Recent content by Oggy

In the sum i goes from 0 to n. And it's (-1)^(n-i). Sorry for the mistakes.

Corrected now, thanks :) (It's (n-i)!)

Let A_i = \frac{1}{n}\cdot \frac{(-1)^{n-i}}{i!\cdot(n-i)!} \int_{0}^{n} \frac{t(t-1)...(t-n)}{t-i}dt I need to prove \sum_{i=0}^{n} A_i = 1 . I tried tinkering with the equation but I'm really at a loss what to do with the integral. I'd appreciate any help.

Yep u^2+1 should be in the denominator. Sorry

Well it's not that difficult or ugly... After the substitution i pointed out it is simply: \int{\sqrt{t}(t^{2}+1)dt}=...=\frac{2}{7}(\tan{x})^{\frac{7}{2}}+\frac{2}{3}(\tan{x})^{\frac{3}{2}}+C

Or \tan{x}=t \Rightarrow \frac{dx}{(\cos{x})^{2}}=dt, and remember \frac{1}{(\cos{x})^{2}}=1+(\tan{x})^{2}.

A line integral is independent of path if there exists a function U, that Fdr is it's exact (total) differential. In this case U=x^2y-3x+5y+y^4. dU=(2xy-3)dx+(x^2+4y^3+5)dy=Fdr.

It's V=\int_0^{2\pi}d\phi \int_0^{2}rdr \int_{r^{2}}^{12-2r^{2}}dz=...=24\pi Because the z=x^2+y^2 and z=12-2x^2-2y^2 intersect at r=2, and r increases from 0 to the maximum of 2.

That's exactly what I referred to.

The area is bounded by y=-2x+4 and y=-2x+7, so that's y+2x=4 and y+2x=7, or v=4 to v=7, so v goes from 4 to 7, and the same for u. With this change of variables, you're actually fitting the area of integration to a rectangle.

I understand double and triple integrals and all, but I'm just wondering why is dxdy=|J|dudv\ x=f(u,v)\ y=g(u,v) Where does that derive from? Why is it? (and also for triple integrals)

That's correct.

I=\int_{0}^{+\infty} e^{-x^2} dx, is solved using a trick, and integrating in polar coordinates, in which the Jacobian helps solving it. It is equal to \frac{\sqrt{\pi}}{2}. Your integral is just 2I, I think.

Just write the Lagrange function of the problem: F(x,y,z)=3x-y-3z+\lambda_1(x+y-z)+\lambda_2(x^{2}+2z^{2}-1) Then find the partial derivatives (they have to be 0), and solve the system of equations.

\frac{d}{dx}x^{n}=nx^{n-1} \Rightarrow \int{x^{n-1}dx}=\frac{x^{n}}{n}