Need Help with Integral : undefined function integrand

  • Thread starter Thread starter barnator
  • Start date Start date
  • Tags Tags
    Function Integral
barnator
Messages
2
Reaction score
0
\int_0^{y-x} (y-x)^3f[x] dx

I really don't know what integration method that should be used to solve this. I would really appreciate anyone pointing me in the right direction, this problem is frustrating me! I tried integration by parts but it doesn't really get me anywhere.
 
Physics news on Phys.org
barnator said:
\int_0^{y-x} (y-x)^3f[x] dx

I really don't know what integration method that should be used to solve this. I would really appreciate anyone pointing me in the right direction, this problem is frustrating me! I tried integration by parts but it doesn't really get me anywhere.

What does f[x] mean? No hope of working the problem without knowing.
 
The whole problem makes no sense. In addition to f being undefined, you should not have the "integration variable", x, in the upper limit of integration- it makes no sense to say that x varies from 0 to x- y.
 
OK thanks for the help. I guess the problem I have has some typos.

I have a related question though referring to the Integral theorem under http://www.atp.ruhr-uni-bochum.de/rt1/syscontrol/node145.html

I'm confused about the work at the integration by parts. It looks like he sets u = \int_0^t f[\tau] d\tau and dv = e^{-st} dt, so that he ends up with du = f[t]dt. I'm not following how to go from u to du.
 
Let the antiderivative of f be F.

By the fundamental theorem of calculus,

\int_0^t f(\tau) \; d\tau = F(t) - F(0)

Clearly, the derivative of this result is F'(t) = f(t).
 
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