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1

Here this might help: Okay, so the first thing you do is: Separate variables on either sides of the equation: $$\frac{\ln y dy}{y} = x dx$$ If we integrate notice that if we make $u = \ln y$ $$udu = xdx$$ And it becomes an easy integral: $$\frac{\ln^2 y}{2} = \frac{x^2}{2} + C$$ Multiply both sides by 2: $$\ln^2 y = x^2 + 2C$$ It doesn't matter what ...


2

You were right. We have: $$\frac{\ln^2 y}{2}=\frac{x^2}{2}+C$$ Now multiply both sides by $2$: $$\ln^2 y = x^2+C$$ NOTE: $C$ is any arbitrary constant, so we'll just make $2C=C$ because it doesn't really matter (still some constant). Now you want to isolate for $y$, that's how you'll get the solution. Hence, first take the square root of both sides so ...


1

Hint. By the chain rule, $$W_s=F_uu_s+F_vv_s\ .$$ Make sure you substitute in the correct values of $s$ and $t$.


2

$$\ln^2(y)=x^2+C\to \ln(y)=\pm\sqrt{x^2+C}\to y=\exp(\pm\sqrt{x^2+C})$$


0

If fix $x$ I can treat $F(x,y)$ like a a function of the one variable $y$. Yes, you can. And you can estimate some quantities like $F(x,y_1)-F(x,y_2)$ in this way. But the partial with respect to $y$ does not help you estimate $|F(x,y_0)-F(x_0,y_0)|$. For this quantity, you can use the definition of continuity of $F$ itself. It might help to say what ...


0

I've established some related explicit formulae. Theorem 1. Let $n$ be any positive integer. Set $$ I_{2n}:=\int_{0}^{\pi/2}\! \! x^{2n} \ln \cos x \: \mathrm{d}x $$ Then $$ I_{2n} = - \frac{\pi^{2n+1}\ln 2}{2^{2n+1}(2n+1)} - (-1)^{n}\frac{(2n)!}{2^{2n+1}}\sum_{p=1}^{n} \frac{(-1)^p}{(2p-1)!}\pi^{2p-1}\zeta(2n-2p+3) \tag1 $$ Set $$ ...


2

So most semester-long university Calculus 1 courses cover up through basic integration techniques (chapters 1,2,3,4,5, and 8 in Larson's book), so that's what I would recommend. From chapter 8, I'm mainly looking at the sections on integration by parts and L'Hospital's rule. Although, I'm not sure what exactly is covered in the CLEP exam, so you might also ...


1

Since $f_{x}=\frac{a-2y}{(a-x-y)^2}$ and $f_{y}=\frac{2x-a}{(a-x-y)^2}$, f does not have any critical points in the interior of the closed region bounded by the lines $x=0$, $y=x$, and $y=\frac{a}{2}$. On the boundary of this region, $f(x,y)=0$ on the line $y=x$; $f(x,y)=-1$ on the line $y=\frac{a}{2}$; and on the line $x=0$, ...


1

If x < y <= a/2, then x - y is always negative but a - x - y is always positive. That means f(x,y) must be a negative number. So the best we can do is 0 - epsilon. Notice that the closer you get to x = y, the closer the numerator approaches zero while the denominator is bounded below by a. So the max of this function does not exist (its supremum is ...


1

Since $0\le x<y\le a/2$, we have that $x-y<0$ and $a-x-y>0$. Hence $f(x,y)<0$ everywhere on the given domain. However, for $x+y<a$, we have $\lim\limits_{x\rightarrow y}f(x,y)=0$. This means that $\sup f(x,y)=0$, but this supremum is never achieved.


0

Thats what Wolfram Alpha says: http://www.wolframalpha.com/input/?i=maximize+%28x-y%29%2F%28a-x-y%29+on+0%3C%3Dx%3Cy%3C%3Da%2F2 So the maximum is basically where $$x-y\rightarrow 0$$


1

To showing differentiability, show that $\displaystyle\sum\frac{\cos(kx)}{x}$ converge uniformly on any compact $\subset (0,1)$.


0

You should have gotten $$\frac{1 - (-1) - 2}{1-1} \leadsto \frac 00$$ This is an indeterminate form, which simply indicates that more work needs to be done, e.g., factoring and canceling the common term in the numerator and denominator, or performing L'Hospital.


3

You should have gotten $\displaystyle\int_{0}^{1-a}[(1-a)x-x^2]\,dx = \left[\dfrac{1-a}{2}x^2-\dfrac{1}{3}x^3\right]_{0}^{1-a} = \dfrac{(1-a)^3}{6}$. Then, you have $\dfrac{(1-a)^3}{6} = \dfrac{9}{2} \leadsto (1-a)^3 = 27$ which should be easy to solve.


2

The square $|x|+|y|\leq1$ can be parametrized by $(x,y)=(t-u,t+u),\quad -\frac{1}{2}\leq t\leq\frac{1}{2},\quad -\frac{1}{2}\leq u\leq\frac{1}{2}$. Thus the desired surface can be parametrized by $(x,y,z)=(t-u,t+u,\frac{1+(t-u)^2}{1+(t+u)^2})$ with the same bounds on $t$ and $u$.


0

Consider the derivatives $$ f'(x)=3x^2-8x^3 $$ $$ f''(x)=6x-24x^2 $$ So we have critical points at $x=0$ and $x=3/8$. For $x\in (-\infty,0)$ we see that $f'(-1)=3-8(-1)=11>0$ so $f(x)$ is increasing. $f'(1)=3-8=-5<0$ so for $x\in (3/8, \infty)$ $f(x)$ is decreasing. We also see $f''(3/8)=6(3/8)-24*(3/8)^2=9/4-9/8=18/8-27/8=-9/8<0$ which ...


2

Using $k=\frac{1}{2}$ in the second rule, we get $$\begin{align} \int\limits_{-2}^{2} (x-3) \sqrt{4-x^2} \ dx &=2\int\limits_{-1}^{1} (2x-3) \sqrt{4-4x^2} \ dx \\&=2\int\limits_{-1}^{1} 2(2x-3) \sqrt{1-x^2} \ dx \\&=4\int\limits_{-1}^{1} (2x-3) \sqrt{1-x^2} \ dx \\&=8\int\limits_{-1}^{1}x\sqrt{1-x^2} \ dx-12\int\limits_{-1}^{1}\sqrt{1-x^2}\ ...


0

$\textbf{Hint:}$ Use the second formula for $k=\frac{1}{2}$.You get linear combination of integral $\int_{-1}^{1} \sqrt{1-x^2} dx$ and integral $\int_{-1}^{1} -2x\sqrt{1-x^2} dx$ Integral $\int_{-1}^{1} x\sqrt{1-x^2} dx$ is easy to calculate: note that $((1-x^2)^{\frac{3}{2}})'=-3x\sqrt{1-x^2}$.


0

Your function is continuous and only non-negative on the interval $[0,0.5]$. Thus the function must take a maximum value on this interval (since it's continuous and the interval is closed), and so any positive value greater than the maximum is not achieved by your function so your function is not surjective.


0

If the function is only positive in $[0, 0.5]$, and it's continuous, it must have a maximum there. Hence, any value greater than this maximum is not $f(x)$, for any $x \in \Bbb R$. Finding this maximum is another problem, better dealt with calculus.


0

Yes, you're right: for $0<x<\frac{1}{2}$ $f(x)$ is positive. But for $x \in (0,\frac{1}{2})$ you have: $$x^3-2x^4=x^3(1-2x)<\frac{1}{2^3}(1-2x)$$ It's because $f(x)$ is positive and $0<x<\frac{1}{2}$. Next: $$\frac{1}{2^3}(1-2x)<\frac{1}{2^3}$$ Because for $0<x<\frac{1}{2}$ you have $0<1-2x<1$. So $f$ reach values less than ...


0

The statement to prove is the following. Let $I$ be a non-empty subset of the real numbers, $x_0\in \overline I$ and $f,g\colon I\to \mathbb R$ functions such that $\lim \limits_{x\to x_0}\left(g(x)\right)=+\infty$ and $\lim \limits_{x\to x_0}\left(f(x)\right)=L$, for some $L\in \mathbb R$. In these conditions it holds true that $\lim \limits_{x\to ...


2

Your friend is right. Surjective means $f(\Bbb R) = \Bbb R$, or for any $y\in \Bbb R$ there exists $x\in \Bbb R$ such that $f(x) = y$. Now, if $f(x) = x$ and $g(x) = -x$, they are both surjective (why?). But $h(x) = f(x) + g(x) = 0$ is not surjective, since $h(\Bbb R) = \{0\}$, for example there does not exist $x\in \Bbb R$ such that $h(x) = 1$ since $h(x) = ...


7

Aside from showing that two limits differ as you do for $y=x$ and $y=x^2-x$, you can also do this in one fell swoop: Let $y=x^3-x$, then you get $$\lim_{(x,y)\to (0,0)}{x^2\over x^3}$$ which clearly diverges to infinity.


2

Depends on your knowledge of differentials and derivatives. In most of the cases, you are taught about differentials via derivatives, and for functions of one variable you always get $$ dy(u) = y'(u)du. $$ So you only have to compute the derivative $y'(u)$ e.g. via the derivative rule for fractions, and put $du$ from the right of that.


4

$$\frac{dy}{du}=\frac{d}{du} \left ( \frac{u+1}{u-1}\right )=\frac{u-1-u-1}{(u-1)^2}=\frac{-2}{(u-1)^2}$$ $$dy=\frac{-2}{(u-1)^2} du$$


2

Perhaps a sledehammer: The following holds: If $\ \ \ $ 1) $f$ is continuous on $[a,b]$, $\ \ \ $ 2) $f'$ exists and is finite for all but a countable subset of $(a,b)$, and $\ \ \ $ 3) $f'$ is Lebesgue integrable, then $$ f(x)-f(a)=\int_a^x f'(t)\,dt $$ for all $a\le x\le b$. In particular, $f$ is absolutely continuous. This is Excercise ...


2

Using inequalities $$4=\sqrt[n]{4^n}\leq \sqrt[n]{3^n+4^n} \leq \sqrt[n]{4^n+4^n} = \sqrt[n]{2}\cdot 4$$ so $$4\leq\lim_{n\to\infty}\sqrt[n]{3^n+4^n}\leq \lim_{n\to\infty}\sqrt[n]{2}\cdot 4 = 4$$ therefore $$\lim_{n\to\infty}\sqrt[n]{3^n+4^n}=4$$


7

$$\lim_{n\to\infty}\sqrt[n]{3^n + 4^n}=\lim_{n \to \infty}\sqrt[n]{4^n \left ( \left (\frac{3}{4} \right )^n+1 \right )}=4 \lim_{n \to \infty}\sqrt[n] {\left (\frac{3}{4} \right )^n+1 }=4 \cdot 1=4 $$


12

Hint $$3^n+4^n =4^n \left( 1+ (\frac{3}{4})^n\right)$$ And $1+ (\frac{3}{4})^n \to 1$


7

We have $$3^n=_\infty o(4^n)$$ so $$\sqrt[n]{3^n+4^n}\sim_\infty\sqrt[n]{4^n}=4$$


1

Note $$ \int_0^\infty e^{-xt}\sin tdt=\frac{1}{1+x^2} $$ and hence $$ \frac{d}{dx}\int_0^\infty e^{-xt}\sin tdt=-\frac{2x}{(1+x^2)^2}. $$ Also $$ \int_0^\infty\frac{\cos(\pi x)}{1+x^2}dx=\frac{1}{2}\pi e^{-\pi}. $$ So \begin{eqnarray} I&=&2\int_0^\infty \frac{x\sin(\pi x)}{(1+x^2)^2}dx=-\int_0^\infty \sin(\pi x) \left(\frac{d}{dx}\int_0^\infty ...


1

All integrals of the form $\displaystyle\int_0^\infty\frac{x^{k-1}}{(x^n+a^n)^m}dx$ can be expressed in terms of trigonometric functions as follows: first, let $x=at$, then $u=\dfrac1{t^n+1}$. Now recognize the expression of the beta function in the new integral, and finally use Euler's reflection formula for the $\Gamma$ function in order to arrive at ...


1

You can make it simpler by first solving $$\lfloor t\rfloor^2=1,$$ giving $$t\in[-1,0[\ \cup\ [1,2[.$$ Then considering the pencil of parallel lines $x+y=t$. (If you like, the lines through $(t,0)$ and $(0,t)$).


0

Hint: The revenue is $R=p \cdot x$ p: constant Price Edit: You can also get the equation, if you integrate MR: $\int MR \ \ dx=\int 100 \ dx$ $\int \ R'(x) \ \ dx=\int 100 \ dx$ $R(x)=100x +C$ C is here 0, because you have no revenue, if you donĀ“t sell a unit.


1

The space $PC(a,b)$ of piecewise continuous functions (p.c. functions) on the interval $[a,b]$ is not complete; this is the content of the example in the left picture in the OP. The definition of p.c. functions is given on the top right picture in the OP: we note that an element of $PC(a,b)$ can have a finite number of "jumps" and all of them must be finite. ...


1

$$ \int\frac{dl}{\left(r^2+l^2\right)^{3/2}} = -\frac{1}{r}\frac{d}{dr}\int\frac{1}{\left(r^2+l^2\right)^{1/2}}dl $$ change of variables $v=l/r$ we find $$ -\frac{1}{r}\frac{d}{dr}\frac{1}{r}\int^{\infty}_{0}\frac{1}{\left(1+v^2\right)^{1/2}}rdv = -\frac{1}{r}\frac{d}{dr}\left[\sinh^{-1}v\right]^{\infty}_{0} $$ $$ ...


3

Let $l=r\tan{u}$, then $dl=r\sec^2{u} \ du$. The integral becomes $$\frac{1}{r^2}\int^{\pi/2}_0\frac{\sec^2{u}}{\sec^3{u}}du=\frac{1}{r^2}$$


3

Note that $\displaystyle \frac{x^{2}}{2+3x^{2}} = \frac{x^{2}}{2} \frac{1}{1+\frac{3}{2}x^{2}}$. For all $u \in \mathbb{R}$ such that $\vert u \vert < 1$, we have : $$ \sum_{n=0}^{+\infty} (-1)^{n}u^{n} = \frac{1}{1+u} $$ Substituting $u$ with $\displaystyle \frac{3}{2}x^{2}$ leads to : $$ \frac{1}{1+\frac{3}{2}x^{2}} = \sum_{n=0}^{+\infty} (-1)^{n} ...


1

Since $$\frac{1}{1+\frac{3}{2}x^2} = \sum_{n=0}^\infty (-1)^n \frac{3^n}{2^n}x^{2n},$$ we deduce that $$ \frac{x^2}{2+3x^2} = \frac{x^2}{2} \frac{1}{1+\frac{3}{2}x^2} = \sum_{n=0}^\infty (-1)^n \frac{3^n}{2^{n+1}}x^{2n+2} $$ This expansion holds true provided that the first expansion holds true, namely $\frac{3}{2}x^2<1$.


0

The answer to the second question is no, as $f(x)=\sqrt x$ shows.


0

One computes $$f(x):={1\over(1+2x)^4}-(1-8x)=40 x^2\ {1+4x+6x^2+3.2 x^3\over(1+2x)^4}\ ,$$ which is $>0$ for $|x|\ll1$. For such $x$ the fraction $q$ on the right hand side is $\doteq1$. Therefore in a first round we make sure that $$40x^2<0.1\ .\tag{1}$$ The latter is equivalent with $|x|<0.05$. This is "for all practical purposes" sufficient, ...


1

For any integer $m>0$, the result is $0$. In case $m=0$ the result is $\pi$


0

Using $$\cos2A=\frac{1-\tan^2A}{1+\tan^2A}$$ $$\int\frac{1-\cos x}{\cos x(1+\cos x)}dx$$ $$=\int\frac{2\sin^2\dfrac x2}{\cos^2\dfrac x2-\sin^2\dfrac x2}\frac12\sec^2\dfrac x2dx$$ $$=\int\frac{2\tan^2\dfrac x2}{1-\tan^2\dfrac x2}\frac12\sec^2\dfrac x2dx$$ $$=\int\frac{2u^2-2+2}{1-u^2}du=-2u+2\int\frac{du}{1-u^2}du$$


0

Hint : \begin{align} \int\frac{1-\cos x}{\cos(1+\cos x)}\ dx&=\int\frac{\color{red}{1+\cos x}-2\cos x}{\cos(\color{red}{1+\cos x})}\ dx\\ &=\int\frac1{\cos x}\ dx-\int\frac{2}{1+\cos x}\ dx\\ &=\int\frac{\cos x}{\cos^2 x}\ dx-\int\frac{2}{1+\cos x}\cdot\frac{1-\cos x}{1-\cos x}\ dx\\ &=\int\frac{d(\sin x)}{1-\sin^2 x}-\int\frac{2-2\cos ...


1

Filling in some of the unpleasant details hidden behind the Shabbeh hint. For simplicity, let's use the "standard" parabola $y = x^2$. The arclength $s(t)$ between the points $(0,0)$ and $(t,t^2)$ is given by $$ s(t) = \int_0^t \sqrt{ 1 + \left(\frac{dy}{dx}\right)^2} dx \\ = \int_0^t \sqrt{ 1 + 4x^2} dx \\ = \frac{1}{4}\left(2t \sqrt{ 1 + 4t^2} + ...


4

We have: Integrand = $\dfrac{1}{\cos x} - \dfrac{2}{1+\cos x} = \sec x - \sec^2 (x/2)$, and $\displaystyle \int \sec x - \sec^2 (x/2)dx = \ln (\sec x + \tan x) - 2\tan(x/2) + C$


1

This is the problem of finding an arc-length parametrization of a curve. While this can always be done in theory and is a great tool in several proofs regarding parametric curves, it is rarely possible to find a closed form solution. The general method is: Describe your curve by $\vec{r}(t)$, for $a \leq t \leq b$. Find $\vec{r}\,'(t) = ...


0

Some books include the codomain as part of the definition while others do not. See Wikipedia for a brief discussion on how Bourbaki gives both kinds of definition. Wikipedia says that the definition ignoring the codomain is preferred in set theory. I am more used to the definition without a codomain. For example, my Baby Rudin (Principles of Mathematical ...


0

I do not know how to solve the equations to find the following, but for any $t_0$, an involute of $X, Y$ can be calculated as $$ f(X,Y) = X - \frac{X'}{\def\abs#1{\left|#1\right|}\abs{(X,Y)'}}\cdot \int_{t_0}^t \abs{(X,Y)'}\, dt,\quad g(X,Y) = f(Y,X) $$ where $\abs\cdot$ denotes the standard absolute value on $\mathbb R^2$.



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