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## New answers tagged calculus

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 $I=2\alpha\sqrt{x+2}-2\sqrt 2\int \frac{\sqrt{x+2}\space dx}{(x+2)\sqrt x}$ $I=2\alpha\sqrt{x+2}-2\sqrt ... 1 You got$\int e^{2\theta}\sin(3\theta)d\theta=...=\sin(3\theta)\frac12e^{2\theta}-\frac32\Bigl(\frac12e^{2\theta}\cos(3\theta)+\frac32\int e^{2\theta}\sin(3\theta)d\theta\Bigr)$. I think inside the parentheses you may need a$-$instead of a$+$, but I will not bother with that. Assuming what you got were correct (and perhaps it is, if I am wrong about ... 1 There cannot be such a sequence.$\chi_{\Bbb Q}$is a double limit of continuous functions, cf. the Dirichlet function. Thus, it's a Baire class 2 function. As the article states, it can't be a Baire class 1 function (single pointwise limit of continuous functions) because such functions have a meager set of discontinuities, unlike$\chi_{\Bbb Q}$. For a ... 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 Here is a plain answer for your reference: Let$\simeq$denote the equality sign up to a constant. We have $$\int e^{2x}\sin 3x dx = \int \sin 3x de^{2x}\frac{1}{2} \simeq \frac{1}{2}[ e^{2x}\sin 3x - 3 \int e^{2x} \cos 3x dx ]\\ = \frac{1}{2}e^{2x}\sin 3x - \frac{3}{2} \int e^{2x}\cos 3x dx;$$ we have $$\int e^{2x}\cos 3x dx = \frac{1}{2}\int \cos 3x ... 4 When you do it with integration by parts, you have to go in the "same direction" both times. For instance, if you initially differentiate e^{2 \theta}, then you need to differentiate e^{2 \theta} again; if you integrate it, you will wind up back where you started. If you do this, you should find something of the form$$\int e^{2 \theta} \cos(3 \theta) d ... 1 There is one specific formula where this works, but that is all:$f(x) = -ln(x)f'(x) = -\frac{1}{x}f''(x) = \frac{1}{x^2}$0 For the$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: 2 As indicated by @levap in the comments, you are likely confusing the (right-hand side, RHS) derivative at$0$with the (RHS) limit of the derivative. I will just add details to clarify. Assuming you mean$f(0)=0$. The derivative at$0$is by definition$\displaystyle\lim\limits_{h\to0}\frac{f(0+h)-f(0)}h = \lim\limits_{h\to0}\frac{h^2\sin(\frac1h)-0}h = ...

0

I know this should be in comment part but I don't have enough reputation to make a comment. Hope a moderator can help. Relating to Callculus's answer, there is one thing that I don't understand. He/she said that "dA is the differential. The value of dA tends to zero, but it is not zero.". Could you explain why value of dA should approach to zero? I heard ...

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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 $\dfrac{dx}{du}=-\dfrac{u}{v}\dfrac{d^2v}{du^2}-\dfrac{1}{v}\dfrac{dv}{du}+\dfrac{u}{v^2}\left(\dfrac{dv}{du}\right)^2$ ...

2

Notice that $$\int_{-1}^{1} |f_j - 1|^2 = \| f_j \|_2^2 - 2\|f_j\|_1 + 2 = \mathcal{O}(2^{-j}).$$ By the monotone convergence theorem, we have $$\int_{-1}^{1} \sum_{j=1}^{\infty} |f_j - 1|^2 = \sum_{j=1}^{\infty} \int_{-1}^{1} |f_j - 1|^2 < \infty.$$ Therefore $\sum_{j=1}^{\infty} |f_j - 1|^2$ is finite a.e. and hence $f_j \to 1$ a.e.

1

I proved your statemement up to a subsequence $f_{j_k}$. We have \begin{align} \int_{-1}^1 |f_j(x)-1|^2dx&=\int_{-1}^1 f_j(x)^2dx -\int_{-1}^1 2f_j(x)dx +\int_{-1}^1 1dx \\ &=\| f_j\|_{L^2}^2-2\| f_j\|_{L^1}+2=\| f_j\|_{L^2}^2-2 \end{align} Since $\| f_j\|_{L^2}$ converges to $\sqrt 2$ : $$\int_{-1}^1 |f_j(x)-1|^2dx\longrightarrow 0.$$ Now we can ...

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Your intuition is correct. Swapping $x$ and $y$ mirrors your function across the line $y = x$, which preserves area.

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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 ... 2 $$\frac{\text d x}{\text d \theta}=\frac{\text d (r\cos\theta)}{\text d \theta}=\frac{\text d (4\sin \theta\cos\theta)}{\text d \theta}=\frac{\text d (2\sin 2\theta)}{\text d \theta}=4\cos 2\theta=4 \cos \frac{\pi}{3}=2$$ $$\frac{\text d y}{\text d \theta}=\frac{\text d (r\sin\theta)}{\text d \theta}=\frac{\text d (4\sin^2 \theta)}{\text d ... 1 The cartesian r \theta graph suggests that when \theta =0, r=3 and the only graphs with a positive x-intercept of three are the two tilted ones. It can't be the first one because that one allows r=0 and the cartesian r\theta graph shows that 1<r<5 1 Notice this is not a maximize the area but minimize the perimeter problem. Let x be one side of the parking lot. Then 6,000/x is the other side. For this you use 2x + 2(6,000/x) meters of fence. But you have some fencing parallel to one of the sides to cut it in two. So the perimeter based on side x is P(x) = 3x + 12,000/x. Take the derivative, set ... 0 If x,\ y are the dimensions of the fenced in area, we are required that x\cdot y = C. The amount of fencing will then be: F = 2(x + y). We can use the constraint to define y as a function of x, then take a derivative to help find the minimum amount of fencing required.$$\frac d{dx}F = \frac d{dx}\bigg(2\left(x+\frac Cx\right)\bigg) = 2 - ... 0 Some basic limits can be obtained by repeated use of l'Hospital's Rule easily, \begin{eqnarray*} \lim_{x\rightarrow 0}\frac{\sin x-x}{x^{3}} &=&-\frac{1}{6},\ \ \ \ \ \ \ \ and\ \ \ \ \ \ \ \ \ \lim_{x\rightarrow 0}\frac{\arcsin x-x}{x^{3}}=\frac{1}{% 6} \\ \lim_{x\rightarrow 0}\frac{\sin x-x+\frac{1}{6}x^{3}}{x^{5}} &=&\frac{1}{120}% ,\ \ \ ... 0 This is true for all fractions, not just differential equations. If$xy = n$, then we can rearrange to get$y = \frac{n}{x}$, but only for$x \neq 0$. Thus, the functions are the same for all but a finite number of points when converting multiplications to divisions. If you require the excluded points then you must either leave the function in multiplicative ... 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) ... 5 No.\left(\frac{\text{d}f}{\text{d}x}\right)^2 = \frac{\text{d}f}{\text{d}x}\cdot \frac{\text{d}f}{\text{d}x}$$Whilst$$\frac{\text{d}^2f}{\text{d}x^2} = \frac{\text{d}}{\text{d}x}\left(\frac{\text{d}f}{\text{d}x}\right)$$4 Take the example f(x)=x^2$$ \left( \dfrac{d\left(x^2\right)}{dx}\right) ^2=(2x)^2 $$but$$ \left( \dfrac{d^2\left(x^2\right)}{dx^2}\right)=2 $$Which is a counterexample to your statement 0 The curve x = y - y^2 crosses the x-axis twice, at y = 0 and y = 1, so these will be the endpoints of the object we have by rotating our curve. The volume is$$\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 $$Likewise, in ... 1 Your statement is false, so you cannot prove it. Counterexample:$$a_n = \left\{ \begin{matrix} 1 & n \mbox{ odd} \\ n & n \mbox{ even} \end{matrix} \right. \ \ \ b_n = \left\{ \begin{matrix} n & n \mbox{ odd} \\ 1 & n \mbox{ even} \end{matrix} \right.$$What is true is that either \limsup_{n \to \infty}a_n = \infty or \limsup_{n \to ... 0 Look at it like this: In the first integral arguments of f() vary from a to b. For that to be the case in ii), x+a must vary from a to b too, so x varies from a-a=0 to b-a. In iii) for x+c to vary from a to b, x would have to vary from a-c to b-c, which is not the case. 0 You're right, it's an example it's in any textbook$$ J=\left(\begin{array}{ccc} \sin\theta\cos\phi & r\:\cos\theta\cos\phi & -r\:\sin\theta\sin\phi\\ \sin\theta\sin\phi & r\:\cos\theta\sin\phi & r\:\sin\theta\cos\phi\\ \cos\theta & -r\:\sin\theta & 0 \end{array}\right) $$1 These problems require a bit of visualization. So image the point (x,y). Now imagine rotating it about the y-axis. What shape will it make? A circle What will be its radius? x Now imagine we take a rectangular strip with width x and height \Delta y and rotate it about the y-axis. It will make a cylinder, and It's volume will be:$$\pi ...

0

If $h'(x) = \frac1{f'(h(x))}$ then $h'(x) = \frac 1 {3h^2(x) + 1}$ If $f(x) = x^3 + x = 2 => x = 1 => h(2) = 1$ so $h'(2) = \frac 1 {3h^2(x) + 1} = \frac 1 {3*1 + 1}= \frac 1 4$

1

Here is a direct proof (as suggested by @AndréNicolas in the comments) that if $a_1,a_2,a_3\cdots\to g$ then $a_1,g,a_2,g,a_3,g\cdots\to g$. Note that the latter sequence is $b_1,b_2,b_3\cdots$ , where $b_{2n}=g$ and $b_{2n-1}=a_n$. So take $\varepsilon>0$, then there is $N$ such that if $n>N$ then $|a_n-g|<\varepsilon$. Let $K=2N-1$. Then if ...

1

You have $f(1) = 2$ so $h(2) = 1$. Put $y = h(x)$ and use implicit diff; $$y^3 + y = x$$ so $$(3y^2 + 1)y' = 1$$ Now put $x = 2$ and you have $$(3 + 1)y'(2) = 1$$ so $y'(2)= 1/4.$

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 ...

9

Your first equation $$m={\sin 2\theta-\sin \theta \over \cos 2\theta-\cos \theta}$$ isn't the slope of the tangent. It's slope of the path of the reflected ray. Since the ray must be perpendicular to the plane of incidence, it better be orthogonal to the tangent of the cardiod. $$m \frac{dy}{dx} = -1$$ And so it is! Obligatory cool picture follows. In ...

0

The Big O notation is specific and the little o notation is more generic. For example, There may be infinitely many classes of functions that are $o (f)$ each of which has distinct $O (g_n)$.

1

Do not substitute any of them because they will change the result drastically. If I'm allowed to do so, I'd suggest you make a substitution $y = x + 1/2$ and your integral will become $$\int_1^\infty \frac{dx}{x^2e^{x^2+x} } = \int_{3/2}^\infty \frac {e^{1/4} dy} {(y - 1/2)^2 e^{y^2}}$$ Hope I've been of help.

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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 ...

1

I want to challenge your thought by asking some questions and hope that it helps you to figure it out. :) What is your definition of behaving? I mean how do you define two function behaving like each other at the neighborhood of some point $x_0$. Can you give some definition for it like the one for asymptotic equivalence? We should be elaborate and exact ...

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 ...

1

$\int_1^\infty \frac{dx}{x^2e^{x^2+x}}$ is a finite integral. we have $\int_1^\infty \frac{dx}{x^2e^{x^2+x} } <\int_1^\infty \frac{e^{-x^2}}{x^2}dx < \int_1^\infty \frac{e^{-x}}{x^2}dx$ ...

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 ...

1

Try this: Since $a_{n}$ is bounded, there is some $M > 0$ so that for all $n$, we have $|a_{n}| < M$. Now, since $a_{n} + b_{n}$ converges, then by definition that means there is some element $c$ so that for every $\epsilon > 0$, we can find a point $N$ in the sequence so that all later points $n \geq N$ satisfy $|a_{n} + b_{n} - c| < ... 2 By assumption there exists$x=\lim_n(a_n+b_n)$. In particular for each given$\varepsilon>0$then$|a_n+b_n-x|<\varepsilon$whenever$n\ge n_\varepsilon$. Now, if$|a_n|\le M$for all$n$, this implies that for all$n\ge n_\varepsilon$it holds $$|b_n|=|a_n+b_n-a_n| \le |a_n+b_n|+|a_n| \le |x|+\varepsilon+M$$ It follows that for all$n$we have $$... 1 Hint: Let \lim a_n + b_n = L and M be a bound on the a_n. Then certainly there exists an N such that$$n > N \ \Rightarrow \ |a_n + b_n - L | < 1 \ \Rightarrow \ |a_n + b_n| - |L| < 1$$and for all such n we also have that |a_n| < M. Now try applying the triangle inequality to find a bound on |b_n|:$$|b_n| = |a_n + b_n - ... 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.

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