# Tag Info

3

This is a Riemann sum question. It's easy to see that a local minimum occurs at x = 4. Thus if you approximate the area under the curve by the rectangle with height f(4) you get a lower bound. Similarly. x = 3 gives a maximum (at least on the relevant interval [3,4]). Thus, approximating the area with the rectangle of height f(3) gives an upper bound.

2

How about $r:=\sqrt{x^2+y^2}$ and $s:=\sqrt{z^2+w^2}$? Let $\theta$ and $\phi$ be the usual angular coordinates of the $xy$-plane and the $zw$-plane, respectively. Then, $\text{d}x\,\text{d}y=r\,\text{d}r\,\text{d}\theta$ and $\text{d}z\,\text{d}w=s\,\text{d}s\,\text{d}\phi$. The ball $B$ is then given by $r^2+s^2\leq 1$, $0 \leq \theta \leq 2\pi$, and ...

2

What you're saying is $$\int_1^2 f(x) \, \mathrm{d}x = f(2) - f(1).$$ Note that the theorem you stated in the question says $$\int_1^2 f(x) \, \mathrm{d}x = F(2) - F(1)$$ where $F = \int f \, \mathrm{d}x \neq f$. You'd need to evaluate the anti-derivative first to get $$\int^2_1\frac{\mathrm{d}t}{8-3t}=-\frac13\int_1^2\frac{\mathrm{d}t}{t-\frac83}=-\frac13 ... 2 You may just write$$ \int^2_1\frac{dt}{8-3t}=-\frac13\int_1^2\frac{1}{t-\frac83}dt=-\frac13 \left[\ln \left| t-\frac83\right| \right]_1^2=\frac13 \ln\left(\frac52\right). $$2 Hint: substitute$${ { e } }^{ -t }=\frac { 1 }{ \sqrt { 5 } } \tan { x }$$1 (f,af) \in \mathcal{G}(A) iff \sqrt{1+|a|^{2}}f \in L^{2}_{\mu}. Furthermore, for any such f,$$ \|(f,af)\|_{\mathcal{G}(A)}=\|\sqrt{1+|a|^{2}}f\|_{L^{2}_{\mu}}=\|f\|_{L^{2}_{d\nu}}, $$where d\nu = \sqrt{1+|a|^{2}}d\mu. So the graph is a Hilbert space. 1 We have:$$ \mathcal{L}\left(e^{-1/x}\right)= \frac{2}{\sqrt{s}}\,K_1(2\sqrt{s}) $$where K_1 is a modified Bessel function of the second kind. Your integral is so an "incomplete modified Bessel function of the second kind", not very nice or elementary. 1$$\begin{align} \int_0^{\infty}e^{-\frac12\left(\frac{x-c}{b}\right)^2}dx&=b\sqrt{2}\int_{-c/b\sqrt{2}}^{\infty}e^{-x^2}dx\\\\ &=b\sqrt{2}\left(\int_{-c/b\sqrt{2}}^{0}e^{-x^2}dx+\int_{0}^{\infty}e^{-x^2}dx\right)\\\\ &=\frac{b\sqrt{2\pi}}{2}\left(\text{erf}(c/b\sqrt{2})+1\right) \end{align}$$1 Yes. First prove it for characteristic functions 1_A where A\in\mathcal R'. This is obvious: \mu(A)=(\mu|_{\mathcal R'})(A). Then use linearity of integrals to prove it for (step)functions that are measurable w.r.t. \mathcal R' and take a finite number of values. Finally for every f measurable w.r.t. \mathcal R' with we can construct a ... 1 This uses the cyclical invariance of the trace: \mathrm{tr}\,ABC=\mathrm{tr}\,CAB, which you can derive by writing it out with indices and noting that the trace "completes the cycle" of pairwise index summations. That, together with the fact that you can insert a trace on the left-hand side because it's a scalar, explains the first equals sign, and then ... 1 First note that since$$ |f_n -f| \leq |f_n| + |f| $$then |f_n -f| \in L^1. Also by (i) we have that  |f_n| + |f| \to 2|f| almost everywhere. Hence hypothesis (ii) gives that$$ \int |f_n| + |f| d\mu \underset{n \to \infty}{\longrightarrow} \int 2|f| d\mu $$Again (i) gives that |f-f_n| \to 0  almost everywhere, thus by a generalized version of the ... 1 The Thomae function is Riemann integrable, while the computation of its Riemann integral is harder than that of the Lebesgue integral: the function is zero almost everywhere, so its Lebesgue integral is zero. 1 In cylindrical coordinates, i.e. (r,\theta,z) with x=r\cos\theta,y=r\sin\theta we have$$ x'(t)=r'(t)\cos\theta(t)-r(t)\theta'(t)\sin\theta(t),\quad y'(t)=r'(t)\sin\theta(t)+r(t)\theta'(t)\cos\theta(t). $$Therefore$$ (x'(t))^2+(y'(t))^2=[(r'(t))^2+r^2(t)(\theta'(t))^2]\cdot[\cos^2\theta(t)+\sin^2\theta(t)]=(r'(t))^2+r^2(t)(\theta'(t))^2,  and the ...

Only top voted, non community-wiki answers of a minimum length are eligible