How about this:
$$\int_{x_0-\varepsilon}^{x_0+\varepsilon} g'(x)\varphi(x)dx
= g(x)\varphi(x)\vert_{x_0-\varepsilon}^{x_0+\varepsilon}-\int_{x_0-\varepsilon}^{x_0+\varepsilon} g(x)\varphi'(x)dx$$ and $$\int_{x_0-\varepsilon}^{x_0+\varepsilon} g(x)\varphi'(x)dx=0$$ if $$\varphi$$ is...
(1+\frac{1}{n})^{n}<e
for n=1
\frac{2}{1}<e
for n =2
\frac{3^2}{2^2}<e
for n =3
\frac{4^3}{3^3}<e
and so on so forth... Multiply them together, we then have
\frac{(n+1)^n}{n!}<e^n
Dick, my original proposal is a dead end cause even I managed to prove it monotonic, I won't be able to evaluate its limit -- FAILED ATTEMPT:cry:. Let me try again.
It's possible to prove (1+\frac{1}{n})^n monotonic increase. As a fact, it converges to e. Hence, (1+\frac{1}{n})^{n}<e
After...
Not necessary.
Good start:) What does it look like after factoring out the term?
It just means -x, or (-1)x. And I'm sure you know -b+a=a-b.
Do you remember how to add/subtract fractions with different denominators?
You are welcome! This happens a lot in self-studying.
We don't actually integrate the functional because u,v, and their partial derivatives are unknown. Or did I miss your question?
We are dealing with the terms related to first variation of \frac{\partial u}{\partial x}. y is treated as constant.
You might not have *fully* understood the derivation of Euler-Lagrange equation for the classic example of
S(t)=\int_{a}^{b}L(t,x,\dot{x})dt. I'm using the classical mechanics convention here where x is function of t, and \dot{x}=dx/dt, L is Lagrangian.
The integration by part of the selected...
Had you made only one mistake, you won't get the right answer. Kinda like (-1)*(-1)=1.
You have made TWO errors.
1. antiderivative of sin(t2) is wrong as Mark suggested.
2. \frac{d}{dx}\left(\frac{cos(x^2)}{2x}\right)\neq -\sin x^2. Review quotient rule.