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I am not sure about it,it seems graph of ellipse,which we can find half area of ellipse half area of ellipse $=\frac { \pi \cdot 1\cdot \left( 0.1013 \right) }{ 2 } \approx 0.159122$ which is close to $1/6$

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$\frac{dP}{dt}$ measures the rate of increase of $P$ in currency units per year At any instant the total in the account is increasing at a rate of $rP$ per year and decreasing at a rate of $w$ per year.

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$r$ is a rate of interest, meaning a percentage per unit time. So say $r$ is 5% per year, the instantatneous rate of growth of the account is 5% of the amount in the account per year: $rP$. Similarly you have to interpret $w$ as the withdrawal rate per unit time, for example, $1,000 per year. You are right that the equations are meaningless if yo take ... 0$w$is the (continuous, constant) rate of withdrawal, and$rP$is the (continuous, proportional) rate of interest accrual. 1 Look at the middle entry in the Jacobian: At the first fixed point it is not$-2\ell$, it is$$\ell \left( 1-\frac{c}{a} \right)$$ not$-2\ell$. Similarly the bottom right is $$\ell \left( 1-\frac{b}{a} \right)$$. The latter is not negative, because$\frac ba < 1$, so this point is not a stable point, it is a saddle. -1 Idea Knowing that $$1 + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + \cdots = \frac{\pi^2}{6},$$ I suspect a relation with this integral... 0$\int_0^2\pi x^2(\frac14x^2)dx=\frac14\left[\frac{x^5}5\right]_0^2=\frac85$1 Because$h\int_{0}^{\infty}e^{-hx}dx=1$, you can write $$\frac{1}{h}\int_{0}^{\infty}e^{-hx}f(x)dx-f(0)=\frac{1}{h}\int_{0}^{\infty}e^{-hx}\{f(x)-f(0)\}dx$$ If$M$is a bound for$f$on$[0,\infty)$and$r > 0$, then, as$h\rightarrow\infty, $$\left|h\int_{r}^{\infty}e^{-hx}\{f(x)-f(0)\}dx\right| \le ... 2 Note that on the numerator of the LHS of the equation, the [...]^2 operation is actually taking the modulus squared. That is, |a+bi|^2 = a^2+b^2. Work it out. The conclusion is trivial after you note that. Complex modulus is the notion of "length" in the complex plane (http://mathworld.wolfram.com/ComplexModulus.html) 2 If f(a)=c and f(b)=d, then$$\begin{align} \int_a^b f(x) \,\,dx+\int_c^d f^{-1}(y) \,\,dy &=\int_a^b f(x) \,\,dx+\int_a^b f^{-1}(f(x)) f'(x) \,\,dx\\\\ &=\int_a^b f(x) \,\,dx+\int_a^b x f'(x) \,\,dx\\\\ &=\int_a^b \left(f(x)+x f'(x)\right) \,\,dx\\\\ &=\int_a^b (xf(x))' \,\,dx\\\\ &=bf(b)-af(a)\\\\ &=bd-ac \end{align}$$Now, let ... 2 If (x_1,y_1),(x_2,y_2) are the points of intersection between the line y=kx+1 and the parabola y=x^2+2x-3, then:$$ \frac{x_1+x_2}{2} = \frac{k}{2}-1$$by Viète's theorem, and:$$ \frac{y_1+y_2}{2} = 1+k\cdot\frac{x_1+x_2}{2} = 1-k+\frac{k^2}{2}.$$There is a well-known relation between the area of the parabolic segment and the area of a specific ... 2 Split the integral in two parts: \int \limits _0 ^1 and \int \limits _1 ^\infty. For the second one, using the fact that |f| \le M, we obtain: h \big| \int \limits _1 ^\infty \Bbb e ^{-hx} f(x) \Bbb d x \big| \le h \int \limits _1 ^\infty \Bbb e ^{-hx} |f(x)| \Bbb d x \le h \int \limits _1 ^\infty \Bbb e ^{-hx} M \Bbb d x = h M \frac {\Bbb e ^{-hx}} ... -3 Try integration by parts and then take the limit. 0 You may try to rewrite first and then take the limit$$ h\int_0^\infty e^{-hx}f(x)\,dx-f(0)=\int_0^\infty e^{-t}\left[f\left(\frac{t}{h}\right)-f(0)\right]\,dt. $$1 first find intercept of y=x^2+2x-3 and y=kx+1$$x^2+2x-kx-4=0 \\x=\frac{k-2 \pm \sqrt{(k-2)^2+16}}{2}$$area=\int_{k_1}^{k_2}((kx+1)-(x^2+2x-3))dx\\ area==$$f(k)= \int_{\frac{k-2 - \sqrt{(k-2)^2+16}}{2}}^{\frac{k-2 + \sqrt{(k-2)^2+16}}{2}}((kx+1)-(x^2+2x-3))dxnow find f'(k) ,f'(k)=0 root(s) 1 Consider the equation f(x) = f'(x^2) + 2x. It can be seen that f(1) = 4, f'(1) = 2 and f^{(n+2)}(1) = 0 for n \geq 0. The demonstration of how these values are obtained is as follows. Since f(x) is differentiable, which is evident from the differential equation, then consider the function in a series expansion given by \begin{align} f(x) = ... 2 Every subsequence of a Cauchy sequence is Cauchy. The condition mentioned in the third form implies the existence of an \epsilon>0 together with a subsequence (x_{n_k})_k that satisfies |x_{n_{k+1}}-x_{n_k}|>\epsilon for each k. Just start with some n_1 and find an index n_2>n_1 with |x_{n_{2}}-x_{n_1}|>\epsilon and repeat this ... 4 Yes, by definition,f(x) \sim g(x) \quad \mbox{as} \quad x \to \infty$$means$$\lim_{ x \to \infty} \frac{f(x)}{g(x)} = 14 As an alternative, we may apply the substitution x\mapsto\sinh^2{x}. \begin{align} \int^1_0\frac{\ln{x}}{\sqrt{x(x+1)}}{\rm d}x &=4\int^{-\ln(\sqrt{2}-1)}_0\ln(\sinh{x})\ {\rm d}x\\ &=4\int^{-\ln(\sqrt{2}-1)}_0\left(\color{green}{\ln\left(1-e^{-2x}\right)}-\ln{2}+x\right)\ {\rm d}x\\ ... 1 Let you function be represented by f(x), thenf'(x)=\frac{1}{4} \left(-\frac{3}{(x+2)^{5/2}}+\frac{1}{(x+3)^{3/2}}+\frac{9}{(x+3)^{5/2}}\right)$$now define$$g(x)=(x+2)^{5/2} (x+3)^{5/2}f'(x)$$simplify a bit to have$$4g(x)=x \Big(x \left(x\sqrt{x+2} +16 \sqrt{x+2}-3 \sqrt{x+3}\right)+52 \sqrt{x+2}-18 \sqrt{x+3}\Big)+48 \sqrt{x+2}-27 \sqrt{x+3}$$Now ... 36 The ceiling function gives one iff \sin(1/x)> 0, otherwise zero. So the integral is just a sum of interval lengths where the sin is positive, i.e.$$ \left(1-\frac{1}{\pi}\right)+\left(\frac{1}{2\pi}-\frac{1}{3\pi}\right)+\left(\frac{1}{4\pi}-\frac{1}{5\pi}\right)+\ldots=1-\frac{1}{\pi}\left(1-\frac12+\frac13-\frac14+\ldots\right)=\\ ... 19 Note thatx \sin(1/x) \in [-1, 1]$when$x \in [0,1]$so your integral is the volume of the set on which$x \sin(1/x) > 0$. This set is $$\left(\frac{1}{\pi}, 1\right) \cup \bigcup_{k=1}^{\infty}\left(\frac{1}{\pi (2k+1)}, \frac{1}{2\pi k}\right)$$ and its volume is $$1 - \frac{1}{\pi} + \sum_{k=1}^{\infty}\left(\frac{1}{2\pi k} - \frac{1}{\pi ... 0 while converting in polar co-ordinate you have directly taken dxdx=drd(theta), you have to use polar transformation Jacobin to do so, that is dxdx=rdr*d(theta), then integrate normally answer will be (ip/2). 0 Switching to polar coordinates, the Jacobian is given by |J| where$$ J = \dfrac{\partial(x,y)}{\partial(r,\theta)} = \begin{vmatrix} \dfrac{\partial x}{\partial r} & \dfrac{\partial y}{\partial r} \\ \dfrac{\partial x}{\partial \theta} & \dfrac{\partial y}{\partial \theta} \end{vmatrix} = \begin{vmatrix} \cos\theta & \sin\theta \\ ... 1 Multiply Jacobian in integrand $$\int_{\theta=\pi/4}^{\theta=3\pi/4}\bigg[\int_{r=0}^{r=2}\bigg(r^3\bigg)dr\bigg]d\theta$$ 2 It may be of some value to write an explicit answer (which is pretty similar to the comment by user251257, only containing more details). Let$e:U\to\mathbb{R}$be differentiable, and let$e':U\to(\mathbb{R}^n)^*$denote the differential of$e$. Let$u\in U$, and let$h\in\mathbb{R}^n$. By differentiability, we ... 2 We look at the interval$(0,1/2]$. In order to deal with positive quantities, we will show instead that$\int_0^{1/2}\frac{-\ln x}{1-x^2}\,dx$exists. Note that in our interval$1-x^2\ge 3/4$. So it is enough to prove that$\int_0^{1/2} \frac{-\ln x}{3/4}\,dx$exists, or equivalently that$\int_0^{1/2}(-\ln x)\,dxexists. This can be viewed as a standard ... 1 The directional derivative is just the derivative of the composition: \begin{alignat*}{2} I\subset\mathbf R&\to U\subset\mathbf R^n &&\to\mathbf R\\ t&\mapsto u+th&&\mapsto e(u+th) \end{alignat*} where the intervalI$is such that$u+th\in $for all$t\in I. Now observe that for a function of one variable, differentiability and ... 2 We have $$\int_{0}^{1}\frac{\log\left(x\right)}{1-x^{2}}dx=\sum_{k\geq0}\int_{0}^{1}\log\left(x\right)x^{2k}dx$$ and by parts $$\int_{0}^{1}\log\left(x\right)x^{2k}dx=\left.\frac{x^{2k+1}\log\left(x\right)}{2k+1}\right|_{0}^{1}-\frac{1}{2k+1}\int_{0}^{1}x^{2k}dx=-\frac{1}{\left(2k+1\right)^{2}}$$ hence ... 2 This is too long for a comment and it could be off-topic. There is no doubt that Ron Gordon's solution is the most elegant for this problem. I have been thinking about sine and cosine integral functions (see my early comment) just because $$\frac x{1+x^2}=\frac 12 \Big(\frac 1{x-i}+\frac 1{x+i}\Big)$$ and $$\int \frac{\sin(x)}{x+a}\,dx=\cos (a) ... 0 OK so you now have three clear explanations how to do this using trig substitution, but you might also find it interesting to consider the integral as the area of part of the inside of the circle x^2+y^2=4 3 Consider the contour integral$$ \oint_C dz \frac{z e^{i z}}{1+z^2}$$where C is a semicircle in the upper half plane of radius R. Then the contour integral is equal to$$\int_{-R}^R dx \frac{x e^{i x}}{1+x^2} + i R \int_0^{\pi} d\theta \, e^{i \theta} \frac{R e^{i \theta} \, e^{i R e^{i \theta}}}{1+R^2 e^{i 2 \theta}}$$As R \to \infty, the ... 1 This integral requires a trigonometric substitution. Let: x = 2\sin(\theta) and dx = 2\cos(\theta) d\theta Plugging this in gives: \int_{-1}^1 \sqrt{4-x^2} dx = \int_{-\frac{\pi}{6}}^\frac{\pi}{6} \sqrt{4-4\sin^2(\theta)} 2\cos(\theta) d\theta = 4\int_{-\frac{\pi}{6}}^\frac{\pi}{6} \sqrt{1-\sin^2(\theta)}\cos(\theta) d\theta = ... 1 Hint: Try making the substitution x=2\sin(u) and find the expressions for dx and u. In general, in the integrals of the form \displaystyle\int\sqrt{a^2-x^2}\,dx you can make the substitution x=a\sin(u). 2 Let x = 2 \sin \theta. Then dx = 2\cos \theta \; d\theta. When x = -1, \theta = -\frac{\pi}{6} and when x = 1, \theta = \frac{\pi}{6}. Therefore$$\int_{-1}^1 \sqrt{4-x^2} \; dx = \int_{-\frac{\pi}{6}}^{\frac{\pi}{6}} \sqrt{4 - 4 \sin^2 \theta} \; 2\cos \theta \; d\theta = \int_{-\frac{\pi}{6}}^{\frac{\pi}{6}} 4\cos^2 \theta \; d\theta.$$Now you can ... 0 Using Green's Theorem, we have$$\iint_D dA=\oint_C xdy \tag 1$$Here, the contour C in (1) is comprised of 3 segments; C_1, C_2, and C_3. The integrals over these separate segments are for C_1, x=y$$\begin{align} \int_{C_1}xdy&=\int_0^{\sqrt{2}} ydy\\\\ &=1 \tag 2 \end{align}$$for C_2, x=-y$$\begin{align} ... 4 By making the Euler substitution\sqrt{x^{2}+x} = x+t, we find \begin{align} \int_{0}^{1} \frac{\log(x)}{\sqrt{x^{2}+x}} \, dx &= 2 \int_{0}^{\sqrt{2}-1} \frac{\log \left(\frac{t^{2}}{1-2t}\right)}{1-2t} \, dt \\ &= 4 \int_{0}^{\sqrt{2}-1} \frac{\log t}{1-2t} \, dt - 2\int_{0}^{\sqrt{2}-1} \frac{\log(1-2t)}{1-2t} \, dt. \end{align} The first ... 2 Suppose that\Omega=\{K\subseteq J: K$is a finite subsets of$J\}$. Then $$\alpha=\sum_{i\in J}a_i=\sup_{k\in\Omega}\sum_{i\in K}a_i.$$ For any$n\in N$there is a finite set$k_n\in\Omega$such that$0\leq\alpha-\sum_{i\in K_n}a_i<\frac{1}{n}$. Set$H=\cup_{n\in N}K_n$, then$His countable. We have 0\leq\alpha-\sum_{i\in H}a_i\leq\alpha-\sum_{i\in ... 1 Let s=\sum_{i\in J}a_i. Let H_n=\{\,i\in J\mid a_i>\frac1n\,\}. Then |H_n|\le ns<\infty. Let H=\bigcup_{n\in\mathbb N} H_n. 1 Since your question is moderately unclear at the moment, I thought I would give you some thoughts that may be helpful in forging ahead. How might I find the largest \delta such that I can challenge \epsilon = 1/2? I think this is the crux of the problem. What do you mean by "challenge" exactly? This is the part that is currently not clear about ... 0 You have \begin{align} \sum_{k=1}^{n} k &= 1 + 2 + 3 + \cdots + n \\ &= n + (n - 1) + (n - 2) + \cdots + 1 \\ \end{align} where in the second line we just reversed the order in which we summed. Now if we add the two lines, notice that term by term we get n + 1, i.e \begin{align} (n) + 1 &= n + 1 \\ (n - 1) + 2 &= n + 1 \\ (n - 2) + 3 ... 1 Thinking aloud, so your limit L=\dfrac1{10} and your function f(x)=\dfrac1{\lfloor x\rfloor}. \forall\varepsilon\exists\delta etc. So, for \varepsilon=\dfrac12, \exists\delta such that if |x-10|<\delta then \bigl|\dfrac1{\lfloor x\rfloor}-\dfrac1{10}\bigr|<\dfrac12. That is \dfrac{-1}2 + \dfrac1{10}< \dfrac1{\lfloor x\rfloor} ... 0 You cannot refute existence of limit based on error margin of ε = \frac{1}{2}, because the function is indeed close to \frac{1}{10} within that error margin in a suitably small interval around 10. If you want to disprove the existence of the limit, you need a smaller error margin. Just explicitly calculate what \frac{1}{\lfloor x \rfloor} is when x ... 1 Notice,\int\left(\frac{f(x)}{x}-\frac{f'(x)}{x^2}\right)dx=\int \frac{d}{dx}\left(\frac{f(x)}{x}\right)dx=\int d\left(\frac{f(x)}{x}\right)=\frac{f(x)}{x}$$Here, for fist part f(x)=e^{2x} & for second f(x)=-e^{-x}$$\int_{0}^{1}\left(\frac{2e^{2x}}{x}+\frac{1}{xe^x}-\frac{e^{2x}}{x^2}+\frac{1}{x^2e^x}\right)dx... 2 \begin{align} \int_0^1(\frac{2e^{2x}}{x}+\frac{1}{xe^x}-\frac{e^{2x}}{x^2}+\frac{1}{x^2e^x})dx&=\int_0^1(\frac{2e^{2x}}{x}-\frac{e^{2x}}{x^2}+\frac{e^{-x}}{x}+\frac{e^{-x}}{x^2})dx\\&=\int_0^1d(\frac{e^{2x}}{x}+\frac{-e^{-x}}{x})\\&=\frac{-e^{-x}+e^{2x}}{x}|_0^1\\&=-e^{-1}+e^2-lim_{x\rightarrow0}\frac{e^{2x}-e^{-x}}{x} ... 0 Notice, this standard formula\int\sqrt{a^2-x^2}dx=\frac{1}{2}\left(x\sqrt{a^2-x^2}+a^2\sin^{-1}\left(\frac{x}{a}\right)\right)$$Hence, we have$$\int\sqrt{16-9x^2}dx=\int\sqrt{(4)^2-(3x)^2}dx$$let 3x=u\implies 3dx=du$$=\frac{1}{3}\int\sqrt{(4)^2-(u)^2}du ... 0 I thought that some of the solutions were a little too long, so here is mine. It is essentially wythagoras's. \begin{array}\\ \int_{-a}^{+a} A^2\cos^2\left(\frac{n\pi a}{2}x\right)dx &=A^2\int_{-\pi a^2}^{+\pi a^2} \cos^2\left(\frac{n}{2}y\right)\frac{dy}{\pi a} \quad (y=\pi a x \text{ so } x = \frac{y}{\pi a})\\ &=\frac{A^2}{\pi a}\int_{-\pi ... 0 It can be demonstrated that: \begin{align}\tag{1} \frac{1}{r^2} \frac{\partial}{\partial r} \left( r^2 \,\frac{\partial f}{\partial r} \right) &= \frac{1}{r^{2}} \, \left( r^{2} \, \frac{\partial^{2} f}{\partial r^{2}} + 2 \, r \, \frac{\partial f}{\partial r} \right) \\ &= \frac{\partial^{2} f}{ \partial r^{2}} + \frac{2}{r} \, \frac{\partial ... 4 Let: f(s)=\int_{0}^{1}\frac{x^s}{\sqrt{1+x}}\,dx. $$We have, by integration by parts:$$ f(s+1)+f(s) = \int_{0}^{1} x^s\sqrt{1+x}\,dx = \frac{1}{s+1}\left(\sqrt{2}-\frac{1}{2}\,f(s+1)\right)$$hence:$$ \left(2s+3\right)\, f(s+1)+(2s+2)\,f(s) = 2\sqrt{2},$$f(0)=2\sqrt{2}-1 and \lim_{s\to +\infty}f(s)=0. We have:$$ f(s)=\sum_{n\geq 0}\frac{(-1)^n ...

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