Help Solving Online Math Problem - Can You Spot the Error?

  • Thread starter Thread starter Exeter909
  • Start date Start date
  • Tags Tags
    Error
Exeter909
Messages
2
Reaction score
0
I have to solve this online problem for homework and was wondering if anyone could help me with it, i can find the answer by graphing and solving, but don't know how to explain the mistake...

http://www.dougshaw.com/findtheerror/FTEusub.html

thanks for the help
 
Last edited by a moderator:
Physics news on Phys.org
If you have an integral from a to b (in this case, -1 to 1), u has to be a 1-1 continuously differentiable function of x with non-zero derivative. That is, u has to be a function of x defined on all of [-1, 1]. It has to be 1-1, so u has to take on a different value for every x. u has to be continuous, differentiable (with respect to x, i.e. you can find du/dx), and du/dx has to be continuous. Moreover, du/dx can never be 0, otherwise dx = du/u', so and if u' is ever 0, then when you change your variables, the integrand will have 0 in the denominator at some point, which isn't allowed. Your u substitution fails about half of these criteria.

For one, it is not defined when x = 0. Second, 1/x4 = 1/(-x)4, so u repeats the value 1/24 at both x = 1/2 and x = -1/2.
 
thank you

thank you!
 
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