New answers tagged

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 be $I$. 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: $$\int_{x=0}^{x=2}\int_{y=0}^{y=\sqrt{1-(x-1)^2}}\sqrt{4-x^2-y^2}\ \mathrm dy\ \mathrm dx$$


1

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.



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