New answers tagged

3

You cannot, because your inequality does not hold: just take $a=4$ and $b=3.5$. If we take the function $$ \psi: \xi \mapsto \int_{0}^{+\infty}\cos(\xi x)e^{-x^4}\,dx $$ this is its graph over $[0,10]$: $\hspace{2cm}$ and it is not monotonic in absolute value, even if fast-decaying.


2

The reason that the dog bone contour is inapplicable here is that branch cuts from, say, $0$ to $\infty$ and $1$ to $\infty$ do not "collapse" into the "slit" from $0$ to $1$. To see this, we note that for $z=x+iy$, $x>1$ and $y\to 0+$ we have $\arg(z)=0$ and $\arg(1-z)=-\pi$, while for $x>1$ and $y\to 0^-$, we have $\arg(z)=2\pi$ and $\arg(z)=\pi$. ...


1

the answer of Behrouz Maleki is correct (+1). I add a figure that illustrate the solution. The region of integration is a triangle in the plane $(u,v)$ delimited by the lines $$ v=s \qquad u=t \qquad v=u $$


4

Take $\log\left(x\right)=v$. We have \begin{align} I&=\int_{0}^{\infty}\frac{\exp\left(-x^{n}\right)-\exp\left(-x^{m}\right)}{x\log\left(x\right)}\ dx\\[10pt] &=\int_{-\infty}^{\infty}\frac{\exp\left(-e^{vn}\right)-\exp\left(-e^{vm}\right)}{v}\ dv\\[10pt] &=\int_{0}^{\infty}\frac{\exp\left(-e^{vn}\right)-\exp\left(-e^{vm}\right)}{v}\ dv-\int_{0}^{...


1

Hint $$I=\int\limits_s^t\int\limits_s^u e^{-\lambda(t-v)}(u-v)^{-\beta-1}dvdu=\int_{s}^{t} \int_{v}^{t}e^{-\lambda(t-v)}(u-v)^{-\beta-1}dudv$$ $$I=-\frac{1}{\beta}\int_{s}^{t}(t-v)^{-\beta}e^{-\lambda(t-v)}dv=-\frac{1}{\beta}\int_{s-t}^{0}(-v)^{-\beta}e^{\lambda v}dv=-\frac{1}{\beta}\int_{0}^{t-s}v^{-\beta}e^{-\lambda v }dv$$


2

First Integral $$\int_{0}^{1}\int_{0}^{1-x}\int_{0}^{xy}x\,dz\,dy\,dz=\int_{0}^{1}\int_{0}^{1-x}x^2y\,dy\,dx=\frac{1}{2}\int_{0}^{1}x^2 (1-x)^2 dx=\frac12\beta(3,3)=\frac1 {60}$$ Second Integral $$x=r\cos\theta\quad,\quad y=r\sin\theta$$ $$\left|\frac{\partial(x,y)}{\partial(r,\theta)}\right|=r$$ $$\int_{0}^{2\pi}\int_{0}^{2}\int_{0}^{3}r^2\,z\,dz\,dr\,d\...


1

Since you want the sine series (if I understand you correctly), you first extend your function to an odd function: $$ \tilde f(x)= \begin{cases} -x & -2\pi<x<-\pi\\ x & -\pi\leq x\leq \pi\\ -x & \pi<x\leq 2\pi \end{cases} $$ Then calculate the Fourier coefficients of this $4\pi$-periodic function as usual: $$ a_n=\frac{2}{4\pi}\int_{-2\...


7

Hint: $$\int r^3\sqrt{8-r^2}dr= \frac12\int r^2\sqrt{8-r^2}d(r^2)=\frac12\int t\sqrt{8-t}\,dt.$$ You can integrate by parts on the second factor. This substitution will work for all integrals of the form $$\int P_o(r)(a^2-r^2)^\alpha dr$$ where $P_o$ is a polynomial with odd powers, and successive integrations by part will progressively lower the ...


6

Substitute $r=2\sqrt{2} \sin\theta$; $dr=2\sqrt{2}cos\theta\ d\theta$ $$(2\sqrt{2})^5\int\sin^3\theta\ \cos^2 \theta \ d\theta=(2\sqrt{2})^5\int\sin^3\theta\ (1-\sin^2 \theta )\ d\theta$$$$(2\sqrt{2})^5\int(\sin^5 \theta- \sin^3 \theta )\ d\theta=(2\sqrt{2})^5\left(\int((1-\cos^2\theta)d(\cos\theta)-\int(1-\cos^2\theta)^2 d(\cos\theta)\right)$$


0

I will solve it vertically. I use sigma notation instead of integral because I am not familiar with manipulating integral form. Solve $x^2+16y^2=144$ in terms of $x$. $$x=\sqrt{144-16y^2}$$ Since it is rotated around the $y$-axis, we can use the $x$ value as the radius for circle cross sections. The area of these cross sections: $$Area=\pi r^2$$ $$Area=\pi(\...


1

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0

For $\int e^{\cos(x-a)}~dx$ , $\int e^{\cos(x-a)}~dx$ $=\int\sum\limits_{n=0}^\infty\dfrac{\cos^{2n}(x-a)}{(2n)!}dx+\int\sum\limits_{n=0}^\infty\dfrac{\cos^{2n+1}(x-a)}{(2n+1)!}dx$ $=\int\left(1+\sum\limits_{n=1}^\infty\dfrac{\cos^{2n}(x-a)}{(2n)!}\right)dx+\int\sum\limits_{n=0}^\infty\dfrac{\cos^{2n+1}(x-a)}{(2n+1)!}dx$ For $n$ is any natural number, $\...


1

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0

Clearly, $g(r_M) \ge \frac{1}{2}\int_{0}^{1}x f(x) dx \overset{\text{def}}{=} \frac{1}{2}\mathbb{E}X$. Choose $c = 1/2$. Then \begin{align*} c \mathbb{E}X = \frac{1}{2} \mathbb{E}X \le \frac{1}{2} (\mathbb{E}X)(2 - F(r_M)) \le g(r_M)(2 - F(r_M)) = \int_{r_M}^{1}x f(x) dx \end{align*} Hence we found such a $c$, independent of our distribution.


0

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0

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0

Using your said integral, I will present how I would write down the solution with pen and paper (which of course is similar to yours): $$\int_0^1\int_x^\sqrt{x} (x+y^3) dy\quad\!\!\!\!dx=\int_0^1\left(xy+\frac{y^4}{4}\right){\huge{|}}_x^\sqrt{x}dx$$ $$=\int_0^1x\sqrt{x}+\frac{x^2}{4}-\left(x^2+\frac{x^4}{4}\right)dx=\int_0^1x^{\frac{3}{2}}-\frac{3x^2}{4}-\...


3

Since the number $\sqrt{2}-1$ is invariant under the transformation $x \mapsto \frac{1-x}{1+x}$, it is natural to make that substitution. Doing so, we find: $$\begin{align} I &= \int_0^{\sqrt{2}-1} \frac{\ln(1+x^2)}{1+x} dx \\&= \int_{\sqrt{2}-1}^1 \cfrac{\ln(1+x^2)+\ln\left(\frac2{(1+x)^2}\right)}{1+x}dx \\&=\frac12 I + \frac12 \int_{\sqrt{2}-1}...


0

The following is the solution to solve it horizontally. Solve for $x$ in terms of $y$ to get $x=\sqrt{144-16y^2}$. This is the right half. We can use method of rings to solve this problem, where we integrate from $y=0$ to $3$ and we use $\sqrt{144-16y^2}$ as our radius: $$\int_{-3}^3 \pi\left(\sqrt{144-16y^2}\right)^2dy=\int_{-3}^3 \pi\left(144-16y^2\right)...


2

You can use $\displaystyle\int_0^2\int_0^{\sqrt{4-2y}}\frac{1}{\sqrt{2y-y^2}}dx\,dy=\int_0^2\frac{\sqrt{4-2y}}{\sqrt{2y-y^2}}dy=\int_0^2\frac{\sqrt{2}}{\sqrt{y}}dy=\lim_{t\to0^+}\int_t^2\frac{\sqrt{2}}{\sqrt{y}}dy$ $\hspace{.6 in}\displaystyle=\lim_{t\to0^+}\sqrt{2}\left[2\sqrt{y}\right]_t^2=\lim_{t\to0+}2\sqrt{2}(\sqrt{2}-\sqrt{t})=4$ Alternate method: ...


1

First, as your friend likely noticed, we can rewrite the main curve as $y=2\cosh 10x$. Also, we are going from $x=0$ to $x=6$, so we integrate from $0$ to $6$. This solid of revolution can be solved using method of rings, where our radius is simply $y=2\cosh 10x$, so we get: $$\int_0^6 \pi(2\cosh 10x)^2dx$$ Now, if we calculate this out, we get the same ...


1

Since this is between $y=\frac 5 x$ and $y=0$ and we're rotating it around $y=6$, we need to use method of cylinders. Therefore, we're going to take the inverse of this function and say $x=\frac 5 y$. Now, at $x=1$, we have $y=5$ and at $x=5$, we have $y=1$. Therefore, from $y=1$ to $y=5$, the height of the cylinder is $\frac 5 y-1$ (remember to subtract $1$...


10

Hint. Assume we have $n>0,\, m>0$ and set $$ f(n,m):=\int_{0}^{\infty}{e^{-x^n}-e^{-x^m}\over x\ln{x}}dx. $$ One may just derivate with respect to $n$: $$ \frac{\partial}{\partial n}f(n)=-\int_{0}^{\infty}x^{n-1}e^{-x^n}dx=-\frac1n $$ then integrating with respect to $n$, using $f(m,m)=0$, gives $$ \int_{0}^{\infty}{e^{-x^n}-e^{-x^m}\over x\ln{x}}...


1

This is actually one of the standard formulas in a first-year calculus course for the surface area of a surface of revolution. You want to compute the integral $$\int_0^{2\pi}\int_0^r \phi(u)\sqrt{1+\phi'(u)^2}\,du\,d\theta.$$ (This is the surface obtained by rotating the graph $y=\phi(x)$ about the $x$-axis.) Alternatively, you can derive this using the ...


4

For primitives, it is purely notation. For definite integrals, there is a reason. With the Riemann sums, you're multiplying. The integral is, roughly speaking, just the Riemann sum $$\sum_{i=1}^nf(t_i)(x_i -x_{i-1}) $$ as the lengths of each subinterval $[x_{i-1}, x_{i}]$ tend to $0$. In other words, they are all "infinitesimally" small. This is in quotes ...


4

The intuitive understanding of the (Riemann) integral is that you are adding together the areas of a continuum of infinitesimally narrow rectangles. The $dx$ in this notation represents the width of those rectangles, while $f(x)$ is the height of the rectangle with base $[x,x+dx]$. Under this intuition, integrating a function without a $dx$ or with a $dx$ in ...


3

$\int f(x) dx$ is just a symbol defining the integral of $f$. $dx$ tells you that the dummy variable of integration is $x$. This is similar to asking why you cannot divide by $\sin$ to get $$\sin x = y \implies x = \frac{y}{\sin} \text{ (woops!)}$$ That having been said, you might want to look at the Riemann–Stieltjes integral as it can handle the first ...


4

The integrand is periodic, with period $2\pi$, so the second one is independent of $a$. In fact $$ \int_{-\pi}^\pi e^{\cos(x-a)}\,dx = 2\pi I_0(1) \approx 7.9549 $$ where $I_0$ is a certain Bessel function. In the integral form for $I_\alpha(x)$ given on that page, put $x=1$ and $\alpha=0$.


2

Here is the answer to the question (after putting the substitution of Did) $$-\frac{1}{2}c^2-\frac{(abdc^m)^{\frac{2}{m}}}{m}\{\Gamma(\frac{-2}{m})-\Gamma(\frac{-2}{m},abd)\}$$ Thanks Did.


0

using Differentiation under the integral sign and since $cos$ is even function we can write it $\cos \left( -5t^{ 2 }-2t-4 \right) =\cos \left( 5t^{ 2 }+2t+4 \right) $ so we get$$\frac { d }{ dy } \left( \int _{ y }^{ x } \cos { \left( 5t^{ 2 }+2t+4 \right) } dt \right) ={ x }_{ y }^{ \prime }\cos { \left( 5x^{ 2 }+2x+4 \right) } -{ y }_{ y }^{ \prime }...


9

Let $$I = \int\frac{1}{(x^4+1)^{\frac{1}{4}}}dx$$ Put $x^2=\tan \theta,$ Then $2xdx = \sec^2 \theta d\theta$ So $$I = \int\frac{\sec^2 \theta}{\sqrt{\sec \theta}}\cdot \frac{1}{2\sqrt{\tan \theta}}d\theta = \frac{1}{2}\int\frac{1}{\cos \theta \sqrt{\sin \theta}}d\theta = \frac{1}{2}\int\frac{\cos \theta}{(1-\sin^2 \theta)\sqrt{\sin \theta}}d\theta$$ Now ...


0

For even $n$, just do $x\mapsto x-\pi/2$ and use the fact that your integrand is becoming odd (and you integrate over the symmetric interval $[-\pi/2,\pi/2]$. For odd $n$, inserting $t=ie^{ix}$ into the generating function expansion $$ e^{\frac{z}{2}(t-1/t)}=\sum_{k=-\infty}^{+\infty} J_k(z) t^k, $$ we find the Fourier cosine series $$ e^{iz\cos x}=J_0(z)+\...


14

Hint: This can be written as : $$\int \frac{x^4dx}{x^5\left(1+\frac{1}{x^4}\right)^{1/4}}$$ Now substitute $1+\frac{1}{x^4}=t^4$ $$\implies t^3dt=-\frac{1}{x^5}dx$$ and $$x^4=\frac{1}{t^4-1}$$ to get $$\int \frac{t^2dt}{1-t^4}$$ Now use partial fractions.


0

From what I understood, the double integral asked is equal to the volume of a solid whose base is ${R}$ and the height varies according to $f(x,y)=|y-x^2|$. Since the region is just rectangular (using Fubini's theorem I think), $$\iint_{R} \left|y-x^2\right| d{R}=\int_0^2\int_{-1}^1 \left|y-x^2\right| dxdy.$$


0

The equation in the mentioned Wikipedia page is a special case of the general problem of motion equation for a system with variable mass that comes from the conservation of the momentum of the system. You can see the derivation of the pertinent equation at this page. The pure energetic approach in OP is not correct because the energy is conserved only in ...


1

You are missing some clue. In particular, you did not use any property of $\sqrt{|y-x^2|}$ besides its existence. Hint: Consider the region where $y\geq x^2$ and $x^2 \geq y$ separately. Drawing the rectangle and the curve might help you. Edit: $$\int\int_R \sqrt{|y-x^2|}dxdy=\int_{-1}^1\int_0^{x^2}\sqrt{x^2-y}dydx+\int_{-1}^1\int_{x^2}^{2}\sqrt{y-x^2}...


1

Let $X$ be the number of attempts without a collision. Then, we have $\mathbb{E}[X]$ $= \displaystyle\sum_{k = 1}^{\infty}k\Pr[X = k]$ $= \displaystyle\sum_{k = 1}^{\infty}\sum_{n = 1}^{k}\Pr[X = k]$ $= \displaystyle\sum_{n = 1}^{\infty}\sum_{k = n}^{\infty}\Pr[X = k]$ $= \displaystyle\sum_{n = 1}^{\infty}\Pr[X \ge n]$. The probability that $X \ge n$ is ...


2

No, the calculation of the expected number of attempts before the first repeated value is a completely different problem than the number of attempts needed for the probability of a hit is some given $\rho$. The sum you are doing in the expected value problem is $$1 + \frac{1}{H} + 2 \frac{H-1}{H} \frac{2}{H} + 3 \frac{H-1}{H} \frac{H-2}{H} \frac{3}{H} + \...


1

$\textbf{1)}$ Let $u=4-x^2, du=-2x dx$ to get $\hspace{.2 in}\displaystyle\int\frac{x^3}{\sqrt{4-x^2}}dx=-\frac{1}{2}\int\frac{x^2}{\sqrt{4-x^2}}(-2x) dx=-\frac{1}{2}\int\frac{4-u}{\sqrt{u}}du=-\frac{1}{2}\int\left(4u^{-1/2}-u^{1/2}\right)du$ $\hspace{.2 in}\displaystyle=-\frac{1}{2}\left(8u^{1/2}-\frac{2}{3}u^{3/2}\right)+C=-4(4-x^2)^{1/2}+\frac{1}{3}(4-...


2

This is called differentiating under the integral sign. It is valid if the function in the integrand, as well as the partial you are interested in is continuous in the argument with respect to which you are integrating. Since that was a mouthful, with your notation: $$\int{\dfrac{\partial}{\partial t}f(x,t)\,\mathrm dx} = \dfrac{d}{dt}\int{f(x,t)\,\mathrm ...


3

Let $\theta=b$ and $B = -a \lt 0$. Then the first integral is a generalization of this integral. Using the same substitutions: $u=a x+b/x$. Then $a x^2-u x+b = 0$, and therefore $$x = \frac{u}{2 a} \pm \frac{\sqrt{u^2-4 a b}}{2 a}$$ $$dx = \frac1{2 a}\left (1 \pm \frac{u}{\sqrt{u^2-4 a b}}\right ) du $$ Now, it should be understood that as $x$ ...


0

You are on the right track, but the integrals involving $v(x)$ and $v_x(x)$ have to be expressed explicitly to go further :


0

$$\int \frac { x^{ 3 } }{ \sqrt { 4-x^{ 2 } } } dx=\int { \frac { { x }^{ 2 }x }{ \sqrt { 4-x^{ 2 } } } } dx=-\frac { 1 }{ 2 } \int { \frac { \left( 4-{ x }^{ 2 }-4 \right) }{ \sqrt { 4-{ x }^{ 2 } } } d{ x }^{ 2 } } =$$ $$-\frac { 1 }{ 2 } \int { \sqrt { 4-{ x }^{ 2 } } d{ x }^{ 2 } } +2\int { \frac { d{ x }^{ 2 } }{ \sqrt { 4-{ x }^{ 2 } } } } =$$ $...


0

Hint. One may first make the change of variable $u=x^2$, $du=2x\,dx$, giving $$ \int \frac {x^3dx}{\sqrt{4-x^2}}=\frac12\int \frac {u\:du}{\sqrt{4-u}} $$ then $v=\sqrt{4-u}$, $u=v^2+4$, $du=2v\,dv$, thus $$ \int \frac {x^3dx}{\sqrt{4-x^2}}=\int (v^2+4)\:dv $$ Can you take it from here?


2

I think as others are getting at, the confusion probably arises from the (great) question: what is $dx$ anyway? I will try and give an intuitive answer that should help a little with the confusion. The rigorous answer is that $dx$ is a differential 1-form, which you will learn about if you do real analysis, but I think a perfectly good intuitive response is ...


1

An operator is an object that acts on another object, typically to the right, though proper definition of the space you're working in will clarify that. We require that operators and the objects they act on fulfill certain requirements. Integrals and derivatives acting on functions fulfill the requirements for a vector space of functions. When an integral is ...


2

\begin{align} & x=\rho \cos \theta \sin \phi \\ & y=\rho \sin \theta \sin \phi \\ & z=\rho \cos \phi \\ \end{align} $$\left| \frac{\partial (x,y,z)}{\partial (\rho ,\theta ,\phi )} \right|={{\rho }^{2}}\sin \phi $$ $${{z}^{2}}={{x}^{2}}+{{y}^{2}}\,\,\Rightarrow \,\,{{\rho }^{2}}{{\cos }^{2}}\phi ={{\rho }^{2}}{{\sin }^{2}}\phi \...


2

The hyperreal view helps clarify the picture with regard to the question whether $f(x)dx$ in the expression $\int_a^b f(x)dx$ is a product or not, as well as the issue of commutativity. From the hyperreal viewpoint, the integral is defined as the standard part of an infinite Riemann sum $\sum_i f(x_i)\Delta x$. In the Riemann sum, the term $f(x_i)\Delta x$ ...


0

Partial answer from another angle: From the viewpoint of ring theory, the following is (obviously) true. Proposition. Let $R$ denote a ring. Suppose $f,g$ and $\epsilon$ are elements of $R$. If $\epsilon$ is quasiregular, then $$f = g + \epsilon f \;\rightarrow\; f = \frac{1}{1 - \epsilon} g.$$ If we can find a name for this proposition,...


2

You can check that (since there is a selected answer, I don't have the energy/motivation to give all details, but start with $x\mapsto x^2$) there is a primitive $$ \frac{1}{2}\sqrt{(x^2-Q^2)(1-x^2)}+\frac{1}{2}(1+Q^2)\arctan\sqrt{\frac{Q^2-x^2}{1-x^2}} +Q\arctan\biggl(Q\sqrt{\frac{1-x^2}{x^2-Q^2}}\biggr) $$ Insert limits (you have to use limits) to find ...



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