# Tag Info

0

The idea here is that all but finitely many of the points in $\{ 1/n : n \in \mathbb{N} \}$, where the discontinuities in $f$ are, are very close to $0$. So you can work this way. Let $\varepsilon \in (0,1)$. Choose $N \in \mathbb{N}$ such that $1/(N+1)<\varepsilon/2$. Then make the first two points of the partition be $0$ and $1/(N+1)$. Now you just ...

0

Let $\alpha=\arctan\sqrt{\frac{x}{2}}$ $$I=\int \frac{\alpha dx}{\sqrt{x+2}}=\int\alpha d(2\sqrt{x+2})=2\alpha\sqrt{x+2}-2\int \sqrt{x+2}\space d\alpha$$ The calculation gives $$d\alpha=\frac{\sqrt {2} dx}{\sqrt x(x+2)}$$ Hence $I=2\alpha\sqrt{x+2}-2\sqrt 2\int \frac{\sqrt{x+2}\space dx}{(x+2)\sqrt x}$ $I=2\alpha\sqrt{x+2}-2\sqrt ... 0 I think that the integral is equivilant to $$2\sum_{n=1}^\infty\left[\frac{(-1)^{n+1}}{4n+3}\sum_{k=1}^n\frac{1}{n(2k-1)}\right]$$ but I have no idea how to tackle this. The numerical estimates are the same. (For some reason I am unable to post this as a comment because the mathjax will not display.) 3 $$\frac{\pi^2}{24}-\frac{2\pi}3+\frac1{36\sqrt{2}}\left[5\pi^2+12\left(4+\ln\left(\frac{1+\sqrt2}2\right)\right)\left(\pi-2\ln\left(1+\sqrt2\right)\right)-48\operatorname{Li}_2\left(\sqrt2-1\right)\right]$$ 0 Your formula is incorrect. Here is the correct formula, where$f(x)$is the outer function and$g(x)$is the inner one: $$π\int_a^b\ (f(x))^2-(g(x))^2\,dx$$ As for why this works, understand what an integral is actually doing to solve the problem. An integral is actually an infinite sum of infinite tiny things. Consider, for instance, rotating a line from ... 0 For the$y<0$case your DE is evaluated incorrectly: $$\frac{dy}{dx}=\sqrt{-y}$$ $$(-y)^{-\frac{1}{2}}\frac{dy}{dx}=1$$ $$-2\sqrt{-y}=x+c$$ $$-4y=(x+c)^2$$ $$y=-\frac{(x+c)^2}{4}$$ This should hopefully help you to visualize the direction field now. Something like this: 0 Hint: Let$y=u+3$, Then$\dfrac{dy}{dx}=\dfrac{du}{dx}\therefore\dfrac{du}{dx}=\dfrac{u}{(u+3)^2+x^2}\dfrac{dx}{du}=\dfrac{x^2}{u}+\dfrac{(u+3)^2}{u}$Let$x=-\dfrac{u}{v}\dfrac{dv}{du}$, Then$\dfrac{dx}{du}=-\dfrac{u}{v}\dfrac{d^2v}{du^2}-\dfrac{1}{v}\dfrac{dv}{du}+\dfrac{u}{v^2}\left(\dfrac{dv}{du}\right)^2$... 0 I've got$~\dfrac1{\sqrt2}\displaystyle\int\frac{y\sin y}{\cos^4 y}~dy,~$but I'm not entirely sure this is a good thing. Of course it is ! Integrate by parts with regard to$f'(y)=\dfrac{\sin y}{\cos^4y}=-\dfrac{\cos'y}{\cos^4y}=\bigg(\dfrac1{3\cos^3y}\bigg)'.$Then rewrite$~\dfrac1{\cos^3y}~$as$~\dfrac{\cos ...

1

$$\int_C x dz=\int_C \frac{z+\overline z}{2} dz\\ \left(z=e^{i\theta}, dz=izd\theta\right)\\ =\int_0^{2\pi}\frac{e^{i\theta}+e^{-i\theta}}{2}ie^{i\theta}d\theta =\int_0^{2\pi}i\frac{e^{2i\theta}+1}{2}d\theta\\ =\left[\frac{e^{2i\theta}}{4}+i\frac{\theta}{2}\right]_0^{2\pi}=\pi i$$

0

We want to find $\int_0^1 x\log(1+x)\,dx-\int_0^1 x\ln(1-x)\,dx$. Since the second integral is under suspicion, let us look at it. Let $u=\ln(1-x)$ and $dv=x\,dx$. Then we have $du=-\frac{1}{1-x}\,dx$, and we can take $v=\frac{x^2}{2}-\frac{1}{2}$. (We used a little trick here!) So our antiderivative is $$\ln(1-x)\left(\frac{x^2}{2}-\frac{1}{2}\right) ... 1 First, just calculate the integral on [0,1-\varepsilon]:$$ \lim_{\varepsilon\to 0}\int_{0}^{1-\varepsilon}x\ln\frac{1+x}{1-x}\mathrm{d}x=\boxplus $$Since:$$ \int (1+x)\ln(1+x)=\frac{1}{2} (1+x)^2 \log (1+x)-\frac{(1+x)^2}{4}+C  \int \ln(x+1)=(x+1)\ln (x+1)-(x+1)+C $$Using the logarithm-identities:$$ \boxplus=\lim_{\varepsilon\to ...

0

Is it even possible to evaluate it analytically ? Yes. The integral evaluates to $\dfrac{2K(a)+3E(a)}{240},~$ where $a=\dfrac{\sqrt5}3.~$ See elliptic integrals for more information. I only want to know if there's a relatively easy way to solve this. Of course. Just let $x=2t$, and use $1-\cos2t=2\sin^2t$, in conjunction with the definition ...

2

Yes. Let's assume that $L=0$. Then we cannot have $|f|\ge 1$ at arbitrarily large $t$ values. If we did, then we could find disjoint intervals $I_n$ such that $f=1$ (or $=-1$, which is analogous, of course) somewhere on each $I_n$, and $|f|\ge 1/2$ on all of $I_n$ and $f=1/2$ at the endpoints. (We find arbitrarily many of these because $|f|$ cannot stay \ge ... 2 Hint. One may recall that, using Frullani's integral, we have $$\int_{0}^{1}\frac{x^{a-1}-x^{b-1}}{\log x}\:dx=\log\frac ba \quad (a>0,b>0). \tag1$$ Considering a finite sum in the integrand, we get \begin{align} &\int_{0}^{1}{\dfrac{1-x}{\log(x)}(x+x^{2}+x^{2^{2}}+\cdots+x^{2^N})}\:dx\qquad (N=0,1,2,\cdots) ... 0 The curve x = y - y^2 crosses the x-axis twice, at y = 0 and y = 1, so these will be the endpoints of the object we have by rotating our curve. The volume is\pi \int^{1}_{0} (y - y^2)^2 dy = \pi \left (\frac {y^3} {3} - \frac {y^4} {2} + \frac {y^5} {5} \right) |^1_0 = \frac {\pi} {30}.$$0 The answer is (A) To show that, let's denote the primitive of f as F on the interval we integrate on. In case I. we have:$$ I=\int_{a}^{b}f(x)dx=F(b)-F(a) $$In case II. we have through the substitution u=a+x, du=dx and changing the upper and lower bounds accordingly, b and a respectively$$ II=\int_{a}^{b}f(u)du=F(b)-F(a)=I $$Likewise, in ... 0 Look at it like this: In the first integral arguments of f() vary from a to b. For that to be the case in ii), x+a must vary from a to b too, so x varies from a-a=0 to b-a. In iii) for x+c to vary from a to b, x would have to vary from a-c to b-c, which is not the case. 1 These problems require a bit of visualization. So image the point (x,y). Now imagine rotating it about the y-axis. What shape will it make? A circle What will be its radius? x Now imagine we take a rectangular strip with width x and height \Delta y and rotate it about the y-axis. It will make a cylinder, and It's volume will be:$$\pi ... 0 (I've changed notation to conform to the mathematics convention of spherical coordinates.) For definiteness, the issue is that when the ellipse $$\frac{x^{2}}{a^{2}} + \frac{z^{2}}{b^{2}} = 1$$ is rotated through an angle\phi_{0}$and revolved about the$z$-axis, the "profile" intersects itself after half a turn. As indicated by the radial ... 1 I think you're trying to prove the following. Your definition of "volume" is Jordan measure: $$\mbox{vol}(A) = \int_{\mathbb{R}^n} \mathbb{1}_A,$$ where we are taking the Reimann integral of the indicator function $$\mathbb{1}_A(x) = \begin{cases} 1 & \mbox{if } x \in A \\ 0 & \mbox{if } x \notin A \end{cases}$$ of a bounded set$A \subseteq ...

1

Well, the curve you have described looks like this And the idea in these kind of problems, is that you take that plot and rotate it around the indicated axis. The 'tail' left by the curve in the rotation, is the surface that encloses the solid. If that is not intuitive, here is a picture So, how to write the region of integration?. Using cylindrical ...

0

The proof given by Angel is the usual one (you can use just density of step functions and use step functions instead of smooth-functions). Another way to see the problem is as $Tg$, where $T$ is a convolution operator (or more specifically, a convolution operator minus Id). By using the Hausdorff-Young inequality, you end up studying a suitable multiplier, ...

1

Notice $\frac{1}{3^x - 8} = -\frac{1}{8 - 3^x} = - \frac{1}{8} \frac{8 - \mathbf{3^x} + \mathbf{3^x} }{(8-3^x)} = -\frac{1}{8} \left( 1 + \frac{3^x}{8-3^x} \right)$ and so $$\int \frac{1}{3^x - 8} = -\frac{1}{8} \left( x + \int \frac{3^x dx}{8 - 3^x} \right) = -\frac{x}{8} +\frac{1}{8 \ln 3} \int \frac{d(8-3^x)}{8 - 3^x} = -\frac{x}{8} + \frac{\ln ... 0 As a way of getting around your error, use the fact that$$\frac 1{3^x-8}=\frac{3^{-x}}{1-8\cdot 3^{-x}}$$then let u=1-8\cdot3^{-x}. 4 Let 3^x-8=u\implies 3^x\ln(3)\ dx=du or dx=\frac{1}{(u+8)\ln (3)}\ du$$\int \frac{1}{3^x-8}\ dx=\int \frac{1}{u(u+8)\ln 3}\ du =\frac{1}{\ln 3}\int \frac{1}{u(u+8)}\ du =\frac{1}{8\ln 3}\int\left( \frac{1}{u}-\frac{1}{u+8}\right)\ du =\frac{1}{8\ln 3}\ln\left| \frac{u}{u+8}\right| +C=\frac{1}{8\ln 3}\ln\left| \frac{3^x-8}{3^x}\right| ...

0

May I expect the closed-form of this integral ? Yes, you may. In fact, the answer is $0$, due to the parity of the sine and cosine functions. my Mathematica couldn't make the result even when I tried to put $n=2$ and $n=3$. Mathematica has no problem evaluating the integral, even in its hypergeometric form, once the two sine terms have been ...

0

\begin{align} u & = \arctan \sqrt{\frac x 2} \\[10pt] du & = \frac{dx/2}{\left(1+ \dfrac x 2\right)2\sqrt{\dfrac x 2}} = \frac{dx}{(2+x)\sqrt{2x}} \\[10pt] dv & = \frac{dx}{\sqrt{x+2}} \\[10pt] v & = 2\sqrt{x+2} \end{align} \begin{align} \int u\,dv & = uv - \int v\,du = 2\sqrt{x+2} \arctan \sqrt{\frac x 2} - \int \frac{\sqrt ...

0

the taylor series of $$\frac{\sin y}{\cos^4y}=\sum_{n=1}^{\infty }\frac{(2(2n-1)!-1)y^{2n-1}}{(2n-1)!}$$ so $$\frac{1}{\sqrt2}\int\frac{y\sin y}{cos^4 y}dy=\frac{1}{\sqrt2}\int\sum_{n=1}^{\infty }\frac{(2(2n-1)!-1)y^{2n}}{(2n-1)!}dy$$ $$=\frac{1}{\sqrt2}\sum_{n=1}^{\infty }\frac{(2(2n-1)!-1)y^{2n+1}}{(2n+1)(2n-1)!}+C$$

0

Notice, $$\int \frac{\tan^{-1}\sqrt{\frac{x}{2}}}{\sqrt{x+2}}\ dx=\int \frac{\tan^{-1}\sqrt{\frac{x}{2}}}{\sqrt 2\sqrt{\frac{x}{2}+1}}\ dx$$ now, let $\frac{x}{2}=\tan^2\theta\implies dx=2\tan\theta\sec^2\theta \ d\theta$, $0\le \theta\le \pi/2$ $$=\frac{1}{\sqrt2}\int \frac{\tan^{-1}\left(\tan\theta\right)}{\sqrt{\tan^2\theta+1}}(2\tan\theta\sec^2\theta \ ... 2 Let x=2 u^2:$$2 \sqrt{2} \int du \, \frac{u}{\sqrt{1+u^2}} \arctan{u} = 2\sqrt{2} \sqrt{1+u^2} \arctan{u} - 2 \sqrt{2} \int \frac{du}{\sqrt{1+u^2}}$$The latter integral is easily done using the sub u=\sinh{v}, so we have as the integral$$2 \sqrt{x+2} \arctan{\sqrt{\frac{x}{2}}} - 2 \sqrt{2} \log{\left (\sqrt{\frac{x}{2}}+\sqrt{1+\frac{x}{2}} \right ...

0

I can prove this inequality if $f (x) /x$ were non-decreasing. All we need to prove is $$\int_{0}^{t} \frac {f (s)} {s} ds \leqslant f(t).$$ Take $g (x) = f (x) / x$ wherever defined, then we should prove $$\int_{0}^{t} g (s) ds \leqslant t g(t).$$ Since $g$ is non-decreasing, we are done.

1

Since the function is not defined for $x=0$, it's not really meaningful to have a single constant of integration for the whole thing. The most general function $F$ (not defined at $0$) for which, at each point $x\ne0$, $F'(x)=\frac{1}{\sqrt{|x|}}$, is $$F(x)=\begin{cases} -2\sqrt{-x}+c_1 & \text{if x<0}\\ 2\sqrt{x}+c_2 & \text{if x>0} ... 3 Substitute \cosh (7x)=\large \frac{e^{7x}+e^{-7x}}{2},$$\int \cosh^4 (7x)\ dx=\int \left(\frac{e^{7x}+e^{-7x}}{2}\right)^4\ dx=\int \frac{e^{28x}+e^{-28x}+e^{14x}+e^{-14x}+6}{16}\ dx=\frac{1}{16}\left(\frac{e^{28x}}{28}-\frac{e^{-28x}}{28}+\frac{e^{14x}}{14}-\frac{e^{-14x}}{14}+6x\right)+C=\frac{1}{16}\left(\frac{\sinh ...

0

Hint: As for trigonometric functions, the solution is to linearise first: \begin{align*}\cosh^4u&=\frac1{16}(\mathrm e^u+\mathrm e^{-u})^4=\frac1{16}(\mathrm e^{4u}+4\mathrm e^{2u}+6+4\mathrm e^{-2u}+\mathrm e^{-4u})\\ &=\frac18\cosh4u+\frac12\cosh2u+\frac38. \end{align*}

0

HINT...use the "double angle" identity $$\cosh^2A=\frac 12(1+\cosh 2A)$$ twice to get the integrand in terms of multiple angles

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Hint: $$\cosh^2x=\frac{1+\cosh2x}{2}$$

0

Not an answer, but an attempt to push you in a good way for it First of all, let's simplify a bit the integral by collecting $2$ at the numerator and $4$ at the denominator: $$\int_0^{\pi} \frac{2(5 - \cos\theta)}{\sqrt{[4^3\cdot (26 - 10 \cos\theta)^3]}} \text{d}\theta$$ namely $$\frac{1}{4} \int_0^{\pi} \frac{(5 - \cos\theta)}{\sqrt{ (26 - 10 ... 1 By doing partial integration twice,$$\int_0^\pi f(t) \sin t dt=-f(t)\cos t + \int_0^\pi f'(t) \cos t dt\\ =-f(t)\cos t + f'(t)\sin t - \int_0^\pi f''(t) \sin t dt \therefore \int_0^\pi f(t) \sin t dt+\int_0^\pi f''(t) \sin t dt=\left[-f(t)\cos t + f'(t)\sin t\right]_0^\pi\\ =-f(\pi)\cos \pi + f'(\pi)\sin \pi=f(\pi)=3 $$1 Lets focus on the second integral:$$ \int_0^\pi f''(t)\sin t\ \mathrm dt $$and integrate by parts:$$ \int_0^\pi f''(t)\sin t\ \mathrm dt=\left[f'(t)\sin(t)\right]_0^\pi-\int_0^\pi f'(t)\cos(t)\mathrm dt=-\int_0^\pi f'(t)\cos(t)\ \mathrm dt $$Integrate by parts again$$ \int_0^\pi f''(t)\sin t\ \mathrm dt=\left[-\cos(t)f(t)\right]_0^\pi-\int_0^\pi ...

0

You seem to be using the Riemann integral, according to the tags you are using. So I assume that $f$ being integrable means, among other properties, that $f$ is bounded. In this case the reasoning is rather easy. Since $\int_a^b f -\int_s^b f= \int_a^s f$ you can bound the right hand side of this equation easily (using step functions) by $(s-a)\max{|f| }$ ...

1

$$\int\ln(2x+1)\ dx$$ Using integration by parts, we have $$u=\ln(2x+1)\Rightarrow du=\frac{2}{2x+1}dx$$ $$dv=dx\Rightarrow v=x$$ Which yields $$x\ln(2x+1)-\int\frac{2x}{2x+1}dx$$ Using substitution, we have $$s=2x+1\Rightarrow \frac12\ ds=dx$$ Therefore $$x\ln(2x+1)-\frac12\int\frac{s-1}{s}ds$$ $$=x\ln(2x+1)-\frac12\int\left(1-\frac{1}{s}\right)\ ds$$ ...

0

Ampere's law states that $$\int_C \vec{B}\cdot d\vec{r} = \mu I.$$ If $C$ is a circle of radius $R$, it can be parametrized as follows: $$\vec{r}(t)=R\cos(t)\vec{i}+R\sin(t)\vec{j},\quad 0\le t \le 2\pi,$$ so Ampere's law can be rewritten as $$\int_0^{2\pi} \vec{B}(\vec{r}(t))\cdot \vec{r}'(t)dt = \mu I.$$ But if the magnetic field $\vec{B}$ is ...

-2

the derivative of the function $x[x]$ is $[x]$ where $[x]$ is the greatest integer function. I has periodic discontinuity and is actually very very VERY difficult to integrate in a way that satisfies the second fundamental theorem of calculus (as while the slope relationship is right, the discontinuity throws off the area sum). In general, the greatest ...

0

I've just written a paper on this subject for class. My answer is a bit different from yours, because I am allowing circles to stack on each other. Rectangular Riemann sums take the form $$\sum_{i=0}^n f(a+i\Delta x)\Delta x$$ when given a function $f(x)$, divided into $n$ partitions, bounded between $a \text{ and } b. \Delta x = \frac{b-a}{n}$. ...

0

Got it. In this context it means that $\alpha(t) \in L$, ie subset of L as example $\alpha(t)=(t^3,t^3)$ for $t \in \mathbb{R}$, or $\alpha(t)=(sin(t),sin(t))$ for $t \in [-1,1]$ And original question regarding proof is then $int_{\alpha} F \cdot d\alpha=\int_{\alpha} (-y,x)\cdot (\alpha_x'(t),\alpha_y'(t))$ Since $\alpha(t) \in L$, we can say it is of ...

0

\begin{align} \int\underbrace{\ln(2x+1)}_\text{This is $u$.} \, dx & = \int u\,dx = ux - \int x\,du \\[10pt] & = x\ln(2x+1) - \int x\cdot \frac{2}{2x+1} \,dx \quad\text{etc.} \\[10pt] & = x\ln(2x+1) - \int \left( 1 - \frac 1 {2x+1} \right) \,dx \qquad \text{etc.} \end{align}

1

This is a really tricky problem to see for the first time, but it does fit with the guidelines you give. The first choice if possible for $u$ is the logarithmic function. However, we also need another function in order to use integration by parts. What are we to do? Well, although it doesn't look like much, we can realize that $\ln(2x+1)=1\cdot\ln(2x+1)$, ...

9

Using $\operatorname{sgn}(x)$ is just a (half-dirty) trick to put the two cases into one. Put in $-1$ vs. $+1$ for $\operatorname{sgn}(x)$ and your eyes will be open.

0

for $y>0$ $$(y)^{-\frac{1}{2}}y'=1$$ $$2y^{\frac{1}{2}}=x+c$$ $$y^{\frac{1}{2}}=\frac{1}{2}(x+c)$$ $$y=\frac{1}{4}(x+c)^2=(\frac{x}{2}+\frac{c}{2})^2=(\frac{x}{2}+k)^2$$ this answer same your answer when the $y<0$ $$(y)^{-\frac{1}{2}}y'=-1$$ the solution is $$y=(-\frac{x}{2}+k)^2$$

0

Note first that $\|f(\cdotp+h)\|=\|f(\cdotp)\|$. Prove it first for continuous functions with compact support (this space is dense in $L^p(\mathbb{R}^n)$): let $g$ be a such function, let $K=supp(g)$ and let $U$ be an open set of $\mathbb{R}^n$ containing $K$ such that $\mu(U)<\infty$. Since $g$ is uniformly continuous, given $\varepsilon>0$ there ...

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