# Tag Info

## New answers tagged integration

0

It seems not true in general. In $\mathbb{R}^2$, Consider $S$ as the set of all $(x,y)$ in $[0,1] \times [0,1]$ such that atleast one of $x$ or $y$ is irrational and $Q$ to be any rectangle containing $[0,1] \times [0,1]$ . These $S$ and $Q$ satisfies the given condition and $S$ contains no rectangle at all.

1

Use the residue theorem! We first assume that: $$\exp\sqrt{x}=z$$ Then your formula turns out to be: $$\int_{1/\sqrt{2}}^{1} \frac{\arccos\frac{z}{\sqrt{2}}} {1-z^2} \,\mathrm{d}x = -\frac{\pi^{3}}{192}$$ The residue theorem gives $$\int_{1/\sqrt{2}}^{1} \frac{\arccos\frac{z}{\sqrt{2}}} {1-z^2} \,\mathrm{d}x ... 0 sin(x\cdot sint) = x\cdot sint - \dfrac{(x\cdot sint)^3}{3!} + ..., and integrate term by term should give the answer. 0 Hint: Where does h(t) achieve its maximum? 1 Yes, the Max function of two Riemann -integrable functions is Riemann -integrable. This is because the Max of two a.e -continuous bounded functions is also a.e -continuous and bounded. For f,g continuous, the function Max{f,g} is continuous. This implies that (since a Riemann-integral must be a.e -continuous) , that Max{f,g} is a.e -continuous; ... 0 Given  \varepsilon> 0  is  k \in \ N  such that \varepsilon >1/k consider a 'homogeneous' partition of  [0,1] \times [0,1] , i.e., all sub- rectangles of partition has area 1/k^2, we know that  m_B = 0 and  m_B = 1 if B intection the line  y = x  is not empty set, so  S (f, P)-s (f, P) = \sum_ {B \ in P} (m_B-m_B) vol (B) = \sum_ ... 0 Note that the product of bounded functions is bounded, so since f and g are Riemann integrable, fg is bounded too. Furthermore, if f and g are continuous at x, so is fg; hence we have the inclusion$$\{x : fg \text{ is not continuous at } x\} \subseteq \{x : f \text{ not continuous}\} \cup \{x : g \text{ not continuous}\}$$Since f is ... 0 Hints: (1) What could you say if g(x)=\int_a^xy(t)dt? (2) Note that \int_{u(x)}^{v(x)}y(t)dt =\int_{a}^{v(x)}y(t)dt - \int_{a}^{u(x)}y(t)dt. 5 Let  \displaystyle I(z) = \int_{0}^{\infty}\frac{\arctan \frac{x}{z}}{e^{\pi x}-1} \ dx. Then$$ I(z) = \int_{0}^{\infty} \int_{0}^{\infty}\frac{1}{e^{\pi x}-1} \frac{\sin xt}{t}e^{-zt} \ dt \ dx = \int_{0}^{\infty} \frac{e^{-zt}}{t} \int_{0}^{\infty} \frac{\sin tx}{e^{\pi x}-1} \ dx \ dt = \frac{1}{2}\int_{0}^{\infty} \frac{e^{-zt}}{t} ...

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Note that $f$ is bounded on $[\epsilon,1]$, so you just need to show that the integral of a bounded function which vanishes outside the Cantor set is zero. Use the fact that the Cantor set is covered by a finite union of closed intervals of arbitrarily small measure.

1

It looks to me that your problem is: $$\iiint_{D}x^{2}\:\mathrm{d}x\:\mathrm{d}y\:\mathrm{d}z,$$ Where $D$ is the region described by: $x^{2}+2y^{2}+z^{2}\leq 2$. Which is a triple integral, rather than the double integral that you describe, however, the process is much the same. We first describe our transformation from Cartesian co-ordinates into the new ...

0

\begin{align} \frac{1}{1+\sin{(2\, \theta)}\, \cos{\alpha}} &=\frac{1}{1+2\sin{\theta}\cos{\theta}\, \cos{\alpha}}\\ &= \frac{\sec{(\theta)}^2}{\sec{(\theta)}^2+2\tan{\theta}\, \cos{\alpha}}\\ &= \frac{\sec{(\theta)}^2}{1+\tan{(\theta)}^2+2\tan{\theta}\, \cos{\alpha}}\\ &= ...

1

Parts (ii), (iii), and (v) are pure linear algebra. We are given a linear map $$L:\quad C^1(I)\to C^0(I),\qquad f\mapsto f'+a\> f\ .$$ Solving the ODE $(2)$ means finding the $f$'s in $C^1(I)$ with $Lf=b$, where $b\in C^0(I)$ is given in advance. Denote the set of these $f$'s by ${\cal S}$. (ii) When $f$, $f_0\in{\cal S}$ then $L(f-f_0)=Lf ... 1 Well, your function does not depend on$\phi$for the very beginning therefore it is correct that you have no dependancy in the end :) Some more details that I noticed: "$f(\theta, \phi) = u_0$for the range$[0 \le \theta < \pi/2]$" <- but you have to integrate over range$[0 \le \theta < \pi]$, what's the$f(\theta, \phi)$for$[\pi/2 \le \theta ...

0

$\int_{0}^{\frac{\pi}{2}}\frac{1}{2(\sin(2\theta)\cos(\alpha)+1)}d\theta=\frac{1}{2}\int_{0}^{\frac{\pi}{2}}\frac{1}{2\sin(\theta)\cos(\theta)\cos(\alpha)+1}d\theta=\frac{1}{2}\int_{0}^{\frac{\pi}{2}}\frac{\sec^{2}(\theta)}{2\tan(\theta)\cos(\alpha)+\sec^{2}(\theta))}d\theta$ ...

1

You need to show that there is linear subspace $U\subset C^1(I,\mathbb{C})$ and a vector $v_0\in C^1(I,\mathbb{C})$ such that every solution of $(2)$ lies in $v+U$. We claim that $U$ is a set of solutions of $(1)$, $v_0=f_0$. Indeed, by paragraph $(ii)$ we already know that every solution of $(2)$ is a sum of $f_0$ and some solution of $(1)$. Thus we ...

11

Let the desired integral be denoted by $I$. Noting that $1-(2t-1)^2=4t(1-t)$ we see that $$I=4\int_{\square}\frac{-\log(1-(2x-1)^2)+\log(1-(2y-1)^2)}{(2x-1)^2-(2y-1)^2}dxdy\quad\hbox{with \square=[0,1]^2}.$$ But, for $t\in(-1,1)$, we have $$-\log(1-(2t-1)^2)=\sum_{n=1}^\infty\frac{(2t-1)^{2n}}{n}$$ So the integrand becomes, for $(x,y)\in(0,1)^2$, ...

0

Or you could simply do the following: $$\text {Area} = 9\times 4 -\int_1^9 \sqrt x dx$$

0

Note \begin{align}\int{\sqrt{26x-x^2}dx}& = \int{\sqrt{169-(x^2-26x+169)}dx} \\ &=\int{\sqrt{13^2-(x-13)^2}dx} \end{align} It can be evaluated using $x-13=13\sin t,0\leq t \leq \pi/2$.

0

Using Trigonometric Substitution, $$2ax-x^2=a^2-(x-a)^2$$ Set $$x-a=a\sin\theta$$ Here $a=13$

1

Hint: First notice that $26x-x^{2}=-(x^{2}-26x)=-((x-13)^{2}-169)=169-(x-13)^{2}$

1

From the picture it is clear that if you want to change the order of integration you should integrate over the region D. Region D can be divided in two parts: For D1 we have: integrate from $x=1$ to $x=y^2$, and $y\in(1,3)$, and for D2 we integrate from $x=1$ to $x=9$ and $y\in(3,4)$. Calculus for D1: $$D_1=\int^3_1 dy \int^1_{y^2} dx$$ and for ...

0

I think it looks fine. Just mention $U$ is upper sum and $L$ is lower sum.

1

I'd suggest using the following theorem: $$\psi_1(x) = \frac{1}{2\pi i} \int_{c - i\infty}^{c + i\infty} \frac{x^{s + 1}}{s(s+1)} \left( \frac{-\zeta'(s)}{\zeta(s)} \right) \mathrm{d}s$$ where $c > 1$. A proof of this equality can, for example, be found in Complex Analysis by Stein and Shakarchi. It is on page 191 being proposition 2.3 of chapter 7. ...

5

I assume you mean $$\int_0^{\infty} \dfrac{dx}{1+x^2} = 2 \int_0^1 \dfrac{dx}{1+x^2}$$ This is true because $$\underbrace{\int_1^{\infty} \dfrac{dx}{1+x^2} = \int_1^0 \dfrac{-du/u^2}{1+1/u^2}}_{x = 1/u} = \int_0^1 \dfrac{du}{u^2+1}$$ Hence, $$\int_0^{\infty} \dfrac{dx}{1+x^2} = \int_0^{1} \dfrac{dx}{1+x^2} + \int_1^{\infty} \dfrac{dx}{1+x^2} = \int_0^{1} ... 1$$\int_{0}^{\infty} \frac {1}{1+x^2} \, \mathrm{d}x=\lim_{n\,\to\,\infty}\left({\arctan n}\right)-\arctan0={\pi\over 2}$$It is not true that$${\pi\over 2}=\pi$$unless someone made a big mistake and math is all wrong. Perhaps you meant$$\int_{0}^{1} \frac {1}{1+x^2} \, \mathrm{d}x=\int_{1}^{\infty} \frac {1}{1+x^2} \, \mathrm{d}x={\pi\over 4}$$0 It is best practice to use a dummy variable in the integrand itself. For example, we could write$$\int_0^x{t\,dt}=\frac{x^2}{2}$$It does not really make sense to write something like$$\int_0^x{x\,dx} because $x$ is a variable which is changing as the integral is computed. $x$ starts at $0$ and moves toward... toward $x$? That said, I think most people ...

0

Do you have equally spaced values available? ~~-> Simpson or similar Can you fix the values where to compute the function? ~~-> use one of the Gaussian methods Wildly varying function? ~~-> some adaptative method

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