What is the relationship between A and a in this graph?

  • Thread starter Thread starter alingy2
  • Start date Start date
  • Tags Tags
    Area
alingy2
Messages
16
Reaction score
0
There is a graph associated to it. Please look at the screenshot.

Ok, so, here is my process.
I modeled the three functions.
y=x
y=1/a^2 x
and y=1/x

Then, I calculated A using calculus. (Integrals)
Integral of x-1/a^2 x from 0 to 1 + integral of 1/x-1/a^2 x from 1 to a
A=1-1/a^2

Now, how is a=e^A?
 

Attachments

  • Screen Shot 2014-03-30 at 6.02.52 PM.png
    Screen Shot 2014-03-30 at 6.02.52 PM.png
    5.4 KB · Views: 471
Physics news on Phys.org
alingy2 said:
There is a graph associated to it. Please look at the screenshot.

Ok, so, here is my process.
I modeled the three functions.
y=x
y=1/a^2 x
and y=1/x

Then, I calculated A using calculus. (Integrals)
Integral of x-1/a^2 x from 0 to 1 + integral of 1/x-1/a^2 x from 1 to a

Correct integral setups.

A=1-1/a^2

Wrong answer. Recheck your integration or show us your steps.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top