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## New answers tagged integration

0

Another solution: Suppose $\mathcal{P} = x_{0}, x_{1}, \ldots, x_{N}$. We're going to show that $U( \operatorname{id}, \mathcal{P}) - U(f, \mathcal{P}) < \epsilon$ for every $\epsilon > 0$, and so $U(f, \mathcal{P}) = U(\operatorname{id}, \mathcal{P}) \geq \frac{1}{2}$, where $\operatorname{id}$ is the identity map on $[0, 1]$. Since $f \leq ... 0 It does not matter if it is decreasing. Given the set$A=\{1/2,...,1/n\}$, define your partition$P=\{x_0,x_1...,x_{2n+2}\}$such that for every$k\in\{1,...,n\}$, $$\frac{1}{k}-\frac{\varepsilon}{2n}\in P$$ and $$\frac{1}{k}+\frac{\varepsilon}{2n}\in P,$$ where$0<\varepsilon < 1$. Of course,$0\in P$and$1\in P$. Giving$x_i$a proper order, we have ... 0 Let$\mathcal P_0$be a partition of$[0,1]$. Refine this partition if necessary to a partition$\mathcal P$containing$n+1$points so that$x_{i}-x_{i-1}=1/n.\ $Note that$x_0=0$,$x_{n}=1$and in general$x_i=i/n$. We have$U(\mathcal P_0)\geq U(\mathcal P)=\frac{1}{n}\sum_{i=1}^{n}f(x_i^{*})$. Now let$\epsilon >0$and choose, using density of the ... 1 Another way: Let$P = \{x_0,x_1,\ldots,x_n\}$be a partition of$[0,1]$, and let$M_i$denote the supremum of$f$on$[x_{i-1},x_i]$. There must exist a rational number$q$in the interval$[(x_{i-1} + x_i)/2,x_i]$, and so it follows that$M_i \geq (x_{i - 1} + x_{i})/2$. This gives: $$U(f,P) = \sum_{i = 1}^{n}M_i(x_i - x_{i - 1}) \geq \sum_{i = ... 1 Given any small positive real \varepsilon > 0, for any partition P = \{x_0, x_1, \ldots, x_n\} of [0, 1] such that 0 = x_0 < x_1 < \cdots < x_n = 1, since \mathbb{Q} is dense in \mathbb{R}, for each i, there exists q_i \in \mathbb{Q} such that x_i - \varepsilon < q_i < x_i, it then follows that$$M_i = \sup_{I_i} f(x) - ... 1 If$r(x) = (x,f(x))$parametrizes${\rm gr}(f)$, then: $$\ell(r) = \int_a^b \sqrt{1+f'(x)^2}\,{\rm d}x \leq \int_a^b \sqrt{2}\,{\rm d}x = \sqrt{2}(b-a).$$ 0 The length of the curve is given by $$\int_a^b \sqrt{f'^2(x)+1}dx$$ 2 If you're thinking Riemannly: What if your$k^\text{th}$partition is$\left\{\frac{j}{k^k} \mid 0 \leq j < k^k \right\}$? If you're thinking Lebesguely: What do you think of$[0,1] = \{0\} \cup \bigcup_{i \in \Bbb{N}} \left( (\frac{1}{i+1},\frac{1}{i}) \cup \{\frac{1}{i}\}\right)$? 0 Based on Lost1's comment: In the first place, to have conditional expectation, we need integrability. A product of integrable random variables is not necessarily integrable: Let$X, Y \in \mathscr L^{1}(\Omega, \mathscr F, \mathbb P)$. Consider$X$and$X - Y$w/$Xhaving an infinite second moment but finite first moment. Then $$E[X(X-Y)] = E[X^2] - ... 0 No, and your integral is incorrect. The floor function is treated as a constant in order to yield an antiderivitive fulfilling the first fundamental theorem. The actual derivative is a question of removing the discontinuity of any greatest integer function. Such a question has not been answered as of yet and is an unknown "jump series". I.E. The sum of all ... 1 The integrand has an antiderivative in terms of elementary functions and polylogarithms. It can be found using Mathematica after expressing inverse trig functions through logarithms of complex arguments, and can be manually checked for correctness using differentiation. After subtracting its limits at \infty and 0 and simplification, we can get the ... 0 First off, you can integrate the function in a way that fulfills the first fundamental theorem. This is done by integrating as if the greatest integer is a constant. This will yield a greatest integer function that is broken. The second step with integrating is to subtract the jump series which is defined as the sum of the jumps made from 0 to the current ... 0 You can check that$$ \lambda[0;y](x) = \lambda[x,1](y). Thus \begin{align} f(1)- \int_0^1 f'(y)*\lambda[0;y](x)dy & = f(1)- \int_0^1 f'(y)*\lambda[x,1](y)dy \\ & = f(1) - \int_x^1 f'(y)dy \\ & = f(1) - f(1) + f(x) = f(x), \end{align} where the last line is the FTC. 0 Now I did integration by parts with u=\sin(2\pi t)e^{-t^2} and v'=e^{t(n+1)} Why ? :-\ Just use Euler's formula in conjunction with the value of the Gaussian integral. All you have to do is to complete the square. 0 To further illustrate the points in my comment. One partition of [1,2] has partition points \frac 5 4,\frac 3 2, \frac 7 4. Here is how to answer the question for that partition: The partition divides [1,2] into four subintervals: \left[1, \frac 5 4\right], \left[\frac 5 4, \frac 3 2\right], \left[\frac 3 2,\frac 7 4\right], \left[\frac 7 4,2\right] ... 0 Hint: First of all, ~\tan'x=1+\tan^2x=\sec^2x.~ Secondly, ~\cos'x=-\sin x. 1 You know that\int_{-\infty}^{\infty}\sqrt{\frac{1}{(t^2-1)^2}-\frac{(n+1)t^{2n}}{(t^{2n+2}-1)^2}}dt=2\int_{0}^{\infty}\sqrt{\frac{1}{(t^2-1)^2}-\frac{(n+1)t^{2n}}{(t^{2n+2}-1)^2}}dt\tag1$$Now we have ... 2 If you read a real analysis textbook such as Calculus by Spivak, they manage to develop calculus rigorously while avoiding differentials like "dx" and "dy" entirely. This is the standard way to make calculus rigorous -- you just avoid using differentials. And indeed, in undergrad differential equations classes, arguments that involve manipulating dx ... 0 U\left.(t)\right|_{g(x)}^{f(x)} 1 If you know U such that$$ U'=u $$then$$ \int_{g(x)}^{f(x)} u(t) dt=U(f(x))-U(g(x)) $$2 This isn't any different as solving without functions in as limits, just find an antiderivative$$U(t) = \int u(t)dt$$and plug in your limits:$$ \int_{g(x)}^{f(x)} u(t) dt = U(f(x))-U(g(x)) $$0$$\lambda\int_0^{+\infty} e^{ax} \text{d}x$$where a = t-\lambda Solving that ad you will get$$\lambda\left(\frac{e^{ax}}{a}\bigg|_{0}^{+\infty}\right)$$a has to be negative, otherwise the integral diverges. So you have to have t-\lambda < 0 6 If t\ge\lambda, the integral diverges. It is implicit that t<\lambda. 1 looks ok, you can also continue with substitution u=\xi\pm x-\eta = \int\limits_{0}^{t}\int\limits_{0}^{t-x-\eta} g(\xi,\eta) d\xi d\eta + \int\limits_{0}^{t} \int\limits_{0}^{t + x-\eta} g(\xi,\eta) d\xi d\eta = \int\limits_{0}^{t}\int\limits_{\eta+x}^{t} g(\xi-x-\eta,\eta) d\xi d\eta + \int\limits_{0}^{t} \int\limits_{ \eta-x}^{t } g(\xi+ ... 0 If we have$$\lim X_n = c$$let us take the expectation of both sides to get:$$E[\lim X_n] = E[c] = c$$Now do we have$$E[\lim X_n] = \lim E[X_n]?$$Note that boundedness of |X_n|'s means that \exists M > 0 s.t.$$|X_n| \le M \ \forall n \ge 1$$Note that M is integrable \because \ E[|M|] = E[M] = M < \infty By the dominated ... 0 In this case you can get by with the minimum amount of intervals, exactly one. Obviously, h=b-a. 1 Since$$\frac{d}{dx}\int_a^x f(x) \, dx=f(x)$$and$$\frac{d}{dx}\int_x^a f(x) \, dx=-\frac{d}{dx}\int_a^x f(x)=-f(x)$$and similarly$$\frac{d}{dx}\int_a^{g(x)} f(x) \, dx=f(x)g'(x)$$You should have,$$\frac{\partial}{\partial x} \left\{ \int\limits_{x - t + \eta}^{x + t - \eta} F(\xi,\eta) d\xi \right\}= \frac{\partial}{\partial x} \left\{ \int\limits_{x ... 0 Begin by enforcing the substitutionx\to e^x. Then, we have \begin{align} \int_0^\infty x^n\sin(2\pi \log(x))e^{-\log^2(x)}\,dx&=\int_{-\infty}^\infty e^{-x^2+(n+1)x}\sin (2\pi x)\,dx\\\\ &=e^{(n+1)^2/4}\int_{-\infty}^{\infty}e^{-\left(x-\frac{n+1}{2}\right)^2}\sin (2\pi x)\,dx\\\\ ... 0 Knowing that the object is initially at rest you can solve both equations for C and D at t=0. 0 I'm not sure why starting with a substitution, since you have to use integration by parts anyway. However, you should remove x altogether, otherwise you might overlook something: if u=\ln x, then x=e^u and dx=e^u\,du, so you get \int \frac{u}{e^{2u}}e^u\,du=\int ue^{-u}\,du $$Integrating by parts,$$ \int ue^{-u}\,du=-ue^{-u}+\int e^{-u}\,du= ... 0 Yes, this is correct (sort of). However, I have three points to make. Most importantly, you forgot the+C$. Any antiderivative should have a constant on the end, with an unknown value (given other information, it may be possible to find this value). I would be careful with how you write out your calculations where doing$u$-substitution. I tend to write ... 0 So then:$u=t-\frac{n+1}{2}$and$du=dt$. So we get$\int_{-\infty}^{\infty} e^{-u^2+\frac{(n+1)^2}{4}} sin(2\pi(u+\frac{n+1}{2})) \mathrm{d}u$. Because of the angle addition theorems, we can write:$e^{\frac{(n+1)^2}{4}} cos(2\pi \frac{n+1}{2}) \int_{-\infty}^{\infty} e^{-u^2} sin(2\pi u) du + e^{\frac{(n+1)^2}{4}} sin(2\pi \frac{n+1}{2}) ...

2

The region of integration is $$D = \{ (x, y) \ \epsilon \ R^2 \ | \ 0 \le x \le a, x \le y \le a \}$$ See that D is both a type - I and type - II domain and thus can be projected onto the y - axis. i.e. $$D = \{ (x, y) \ \epsilon \ R^2 \ | \ 0 \le y \le a, 0 \le x \le y \}$$So swapping here connotes the following transformation of the double integral ...

3

Hint: $\displaystyle t^2-t(n+1)=(t-\frac{n+1}{2})^2-\frac{(n+1)^2}{4}$. So try to put $\displaystyle u=t-\frac{n+1}{2}$

1

The above answer is very good to understand why the integrand is odd. I will instead focus on how to prove the value of the integral. Note that for all integers $m,n\in\mathbb{Z}$ one have $$\int_0^{2\pi} \sin mx \cos nx \,\mathrm{d}x = 0$$ This can be shown by rewriting $$\sin mx \cos nx = \frac{1}{2}\sin(m+n) x - \frac{1}{2}\sin(m-n) x$$ ...

1

The expression $(16 \sin \phi \cos^2\phi-4\sin^{2}\phi+8\sin\phi)$ is not an odd function of $\phi$. Consider each of the terms one-by-one: $$\int^{2\pi}_{0} 16 \sin \phi \cos \phi\ d \phi = 0$$ because $\sin \phi \cos \phi$ is antisymmetric (odd) about the $\phi=\pi$ line. \int^{2\pi}_{0} -4\sin^{2}\phi\ ...

2

HINT: Notice, $$\int_{1}^{\infty}\frac{dt}{t^4+t^2+1}=\lim_{z\to \infty}\int_{1}^{z}\frac{dt}{t^4+t^2+1}$$ $$=\lim_{z\to \infty}\int_{1}^{z}\frac{\left(\frac{1}{t^2}\right)dt}{t^2+\frac{1}{t^2}+1}$$ $$=\frac{1}{2}\lim_{z\to \infty}\int_{1}^{z}\frac{\left(1+\frac{1}{t^2}\right)-\left(1-\frac{1}{t^2}\right)}{t^2+\frac{1}{t^2}+1}\ dt$$ $$=\frac{1}{2}\lim_{z\to ... 2 Hints: Do decompose the fraction using the given factorization. You will get two fractions with a first degree numerator and second degree denominator. By adding a suitable constant to the numerator, you will turn it to the derivative of the denominator, making the fractions easy. Then you need to compensate with two integrands of the form$$\frac ...

1

$$\lim_{z\to \pi/2^-}\int_0^z(\sec^2x-\sec x \tan x)\,dx=\lim_{z\to \pi/2^-}[\tan z-\sec z+1]=\lim_{z\to \pi/2^-}\frac{\sin x+\cos x-1}{\cos x}=1$$

0

If you want to know what a mathematical thing is, you need two things: How it works - what the rules of using it are. A way of expressing it in terms of mathematical things you know and trust. For differentials, we know what the rules of using them are. And I'm told some clever folks have worked out how to model them (notably Abraham Robinson with ...

2

There is a connection but to make it clear you have to be more precise about the boundaries of the inner integral on the right hand side of your second equation. The primitive function $F$ of $f$ is only determined up to a constant. Assuming that it is chosen so that $F(a)=0,$ then we have $$\int_a^bFg'=\int_{x=a}^b\int_{t=a}^xf(t)dt\ g'(x)dx$$ The ...

0

Your integral can also be written as $\int_0^1{\int_0^1{(x^3+y^3)dx}dy}$. Do you know how to proceed now?

0

The notation $0\le x,y\le1$ is often the lazy form of $0\le x\le1$, and $0\le y\le 1$. In my opinion it is needlessly confusing.

0

HINT: $$\int\frac{1}{\sqrt{a^2-x^2}}\space\text{d}x=\int\frac{1}{a\sqrt{1-\frac{x^2}{a^2}}}\space\text{d}x=\frac{1}{a}\int\frac{1}{\sqrt{1-\frac{x^2}{a^2}}}\space\text{d}x=$$ Substitute $u=\frac{x}{a}$ and $\text{d}u=\frac{1}{a}\space\text{d}x$: ...

0

Let $X_i\sim\mathcal N(1,\sigma_i^2)$ be independent for $i=1,2$, then the expression above is $$\mathbb P(X_1>x) + \mathbb P(X_2>x) = a.$$ After standardizing we find that $$\Phi\left(\frac{x-1}{\sigma_1}\right) + \Phi\left(\frac{x-1}{\sigma_2}\right)=a,$$ where $\Phi$ is the distribution function of the normal distribution with $\mu=0$, ...

0

To avoid the notational mess, the longitude coordinate covers a full turn, from Greenwich to Greenwich, the colatitude coordinate covers a half turn, from the North to the South pole. The Jacobian is weighted by the sine of the colatitude (parallel lines vanish at the poles and are the longest at the equator). So the double integral on the angles will ...

0

We have $a(x)=\beta ^Tx+\beta_0$, and $\nabla a = \beta$, but as $x$ is in terms of $t$, $a$ can be alos seen it terms of $t$, and $a(t)=\beta ^Tx(t)+\beta_0$, and we have $\dot{a}(t)=\frac{da}{dt}= \beta^T \dot{x}(t)$. Again to the integral $(4)$, we have $$d(\chi_A, \chi_B)=\Bigg| \int_0^1\sqrt{ \dot{x}(t)^T\beta \beta^T \dot{x}(t). ... 0 Your formula of mass = volume \times density needs to be a bit modified here since the density is non-uniform. Every bit of volume of the sphere has a different density so you have to integrate it appropriately as follows:$$M=\int_0^1 \text{density}\cdot dV = \int_0^1 (1-\rho^2) \cdot dV$$and we know that V=\frac{4}{3}\pi \rho^3 where \rho is the ... 0 You are wrong. The density is not uniform so you need to integrate the following:$$\int_0^r(4\pi\rho^2)(1-\rho^2)d\rho$$1 You can project the solid onto the xy - plane. You'll find that the projection onto the xy - plane is a right - angled triangle which is given by$$D = \{ (x, y) \ \epsilon \ R^2 \ | \ 0 \le x \le 1, 0 \le y \le x \}$$and therefore the solid can be represented as$$ \{ (x, y, z) \ \epsilon \ R^3 \ | \ 0 \le z \le 1 - x^2, (x, y) \ \epsilon \ D \} Then you ...

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