we have: .$\displaystyle 2\left[\,\arcsin\frac{x}{2} + \frac{x\sqrt{4-x^2}}{4}\,\right]^1_{\text{-}1}$

$\displaystyle 2\left[\arcsin\left(\frac{1}{2}\right) + \frac{1\cdot\sqrt{4-1^2}}{4}\right] -$$\displaystyle 2\left[\arcsin\left(-\frac{1}{2}\right) + \frac{(-1)\sqrt{4 - (-1)^2}}{4}\right]$

. . $\displaystyle = \;2\left(\frac{\pi}{6} + \frac{\sqrt{3}}{4}\right) - 2\left(-\frac{\pi}{6} - \frac{\sqrt{3}}{4}\right)$

. . $\displaystyle = \;\frac{\pi}{3} + \frac{\sqrt{3}}{2} + \frac{\pi}{3} + \frac{\sqrt{3}}{2}$

. . $\displaystyle = \;\frac{2\pi}{3} + \sqrt{3} \;\approx\;3.826446$

7. Originally Posted by Soroban

Integrate: .$\displaystyle 2\left[\theta + \frac{1}{2}\sin2\theta\right]$
At this stage couldn't you x sub 1 and -1 into
$\displaystyle x = 2\sin\theta\$
and get the answer that way
so you end up with
$\displaystyle 2\left[\theta + \frac{1}{2}\sin2\theta\right]^\frac{\pi}{2}_{\frac{3\pi}{2}}$

8. hello, Nath!

Your idea is a good one . . .

At this stage couldn't you x sub 1 and -1 into $\displaystyle x = 2\sin\theta$ and get the answer that way . . . Yes!

so you end up with: $\displaystyle 2\left[\theta + \frac{1}{2}\sin2\theta\right]^\frac{\pi}{2}_{\frac{3\pi}{2}}$ . . . but your limits are incorrect.

We have: .$\displaystyle 2\sin\theta = x$

For $\displaystyle x = -1:\;\;2\sin\theta\,=\,-1\quad\Rightarrow\quad\sin\theta\,=\,-\frac{1}{2}\quad\Rightarrow\quad\theta = -\frac{\pi}{6}$

For $\displaystyle x = 1:\;\;2\sin\theta\,=\,1\quad\Rightarrow\quad\sin \theta\,=\,\frac{1}{2}\quad\Rightarrow\quad \theta = \frac{\pi}{6}$