# Tag Info

0

Using polar coordinates, $x^2+y^2=2x$ gives $r^2=2r\cos\theta$ so $r=2\cos\theta$. For the part of the region inside the circle which is also in the first quadrant, we get $\displaystyle\int_0^{\frac{\pi}{2}}\int_0^{2\cos\theta}\sqrt{4-r^2}\; ... 0 HINT: Note that the integral can be written $$\int \frac{2 \lambda a}{ (e^{at}-1)\lambda \sigma^2+2ae^{-at}}\,dt=\int \frac{2 \lambda a e^{at}}{e^{at} (e^{at}-1)\lambda \sigma^2+2a}\,dt$$ Now, enforce the substitution$x=e^{at}$with$dx=ae^{at}\,dt$and evaluate the integral $$\int \frac{2\lambda}{\lambda \sigma^2x(x-1)+2a}\,dx$$ 2 Sure, you can do it, but Matlab also has an integral function that may be simpler to apply. Here is an example for $$\int_0^1\frac{dx}{1+x^2}=\frac{\pi}4$$ % test.m f = @(t,x) 1./(1+t.^2); % Define integrand a = 0; % Define limits b = 1; y0 = 0; % Initial value is zero [t,y] = ode45(f,[a b],y0); format long; y(end) fun = @(x) 1./(1+x.^2); integral(fun,a,b) ... 0 By Fubini's theorem, you can perform the integrals one at a time in either order (i.e., you can integrate in$x$first then$y$or in$y$first and then$x$). If you do the latter, you will find that for each$x$,$\int_{-b}^b f(x,y) dy = 0$. Indeed, we see $$\int_{0}^b f(x,y)dy = \int^{-b}_0 f(x,-y)(-1) dy = \int^{-b}_0 f(x,y)dy = - \int^0_{-b} f(x,y) dy.$$ ... 2 Let's write$\psi = \dfrac{-1}{\phi} = \dfrac{1-\sqrt{5}}{2}$. The trick is to multiply numerator and denominator of$\dfrac{(x\phi)^n - (-1)^n}{\phi^{2n} - (-1)^n}$with$\phi^{-n}$to get $$\frac{(x\phi)^n - (-1)^n}{\phi^{2n} - (-1)^n} = \frac{x^n - \psi^n}{\phi^n - \psi^n} = \frac{x^n - \psi^n}{\sqrt{5}\,F_n}.$$ Now we can multiply the numerator with ... 2 The Cantor set has measure$0$and has the same cardinality as$[0,1]$, which has measure$1$. 0 I can only assume that your integral, properly formatted, is: $$\iint_{\cal D} \sqrt{x y (x^2 + y^2)^{n/2}} dx/ dy.$$ Let$x^2 + y^2 = r^2$and$x = r \cos(\theta)$and$y = r \sin(\theta )$, and$dx\ dy = r\ dr\ d\theta. Then the integral becomes: $$\int\limits_{r = 0}^2 r^{n+1} \, dr \int\limits_{\theta = 0}^{\pi/2} \sqrt{ \cos(\theta) \sin (\theta)} ... 0 The formula is wrong. Already for n=1 the result is \frac{8\pi R^3}3\ne\frac{8\pi R^3}{3!}. For n=2 it's$$ (8\pi R^3)^2\int_0^1\mathrm dx_1x_1^2\int_0^\sqrt{1-x_1^2}\mathrm dx_2x_2^2=(8\pi R^3)^2\int_0^1\mathrm dx_1x_1^2\left(1-x^2\right)^\frac32=(8\pi R^3)^2\frac\pi{32}\ne\frac{(8\pi R^3)^2}{6!}\;. $$1 We present first a direct approach here that applied Leibniz's Rule for Differentiating Under the Integral. Let I(a) be the function given by$$I(a)=\int_a^\infty e^{-ax^2}\,dx \tag 1$$Enforcing the substitution x \to x/\sqrt{a} yields$$\begin{align} I(a)&=\frac{1}{\sqrt{a}}\int_{a^{3/2}}^\infty e^{-x^2}\,dx\\\\ &=\frac12 ... 1 Hint: The ingredients are $$B\left(a,b\right)=\frac{\Gamma\left(a\right)\Gamma\left(b\right)}{\Gamma\left(a+b\right)}$$the link between Gamma and Polygamma functions ... 8 Recall this useful formula: $$\int_a^b f(x)\ \mathrm dx=\int_a^b f(a+b-x)\ \mathrm dx$$ Let the required value of the integral beI. Now, we have: $$I=\int_0^{\frac{ \pi }{2}} \frac{\sin^{1395}x}{\sin^{1395}x + \cos^{1395}x}\ \mathrm dx = \int_0^{\frac{ \pi }{2}} \frac{\sin^{1395}(\frac\pi2-x)}{\sin^{1395}(\frac\pi2-x) + \cos^{1395}(\frac\pi2-x)}\ ... 2 If you integrate \int xy\,dx = \frac{x^2}{2}y, plug-in (2,1) and (0,0), you would get 2-0. For the other half, \int -y^2\,dy = -\dfrac{y^3}{3}, plug in end points, and get -\dfrac{1}{3}-0. This would yield a line integral value 2-\dfrac{1}{3}=\dfrac{5}{3}. While that would make this an easy computation, it's totally wrong. First, let's get the ... 1 We have:$$x^2+y^2=2x$$Moving everything to the left:$$x^2-2x+y^2=0$$Completing the square:$$(x-1)^2+y^2=1$$Which is a circle with centre (1,0), hence:$$0\le x\le2$$Solving y:$$y=\sqrt{1-(x-1)^2}$$So the integral becomes:$$\int_{x=0}^{x=2}\int_{y=0}^{y=\sqrt{1-(x-1)^2}}\sqrt{4-x^2-y^2}\ \mathrm dy\ \mathrm dx1 This is the most common asked question. If you google it, you will find many links explain different ways to solve such integrals. Let us try ourselves, \begin{align} \int{x^2 e^{(-a x^2)}}dx=\int{x (x e^{(-a x^2)})}dx=x\int{x e^{(-a x^2)}dx}-\int{\int{x e^{(-a x^2)}dx}}dx \end{align} Now we know that \int{x e^{(-a x^2)}}dx=-\frac{1}{2}\,{\frac {{{\rm ... 1 We have:\int_0^ue^{-t^2}\mathrm dt=\frac{\sqrt\pi}2\mbox{erf}(u)$$Letting t=x\sqrt a, u=w\sqrt a:$$\int_0^we^{-ax^2}\mathrm dx=\frac{\sqrt\pi}{2\sqrt a}\mbox{erf}\left(\frac{w}{\sqrt a}\right)$$Also: \displaystyle\quad\int\mbox{erf}(x)\ \mathrm dx \displaystyle=x\mbox{erf}(x)-\int x\ \mathrm d\left(\mbox{erf}(x)\right) ... 1 If a=\frac1{t^H\sqrt[4]2}, b=\frac{t^H}{2^{3/4}}, then we need$$\begin{align}\int_0^{\infty}xe^{-2x}\cdot\frac2{\sqrt{\pi}}\int_0^{ax-b}e^{-y^2}dy&=\left.\frac2{\sqrt{\pi}}\left(-\frac x2-\frac14\right)e^{-2x}\int_0^{ax-b}e^{-y^2}dy\right|_0^{\infty}\\ &+\frac a{\sqrt{\pi}}\int_0^{\infty}\left(x+\frac12\right)e^{-2x}e^{-(ax-b)^2}dx\\ ... 0 This is the simplest method I guess. 4 From $$y = e^{\ln|x|-\ln|x+1|+\ln|2|}$$ you may write $$y = e^{\ln|2x|-\ln|x+1|}=e^{\ln\left|\frac{2x}{x+1}\right|}= \frac{2x}{x+1}.$$ 2 $$(x^3+y^2)3x^2dx+(y^2-x^3)2ydy=0$$ HINT : Obviously, this form suggests the change of variables :\begin{cases} x^3=X \\ y^2=Y \end{cases}$$$(X+Y)dX+(Y-X)dY=0$$ I suppose that you can take it from here. HINT : change of function$Y=XF(X)\quad\to$first order linear ODE. 0 Hint. You can use the relation$|\mathrm{d}z| = \frac{r}{iz} \, \mathrm{d}z$to write $$\int_{|z|=r}\frac{|\mathrm{d}z|}{|z-a|^2} = \frac{r}{i} \int_{|z|=r}\frac{\mathrm{d}z}{(z-a)(r^2-\bar{a}z)}.$$ The denominator has simple poles at$z = a$and at$z = r^2/\bar{a}$. Notice that if$|a| \neq r$, then exactly one of them lies in the circle$|z| = r$. ... 0 The projection into the$xy$-planes shows the outline formed by the planes$y=1$,$y=x$, and$y=-x$. You can see that if$x$is the outer variable, then for$x\le0$,$-x\le y\le1$, whereas when$x\ge0$,$x\le y\le1$. You can determine that by drawing a vertical line (a curve of constant x) through the figure. That means we would have to break up the ... 0 Functions on a finite set would be characterized by a finite set of parameters. The idea of characterizing polynomials of all orders on a non-trivial interval of$\mathbb{R}$using a finite set of parameters is immediately objectionable. You just know that cannot be right, which rules out the idea for more general functions that include such polynomials on ... 0 A Puiseux series is formal Laurent power series in$T^{\frac1n}$for some$n$(the$n$may vary with the series). The set of Puiseux series is a field, denoted$k{\ll} T{\gg}$, and it is the algebraic closure of the field of formal power series$k[[T]]$. 0 "(Hint: Substituting the first few values of n yields a system of linear equations in a, b, c, d, which has a unique solution)." So$\sum_{j=0}^0 j^2 = 0 = a*0^3 + b*0^2 + c*0 + d = d\sum_{j=0}^1 j^2 = 1 = a*1^3 + b*1^2 + c*1 + d = a + b + c +d\sum_{j=0}^2 j^2 = 1 + 4 = 5 = a*2^3 + b*2^2 + c*2 + d = 8a + 4 b + 2c + d\sum_{j=0}^3 j^2 = 1 + 4 + ...

1

For a general Lagrangian $L[t,q,q',q'',\ldots]$ the Euler-Lagrange equation reads $$L_q = [L_{q'}]' - [L_{q''}]'' + [L_{q''}]''' - \ldots = \sum_{n\geq 1} (-1)^{n+1}[L_{q^{(n)}}]^{(n)}$$ If the Lagrangian is time-independent. $L_t = 0$, we have $$\frac{dL}{dt} = L_qq' + L_{q'}q'' + L_{q''}q''' + \ldots = L_q q' + \sum_{n\geq 1}L_{q^{(n)}}q^{(n+1)}$$ and ...

0

$$P\int_\frac{S}{a}^T(0-at)\:\text{dt}$$ $$= -P\int_\frac{S}{a}^Tat\:\text{dt}$$ $$= -Pa \cdot \frac{t^2}{2} |_{t =S/a}^{t =T}$$ $$= -\frac{Pa}{2} \cdot (T^2 - \frac{S^2}{a^2})$$ $aT = Q \Rightarrow T = Q/a$ $$= -\frac{Pa}{2} \cdot (\frac{Q^2}{a^2} - \frac{S^2}{a^2})$$ $$= \frac{P(S^2 - Q^2)}{2a}$$

1

Using the substitution you started with: $$\int \frac{v}{2 + v} \, dv = \int \frac{u-2}{u} \, du = \int \left(1 - \frac{2}{u} \right) \, du = u-2\ln|u| + C =v-2\ln|2+v|+D.$$ Here is another method: $$\int \frac{v}{2 + v} \, dv = \int \frac{2+v-2}{2 + v} \, dv = \int \left (1 - \frac{2}{2+v} \right) \, dv = v-2 \ln|2+v| + C.$$

0

If $f(x)$ is monotonic increasing, then $f(n) < \int_n^{n+1} f(x) dx < f(n+1)$. Therefore $\frac1{n}\sum_{k=0}^{n-1} f(k) < \int_0^n f(x) dx < \frac1{n}\sum_{k=1}^{n} f(k)$ or $\frac1{n}(f(0)-f(n)) < \int_0^n f(x) dx-\frac1{n}\sum_{k=1}^{n} f(k) < 0$. Therefore, if $f(x)$ is monotonic increasing and $\dfrac{f(n)}{n} \to 0$, then ...

0

Let's rewrite $F(x)$ a little bit. Therefore, let $G$ be the antiderivative of $g$, i.e. $G'(x)=g(x)$ for all $x$. Then, using the fund. theorem of calculus, we can write $$F(x) = \int_x^{\sin x} \left(\int_0^{\sin t} (1+u^4)^{0.5} \,du\right)dt =: \int_x^{\sin x} g(t)\,dt = G(\sin x) - G(x).$$ Now it is easier to handle: $$F'(x) = G'(\sin x) \cdot \cos ... 0 It looks like a Cesaro sommation (but Cesaro cared more about series than functions) https://en.wikipedia.org/wiki/Ces%C3%A0ro_summation 2 The function f(x,y,z) = e^y\cos x+z is a potential for the field (-e^y\sin x,e^y \cos y, 1), so by the fundamental theorem of calculus we have$$\int_C -e^y\sin x\,{\rm d}x + e^y\cos x\,{\rm d}y + {\rm d}z = f(\pi,\pi,0)-f(0,0,1).$$2 It does seem a bit on the ugly side in that the region seems to be now down to one square pyramid with vertex at (0,0,2). I think the x-component of \vec\nabla\times\vec V should be (y+1)e^y: check it. For part b), I get your answer now that the domain of w has been corrected. For the Jacobian, I get$$J=\left|\det\begin{bmatrix}\frac{\partial ...

1

Note that the integration region is a triangular region with vertices at $(a,a)$, $(a,x)$, and $(x,x)$ in the $s-\xi$-plane. Thus, if the inner integral is on $s$, we see that for any fixed $\xi$, $s$ begins at $a$ and ends at $\xi$. If the inner integral is on $\xi$, we see that for any fixed $s$, $\xi$ begins at $s$ and ends at $x$. Therefore, we can ...

1

$$\int_0^1 \frac{200\sqrt5(1-x^2)-300(1-x)^2}{ \left[5\sqrt5(1+x)^2-15(1-x^2)+2\sqrt5(1-x)^2 \right]^2}dx=(2\phi+1)(\phi+2)$$

1

If $P=(a_0,b_0)$ is a critical point, i.e. solution of both $f_a=0$ and $f_b=0,$ AND the determinant of the Hessian at $P$ (what you have displayed, but with coordinates of critical point plugged in) is positive, then that tells you there is either a local max or a local min at $P.$ You have to look say at the first partials near $P$ to tell which. On the ...

0

Okay, you have that your segment goes from $(0,0)$ to $(1,2)$. If you let $t$ be a parameter that expresses the value of $1$, then $2$ will be $2t$. So your $x=t$ and your $y=2t$ but they are bounded from $0$ to $1$ since your ending point is $(1,2)$. So, $t \in [0,1]$. Then you get $dy = d(2t) = 2$ and your integral becomes : $\int_C 3xydy = \int_0^1 ... 2 Hint:$\int_0^\infty\dfrac{e^{-x^2}}{\sqrt{t^2+x}}~dx=2\int_0^\infty e^{-x^2}~d\left(\sqrt{t^2+x}\right)=2\int_t^\infty e^{-(x^2-t^2)^2}~dx=2\int_t^\infty e^{-x^4+2t^2x^2-t^4}~dx$Similar to Evaluating$\int_{1}^{\infty}\exp(-(x(2n-x)/b)^2)\,\mathrm dx$0 Am giving a try. Did a bit progress in my way and so want to share it. $$I = \int_{-1}^{1} e^x \cdot (f(x) + f(-x))dx$$ $$= 0 - \int_{-1}^1 e^x \cdot (f'(x) - f'(-x))dx$$ $$= 2f'(c)(e-\frac{1}{e})$$ Now, $$I = \int_{-1}^1 e^x f(x) + \int_{-1}^{1} e^{-x} f(x)$$ $$\implies \int_{-1}^{1} e^{-x} f(x)= 2(e-\frac{1}{e})\cdot(f'(c) - f(1) + f(-1))$$ And am ... 0 Hint: By parts, $$I=\int\sin^2(x)dx=-\cos(x)\sin(x)+\int\cos^2(x)dx=-\cos(x)\sin(x)+J,$$ and $$J=\int\cos^2(x)dx=\int(1-\sin^2(x))dx=x-I.$$ 2 First, note that the derivative of the coefficient of "dx",$10x^4- 2xy^3$, with respect to y, is$-6xy^2$and that the derivative of the coefficient of "dy",$-3x^2y^2$is with respect to x also$-6xy^2$. That tells us that the integral is independent of the path. One method of doing this would be to choose some simple path, say the straight line between ... 1 In a question like this, you first try the fundamental theorem for line integrals and then, if that fails, attempt harder techniques. Summarizing from wikipedia: If$f(x,y)$is a differentiable function on$\mathbb{R}^2$and$\gamma$a path from$(a,b)$to$(c,d)$, then $$f(c,d)-f(a,b)=\int_\gamma f_x(x,y)dx+f_y(x,y)dy.$$ (I am being very lazy with the ... 0 $$\int\frac{1}{\sin (x)} dx=\int\frac{sin(x)}{sin^2(x)}dx=\int\frac{sin(x)}{1-cos^2(x)}$$ Use substitution$t=cos(x) \to dt=sin(x)dx\to dx=\frac{dt}{sin(x)}$$$\int \frac{sin(x)}{1-t^2}*\frac{dt}{sin(x)}=\int \frac{dt}{1-t^2}=arth(t)=arth(cos(x))+C$$ 4 Hint. One may just interchange sum and integration : $$\int_0^\infty\sum_{n=0}^{\infty}\frac{x^n}{2^{(n+1)^sx^{n+1}}+1}dx=\sum_{n=0}^{\infty}\int_0^\infty\frac{x^n}{2^{(n+1)^sx^{n+1}}+1}dx$$ then, by the change of variable $$u=(n+1)^sx^{n+1}, \quad du=(n+1)^{s+1}x^ndx,$$ one gets $$... 3 With t=\dfrac u\lambda+\lambda,$$\int_\lambda^\infty e^{-t^2/2}dt=\int_0^\infty e^{-(u/\lambda+\lambda)^2/2}\frac{du}\lambda=\frac{e^{-\lambda^2/2}}\lambda\int_0^\infty e^{-u^2/2\lambda^2}e^{-u}dt.$$For large \lambda,$$\int_0^\infty e^{-u^2/2\lambda^2}e^{-u}du\approx\int_0^\infty e^{-u}du.$$1$$ \frac{\partial}{\partial t}f(x,t)=\frac{\partial}{\partial t}\int_{0}^{g(x,t)} e^{-u^2} du=e^{-g^2(x,t)}\frac{\partial}{\partial t}g(x,t)\ , $$where one uses the fundamental theorem of calculus \frac{d}{dz}\int_0^z dt\ h(t)=h(z). Taking a second derivative (using the product rule)$$ \frac{\partial^2}{\partial t^2}f(x,t)=\frac{\partial}{\partial ... 2 Almost from definition $$I=\int e^{-\frac{t^2}{2}}\,dt=\sqrt{\frac{\pi }{2}} \text{erf}\left(\frac{t}{\sqrt{2}}\right)$$ So $$J=\int_\lambda^\infty e^{-\frac{t^2}{2}}\,dt=\sqrt{\frac{\pi }{2}} \text{erfc}\left(\frac{\lambda }{\sqrt{2}}\right)$$ Now, look here for the asymptotic expansion for large$\lambda$to get $$J=e^{-\frac{\lambda ^2}{2}} ... 1 One easy way to do it to notice that$$\frac{\partial}{\partial s}\int_0^{\infty}\dfrac{e^{-sk}sinkx}{k}=-\int_0^{\infty}e^{-sk}sinkx=-\dfrac{x}{x^{2}+s^{2}}$$Then you just have to revert the derivative with respect to "s"$$-\int ds \dfrac{x}{x^{2}+s^{2}}=-\arctan(s/x)$$2 Assume x>0,\,s>0. Then by differentiating the following identity with respect to s,$$ f(s)=\int_0^\infty {\frac{e^{−sk}} k}\sin(kx)\,dk $$one obtains$$ f'(s)=-\int_0^\infty e^{−sk}\sin(kx)\,dk=-\frac{x}{x^2+ s^2} $$which gives$$ f(s)=-\arctan \left( \frac{s}x\right)+C. $$Observing that, as s \to \infty, f(s) \to 0, we then obtain ... 8 This may not be what you want, but the shortest way is definite by geometric intuition. The integrand is nothing but a semicircle centered at (0.5, 0.0) with radius 0.5. So the integration, or the area, is 1/2 \cdot \pi (1/2)^2 = \pi/8. 3$$\sqrt{x-x^2} = \frac{1}{2}\sqrt{1-(2x-1)^2},$$so a single substitution of the form$$2x - 1 = \sin \theta, \quad dx = \frac{1}{2} \cos \theta \, d\theta$$immediately yields$$\int \sqrt{x-x^2} \, dx = \frac{1}{4}\int \cos^2 \theta \, d\theta,$\$ and the rest is straightforward.

Top 50 recent answers are included