Integral Definition and 1000 Threads

I want to do this integral in the picture: where r1 and a are constants. I know I can integrate each part separately. There will be an integral with respect to r2 multiplied by integral with respect to theta2 and the last one with respect to phi2. But the term under square root confuses me. Can...

Question: Suppose I have a data file for the acceleration of an object after every ## \Delta t_i##, how do I obtain the displacement of it? Context: Integral in a PID loop, although not exactly what I am asking as one is sum of error: $$\int_0^T \int_0^T \ddot {\vec \theta(t)}dtdt$$ the other...

Hi, PF, there goes the definition of General Riemann Sum, and later the exercise. Finally one doubt and my attempt: (i) General Riemann Sums Let ##P=\{x_0,x_1,x_2,\cdots,x_n\}##, where ##a=x_0<x_1<x_2<\cdots<x_n=b##, be a partition of ##[a,b]##, having norm ##||P||=\mbox{max}_{1<i<n}\Delta...

I just came across this and it seems we do not have a definite answer...there are those who have attempted using integration by parts; see link below...i am aware that ##\cos x## has no closed form...same applies to the exponential function...

Looking to evaluate an integral of the form $$\int_0^{\infty} \frac{p^2 dp}{\mathrm{exp}(a\sqrt{p^2+b^2}) \pm 1} $$Changing to ##x(p) = a\sqrt{p^2 + b^2}## gives $$\frac{1}{a^3} \int_0^{\infty} \frac{\sqrt{x^2-(b/a)^2}}{e^x \pm 1} dx$$Wolfram alpha doesn't tell me anything useful, sadly.

Evaluate the definite integral $$\int_0^\infty \frac{x^2 + 1}{x^4 + 1}\, dx$$

Referring to this link : https://qcdloop.fnal.gov/bubg.pdf Using Mathematica Integrate command to solve it does not give the result stated here but I am unclear as to how they got to the result in the 4th line. It is clear that the integrand (1st line) can diverge for certain values of the...

I have tried WolfarmAlpha but it could help me. Please note this is not a homework exercise. I am a researcher and I am looking to model viscosity development of resin. there I came across with this express :) $$\int{\frac{1}{a\cdot e^{bx}+c\cdot e^{kx}}dx}$$

Given a function ##f##, interval ##[a,b]##, and its tagged partition ##\dot P##. The Riemann Sum is defined over ##\dot P## is as follows: $$ S (f, \dot P) = \sum f(t_i) (x_k - x_{k-1})$$ A function is integrable on ##[a,b]##, if for every ##\varepsilon \gt 0##, there exists a...

We don't need to worry about the n = -1 so we can assume that the function is continuous on any interval [a,b] where a, b are real numbers if I separate my interval into N partitions, then the right side values in my interval are a + \frac{b-a}{N}, a + 2 \frac{b-a}{N}, ... , a + k...

Hi, so I'm trying to find the volume of a shape using integral, I found the equation of one plane in 3D space but the second one is something like that, which I cannot write in integral as a function: ##\frac{2(2x-a)}{a}=-\frac{2(6y-a\sqrt3)}{a\sqrt3}=\frac{2z-a\sqrt3}{a\sqrt3}## In the 3D...

my notebook says that we can rewrite the integral $$\int {75\sin^3⁡(x) \cos^2⁡(x)dx}$$ as $$\int {75 \cos^2(x)\sin(x)dx} - \int {75\sin(x)\cos^4(x)dx}$$ however, i have literally no idea how it got to this point, and i unfortunately can't really provide an "attempt at a solution" for this...

So for this question, I understand the math but just wanted to be clear on a few things. So I had this question on my midterm but instead of tensile and compressive stresses, it was tensile and tensile stress. I initially thought that the delta sigma in the integral was the maximum stress so in...

I'm supposed to do the surface integral on A by using spherical coordinates. $$A = (rsin\theta cos\phi, rsin\theta sin\phi, rcos\theta)/r^{3/2}$$ $$dS = h_{\theta}h_{\phi} d_{\theta}d_{\phi} = r^2sin\theta d_{\theta}d_{\phi}$$ Now I'm trying to do $$\iint A dS = (rsin\theta cos\phi, rsin\theta...

I need compute the integral $$(2\pi)^{-3} \int d^3p e^{-l|p|}e^{i \vec{x} \cdot \vec{p}}$$ The problem does not specified the limits of integration The result is $$\frac{1}{\pi^2} \frac{l}{\sqrt{\vec{x}^2+l^2}}$$I saw the references about t-Student and I had not achieved it. I have tried to...

For this problem, The solution is, However, why have they not included limits of integration? I think this is because all the small charge elements dq across the ring add up to Q. However, how would you solve this problem with limits of integration? Many thanks!

In Greiner's Classical Electromagnetism book (page 126) he has a derivation equivalent to the following. $$\int_V d^3r^{'} \nabla \int_V d^3r^{''}\frac {f(\bf r^{''})}{|\bf r + \bf r^{'}- \bf r^{''}|}$$ $$ \bf z = \bf r^{''} - \bf r^{'} $$ $$\int_V d^3r^{'} \nabla \int_V d^3z \frac {f(\bf z +...

Are both integral on picture below equal zero? I think both are zero, area of zero section under function must be zero. If M=∞, b=∞ , what is reslut? Logically ∞/∞ will be 1..but...

##curl([x^2z, 3x , -y^3],[x,y,z]) =[-3y^2 ,x^2,3]## The unit normal vector to the surface ##z(x,y)=x^2+y^2## is ##n= \frac{-2xi -2yj +k}{\sqrt{1+4x^2 +4y^2}}## ##[-3y^2,x^2,3]\cdot n= \frac{-6x^2y +6xy^2}{\sqrt{1+4x^2 + 4y^2}}## Since ##\Sigma## can be parametrized as ##r(x,y) = xi + yj +(x^2...

Knowing that to be Hermitian an operator ##\hat{Q} = \hat{Q}^{\dagger}##. Thus, I'm trying to prove that ##<f|\hat{Q}|g> = <\hat{Q}f|g> ##. However, I don't really know what to do with this expression. ##<f|\hat{Q}g> = \int_{-\infty}^{\infty} [f(x)^* \int_{-\infty}^{\infty} |x> <x| dx f(x)] dx##...

Author's answer: Recognizing that this integral is simply a vector line integral of the vector field ##F=(x^2−y^2)i+(x^2+y^2)j## over the closed, simple curve c given by the edge of the unit square, one sees that ##(x^2−y^2)dx+(x^2+y^2)dy=F\cdot ds## is just a differentiable 1-form. The...

I found some interesting equations on cosmology and I was wondering how to introduce the integral in an excel sheet: "Paste ( .443s^3+1)^(-1/2) in for the integrand, type in s for the variable and 1 to 2 for the limits. Press submit, then change 2→3→4→5 and repeat." (from the thread...

Can anyone explain to me why the second one is the right? (See the attachment)

I made this exercise up to acquire more skill with polar coordinates. The idea is you're given the acceleration vector and have to find the position vector corresponding to it, working in reverse of the image. My attempts are the following, I proceed using 3 "independent" methods just as you...

Suppose ##f## is holomorphic in an open neighborhood of the closed unit disk ##\overline{\mathbb{D}} = \{z\in \mathbb{C}\mid |z| \le 1\}##. Derive the integral representation $$f(z) = \frac{1}{2\pi i}\oint_{|w| = 1} \frac{\operatorname{Re}(f(w))}{w}\,\frac{w + z}{w - z}\, dw +...

I know the Gauss law for surface integral to calculate total charge by integrating the normal components of electric field around whole surface . but in above expression charge is calculated using line integration of normal components of electric field along line. i don't understand this...

Why should it be ##\int_{a}^{x} f(t)dt## ? Couldn't it be like this: Let F(x) = ##\int f(x)dx## so ##\int_{a}^{x} f(x)dx## = F(x) - F(a) Thanks

I'm trying to solve an improper integral, but I'm not familiar with this kind of integral. ##\int_{-\infty}^{\infty} (xa^3 e^{-x^2} + ab e^{-x^2}) dx## a and b are both constants. From what I found ##\int_{-\infty}^{\infty} d e^{-u^2} dx = \sqrt{\pi}##, where d is a constant and...

Here is my attempt (Note: ## \left| \int_{C} f \left( z \right) \, dz \right| \leq \left| \int_C udx -vdy +ivdx +iudy \right|## ##= \left| \int_{C} \left( u+iv, -v +iu \right) \cdot \left(dx, dy \right) \right| ## Here I am going to surround the above expression with another set of...

My answer is False! I think must stated "in general," in the beginning of the statement. Cause this could be true if f or g = zero. There may be other cases also. Is my answer right?

Greetings all. I just got confused by the following. Consider volume integral, for simplicity in 1D. $$ V(A) = \int_{A} dz. $$ If ##z## can be written as an invertible function of ##x##, i.e. ##z=f(x)##, we know the change of variables formula $$ V(A)=\int_{A} dz= \int_{z^{-1}(A)} |z'(x)|dx...

This brief worked example from a textbook section on the method of images is confusing me. Specifically I am confused about the vector α in the integral on the last line. When α (or θ) is an angle, I've only ever seen the vector quantity α (or θ) as a polar vector in the plane. But here...

∫zds=∫acos(t)*( (acos(2t))^2+(2asin(t))^2+(-asin(t))^2 )^1/2 dt , (0≤t≤pi/2) Simplified : ∫a^2cos(t)*(cos^2(2t)+5sin^2(t) )^1/2 dt , (0≤t≤pi/2) However here i get stuck and i can´t find a way to rewrite it better or to integrate as it is. Can i please get some help in this?

Evaluate ##\displaystyle\int_{0}^{3}\frac{x+3}{\sqrt{x^{3}+1}}dx+5## W|A returned 11.7101 ok subst is probably just one way to solve this so ##u=x^{3}+1 \quad du= 3x^2##

So, I am able to calculate the electric potential in another way but I know that this way is supposed to work as well, but I don't get the correct result. I calculated the electric field at P in the previous exercise and its absolute value is $$ E = \frac {k Q} {D^2-0.25*l^2} $$ This is...

Hello. As is known, we can neglect high-order term in expression ##f(x+dx)-f(x)##. For ##y=x^2##: ##dy=2xdx+dx^2##, ##dy=2xdx##. I read that infinitesimals have property: ##dx+dx^2=dx## I tried to neglect high-order terms in integral sum (##dx^2## and ##4dx^2## and so on) and I obtained wrong...

$$F=G\frac{m_1 m_2}{r^2}$$ is presumably for point masses. If the masses weren't a point masses, then wouldn't you need a version of the formula that sums up the gravity for each infinitesimal portion of the masses? And for my money, "summing up" in physics is integrals, right? So would it be...

But the answer provided is ##\frac{15}{4} ## How is that? What is wrong in the above computation of answer?

Evaluate the surface integral $\iint\limits_{\sum} f \cdot d\sigma $ where $ f(x,y,z) = x^2\hat{i} + xy\hat{j} + z\hat{k}$ and $\sum$ is the part of the plane 6x +3y +2z =6 with x ≥ 0, y ≥ 0, z ≥ 0 , with the outward unit normal n pointing in the positive z direction. My attempt to answer...

Hi. I was reading Lighthill, Introduction to Fourier Analysis and Generalised Functions and in page 17 there is an example/proof where I can't make sense of the following step: $$ \left| \int_{-\infty}^{+\infty} f_n(x)(g(x)-g(0)) \, \mathrm{d}x \right| \le \max{ \left| g'(x) \right| }...

I don't have any idea to answer this question. So, any math help will be accepted. I know ##\nabla fg = f\nabla g + g\nabla f \rightarrow (1) ## But I don't understand to how to use (1) here?

I don't have any idea about how to use the hint given by the author. Author has given the answer to this question i-e F(x,y) = axy + bx + cy +d. I don't understand how did the author compute this answer. Would any member of Physics Forums enlighten me in this regard? Any math help will be...

. Let C be a smooth curve with arc length L, and suppose that f(x, y) = P(x, y)i +Q(x, y)j is a vector field such that $|| f|(x,y) || \leq M $ for all (x,y) on C. Show that $\left\vert\displaystyle\int_C f \cdot dr \right\vert \leq ML $ Hint: Recall that $\left\vert\displaystyle\int_a^b g(x)...

I'm looking at a proof of the asymptotic expression for the Elliptic function of the first kind https://math.stackexchange.com/questions/4064023/on-the-asymptotic-behavior-of-elliptic-integral-near-k-1 and I'm having trouble understanding this step in the proof: $$ \begin{align*} \frac{1}{2}...

Now the steps to solution are clear to me...My interest is on the constant that was factored out i.e ##\frac{2}{\sqrt 3}##... the steps that were followed are; They multiplied each term by ##\dfrac{2}{\sqrt 3}## to realize, ##\dfrac{2}{\sqrt 3}\int \dfrac{dx}{\left[\dfrac{2}{\sqrt...

R is the triangle which area is enclosed by the line x=2, y=0 and y=x. Let us try the substitution ##u = \frac{x+y}{2}, v=\frac{x-y}{2}, \rightarrow x=2u-y , y= x-2v \rightarrow x= 2u-x + 2v \therefore x= u +v## ## y=x-2v \rightarrow y=2u-y-2v, \therefore y=u- v## The sketch of triangle is as...

Is the answer given above correct?

In the book it is mentioned that, in path c, the line integral would be: $$\int \vec{F}\cdot \vec{dr} = A \int_{1}^{0}xy dx = A\int_1^0 x dx = -\dfrac{A}{2}$$. but I think that dx is negative in that case, the result would be positive, right?

Find question here, My approach, using cosine sum and product concept, we shall have; ##\cos (A+B)-\cos (A-B)=-2\sin A\sin B## ##⇒\cos D-\cos C=-2\sin\dfrac{C+D}{2} \sin\dfrac {C-D}{-2}## ##⇒-3[\cos(A+B)-\cos(A-B)]=6\sin A sinB## We are given ##A=4θ## and ##B=2θ##, therefore, ##⇒-3[\cos...

Write a program that uses the Monte Carlo method to approximate the double integral $\displaystyle\iint\limits_R e^{xy}dA$ where $R = [-1,1] \times [0, x^2]$. Show the program output for N = 10, 100, 1000, 10000, 100000 and 1000000 random points. My correct answer: My Java program...