# Tag Info

0

Let $$f(x) = e^{2x+\lfloor x \rfloor}.$$ This has a definite integral between any two real numbers $a$ and $b$, $$\int_a^b e^{2x+\lfloor x \rfloor} dx.$$ As has been mentioned already, there's a technical difficulty applying the standard definition of antiderivative to this function over any interval that contains an integer, but you can define $H(x)$ ...

0

Its always a bit hard to guess what another person might find intuitive, but here are my two cents on the topic. You can interpretate the elements of $\mathbb{R}^n$ as functions from the set $\{1,...,n\}$ to $\mathbb{R}$, where for $f \in \mathbb{R}^{n}$, $f(i)$ would just be the $i$-th component of the vector. We know from linear algebra that any linear ...

2

By rescaling the variable, let us replace the constant $8$ by $1$, for convenience. The equation $y=\sqrt{1-x^2}$ represents the upper-half of the unit circle, and the integral $$\int_{t=0}^x\sqrt{1-t^2}dt$$ is the area of a vertical "slice" between the abscissas $0$ and $x$. You can compute it as the area of a sector of aperture $\theta$ such that ...

0

The integral $\int R(x) \sqrt{\frac{x-a}{x-b}}dx$ in which form yours can be written, where $R(x)$ is a rational function, can be solved by substitution, $y=\sqrt{\frac{x-a}{x-b}}$. This substitution reduces the problem to the integral of a rational function, which can at be solved by partial fractions.

0

HINT: When integrating by parts in the middle a stage comes to integrate $$\int \frac{x\, dx}{\sqrt{1-x^2}}$$ $$= \dfrac12 \int \frac{2 x\, dx}{\sqrt{1-x^2}}= \dfrac12 \int \frac{d\,(x^2) }{\sqrt{1-x^2}}$$ which can be integrated with $x^2 = u$ substitution or directly.

1

If the limit $y = \lim_{n \to \infty} f(x; n)$ exists, wouldn't it be given by $$y = x + \frac1{y} \implies y^2-x y-1=0 \implies y = \frac{x}{2} + \frac12 \sqrt{x^2+4}$$ Then, by dominated convergence, $$\lim_{n \to \infty} \int_0^1 dx \, f(x;n) = \frac12 \int_0^1 dx \, \left (x + \sqrt{x^2+4} \right ) = \frac{\phi}{2} + \log{\phi}$$ This is close to, ...

1

let $$I=\int\sqrt{8-x^2}\ dx\tag 1$$ using integration by parts, $$I=\sqrt{8-x^2}\int 1\ dx-\int \left(\frac{-2x}{2\sqrt{8-x^2}}\right)\cdot x\ dx$$ $$I=\sqrt{8-x^2}(x)-\int \frac{(8-x^2)-8}{\sqrt{8-x^2}} \ dx$$ $$I=x\sqrt{8-x^2}-\int \left(\sqrt{8-x^2}-\frac{8}{\sqrt{8-x^2}} \right)\ dx$$ $$I=x\sqrt{8-x^2}-\int\sqrt{8-x^2}\ dx+8\int \frac{1}{\sqrt{8-x^2}}\ ... 0 Integration by parts is the method for this one, but here is another way of thinking (I don't say it is really different from integration by parts, and I don't suggest to use this instead of integration by parts). Let us think of 0<x<1. Draw the graph of y=\arcsin x. From the fundamental theorem of calculus we know that x\mapsto ... 1 The idea is that we want to get rid of the square root. So we use the fact that 1 - \sin^2 = \cos^2, so that$$\sqrt{1 - \sin ^2 x } = \sqrt{\cos^2 x} = |\cos x|$$which we then can integrate (of course there will be another factor coming from dx but that works out) Same idea when you're faced with \sqrt{1 + x^2}; this time we use the fact that 1 + ... 4 Notice,$$\int \frac{x}{(x^2-3x+17)^2}\ dx=\frac{1}{2}\int \frac{(2x-3)+3}{(x^2-3x+17)^2}\ dx=\frac{1}{2}\left(\int \frac{(2x-3)}{(x^2-3x+17)^2}\ dx+3\int \frac{1}{(x^2-3x+17)^2}\ dx\right)=\frac{1}{2}\left(\int \frac{d(x^2-3x+17)}{(x^2-3x+17)^2}+3\int \frac{1}{\left(\left(x-\frac{3}{2}\right)^2+\frac{59}{4}\right)^2}\ dx\right)$$... 0 Separate the integral into two integrals: \displaystyle\int\frac{u}{u^2 + \left(\sqrt{\frac{59}{4}}\right)^{2}} + \displaystyle\int\frac{\frac{3}{2}}{u^2 + \left(\sqrt{\frac{59}{4}}\right)^{2}} Use another u-substitution on the first integral and then use trig substitution on the second one. 2 One has$$\int \frac{u+\frac{3}{2}}{u^2+\frac{59}{4}} du = \int \frac{u}{u^2+\frac{59}{4}}du + \int \frac{\frac{3}{2}}{u^2+\frac{59}{4}}du.$$The first term can be computed be setting v = u^2+\frac{59}{4} and the second thanks to the \arctan function. 0 Another thing you could try is substituting x=\sin t, so that \mathrm{d}x=\cos t\,\mathrm{d}t.$$\int\sin^{-1}x\,\mathrm{d}x=\int t\cos t\,\mathrm{d}t$$If you can find a way to integrate this without using integration by parts, then more power to you. 0 Notice, use integration by parts as follows$$\int \sin^{-1}(x)\ dx=\int \sin^{-1}(x)\cdot 1\ dx=\sin^{-1}(x)\int 1\ dx-\int \left(\frac{d}{dx}(\sin^{-1}(x))\cdot \int 1\ dx\right)\ dx=\sin^{-1}(x)\cdot(x)-\int \frac{1}{\sqrt{1-x^2}}\cdot (x)\ dx=x\sin^{-1}(x)+\frac 12\int \frac{(-2x)}{\sqrt{1-x^2}}\ dx=x\sin^{-1}(x)+\frac 12\int ...

1

$$\Large \int_{a}^{b} x^{x^{x^{x\,\cdots}}} \, dx = -\Large \int_{a}^{b} \frac{W\left(-\ln(x) \right)}{\ln(x)}\, dx$$ Where W is the Lambert W function. The integral is convergent on $\quad e^{-e}\leq a\leq e^{1/e} \quad \text{and} \quad e^{-e}\leq b\leq e^{1/e}$ There is no closed form with a finite number of standard functions. Example of serie ...

0

Here it is: \begin{align*} \int\frac{dx}{(9+x^2)^2}\,dx &=\frac{1}{81}\int\frac{dx}{(1+(x/9)^2)^2} \;\;\;\;\;\;\;\;\mbox{put now}\;x=9\tan y \\ &=\frac{1}{81}\int\frac{1}{(1+\tan^2y)^2}\frac{9}{\cos^2y}\,dy\\ &=\frac19\int\frac{1}{\frac{1}{\cos^4y}}\frac{1}{\cos^2y}\,dy\\ &=\frac19\int{\cos^2y}\,dy\\ &=\frac1{18}\left[\cos y\sin y ...

0

We can do it integration by parts method as taking f(x)= sin and f(x)2 = 1. Also we can take x=sin(t) as second method. My maths is this level i can solve this sum like this only..but i want a method that we can use directly.

1

Following runaround's excellent hint, the required integral is $$3\int_{-1}^1 z^{2n}\hskip-1.4cm \underbrace{\int\int dx\,dy}_{\mbox{area of disk of radius }\sqrt{1-z^2}} \hskip-1.3cm dz=3\pi \int_{-1}^1 z^{2n}(1-z^2)\,dz={12\pi\over(2n+3)(2n+1)}.$$

1

Hint. You may use $$\sin (2\theta)=\frac{2\tan \theta}{1+\tan^2\theta}$$ and $$\theta= \arctan \frac{x}3.$$

1

Setting $\tan\theta=\frac x3$ & $\theta=\tan^{-1}\left(\frac x3\right)$, one should get $$\frac{3}{324}\left(\sin 2\theta+2\theta\right)+C$$ $$=\frac{1}{108}\left(\frac{2\tan\theta}{1+\tan^2\theta}+2\theta\right)+C$$ $$=\frac{1}{108}\left(\frac{2\cdot \frac{x}{3}}{1+\frac{x^2}{9}}+2\tan^{-1}\left(\frac x3\right)\right)+C$$ ...

0

In spherical coordinates : $$\begin{cases} x=r\cos(\theta)\sin(\phi) \\ y=r\sin(\theta)\sin(\phi) \\ z=r\cos(\phi) \\ \end{cases}$$ $$\iiint_V \ x^{2n} + y^{2n} + z^{2n} \,dx\,dy\,dz = \iiint_V \ \left( \cos^{2n}(\theta)\sin^{2n}(\phi) + \sin^{2n}(\theta)\sin^{2n}(\phi) +\cos^{2n}(\phi) \right) r^{2n+2}\sin(\phi)\,dr\,d\theta\,d\phi$$ in : ...

3

HINT Draw a geometric representation of $\alpha = \arcsin((x-2)/2)$, i.e. a right-angled triangle with hypotenuse $2$ and one side $x-2$. Mark where the angle $\alpha$ would be. Now find the length of the 3rd side using Pythagorean Theorem and compute $\tan \alpha$ directly from the triangle.

3

Let $y=\arcsin\dfrac{x-2}2\implies\sin y=\dfrac{x-2}2$ As $-\dfrac\pi2\le y\le\dfrac\pi2,\cos y\ge0$ $\cos y=+\sqrt{1-\left(\dfrac{x-2}2\right)^2}=\dfrac{\sqrt{4x-x^2}}2$ $\tan\left(\arcsin\dfrac{x-2}2\right)=\tan(y)=\dfrac{\sin y}{\cos y}=?$

1

Another way using reduction formula: $$\int(a^2-x^2)^nx^mdx$$ $$=(a^2-x^2)^n\int x^m\ dx-\int\left(\dfrac{d\{(a^2-x^2)^n\}}{dx}\int x^m\ dx\right)dx$$ $$\implies\int(a^2-x^2)^nx^mdx=(a^2-x^2)^n\cdot\dfrac{x^{m+1}}{m+1}+\dfrac{2n}{m+1}\int(a^2-x^2)^{n-1}x^{m+2}\ dx$$ Setting $n=\dfrac32, m=-6$ ...

6

You should use the symmetry to reduce it to $$3\iiint_V \ z^{2n} \,dx\,dy\,dz,$$ then use spherical coordinates to integrate.

1

Different method here. Consider $$\dfrac{d}{da} \int_0^a \sqrt{a^2-x^2} dx = a \int_0^a \frac{1}{\sqrt{a^2-x^2}} dx + \underbrace{\sqrt{a^2-a^2}}_0$$ by the Leibnitz rule of differentating under the integral. Now, the integral on the left-hand side is just a quarter of the area of the circle of radius $a$ (because the integrand is the $y$-coordinate of ...

0

More thoughts about this. Matrix $A$ can be thought of as a linear operator from $\mathbb{R}^n$ to $\mathbb{R}^n$. In a similar way your integral transform $L$ is an operators from a (Hilbert) space of functions to a different space. Just like you can define characteristics of $A$ (like eigenvalues and eigenvectors), and talk about basis of its image, so ...

3

Notice, you have made mistake while substituting let $3x=4\sin\theta\implies dx=\frac{4}{3}\cos\theta\ d\theta$ $$\int \frac{(16-9x^2)^{3/2}}{x^6}\ dx$$$$=\int\frac{(16-16\sin^2\theta)^{3/2}}{\left(\frac 43\sin\theta\right)^6} \left(\frac{4}{3}\cos\theta\ d\theta\right)$$ $$=\frac{243}{16}\int \frac{\cos^4\theta}{\sin^6\theta}\ d\theta$$ ...

0

You want to use the following theorem: If $f$ is integrable on $[a,b]$, then $f$ is bounded on $[a,b]$ and for each $\epsilon>0$ there exists a partition $P$ of $[a,b]$ such that $$U(P,f)-L(P,f)<\epsilon$$ where $U,L$ denote Darboux upper sum and Darboux lower sum, respectively. Given any $\epsilon>0$, there corresponds a partition ...

1

}(x-t)\right]\left[\frac{1}{2}+\sum_{n=1}^{N}\cos[n(t-x)]\right]=\frac{1}{2}\sin\left[\frac{1}{2}(x-t)\right]+\sum_{n=1}^N\sin\left[\frac{1}{2}(x-t)\right]\cos\left[n(x-t)\right]\\=\frac{1}{2}\sin\left[\frac{1}{2}(x-t)\right]+\frac{1}{2}\sum_{n=1}^{N}\sin\left[\left(n+\frac{1}{2}\right)(x-t)\right]-\sin\left[\left(n-\frac{1}{2}\right)(x-t)\right]\end{align*} ...

3

HINT: Use the property that $$I=\int_a^b f(x) dx =\int_a^b f(a+b-x) dx$$ And then you will get that $$2I=\int_{-\pi}^\pi \frac{\sin nx}{\sin x}dx$$ P.S. You may have to use the property that $\sin x$ is an odd function.

0

In the context of Lebesgue integration, instead of "Stieltjes", we speak of complex measures, say $\mu$. By the Lebesgue-Radon-Nikodym theorem, you can decompose $\mu$ uniquely into two measures $\mu = \mu_a + \mu_s$. $\mu_a$ is absolutely continuous with respect to the Lebesgue measure $m$, and $\int_{0}^{t}gd\mu_a$ is of bounded variation. The ...

4

Notice, $$\int_0^a\frac{1}{\sqrt{a^2-x^2}}\:\mathrm dx$$ Let $x=a\sin\theta\implies \mathrm dx=a\cos\theta\:\mathrm d\theta$, \begin{align}&=\int_{0}^{\pi/2}\frac{1}{a\cos \theta}(a\cos\theta\ \, \mathrm d\theta)\\ &=\int_{0}^{\pi/2}\,\mathrm d\theta \\& =\left(\theta\right)_{0}^{\pi/2}\\&=\frac \pi 2\end{align}

7

Here is a tip, for $a>0$ Factor a out: $$\frac {1} {a}\int_{0}^{a}\frac{1}{\sqrt{1-\dfrac{x^2}{a}}}\, \mathrm dx$$ And then set $t=x/a$ threfore $\mathrm dt=\frac 1 a\,\mathrm dx$ $$\int_{0}^{a}\frac{1}{\sqrt{1-t^2}}\,\mathrm dt\\ =\arcsin\left(\frac x a\right)\bigg|_0 ^a\\$$ for $a<0$ switch the order of the integration bounds, factor ...

0

As it turns out the two results for $\psi$ are actually equivalent as by the dispersion relation: $$\frac{1}{\omega}-\alpha=\frac{1}{\omega}\left(1-\alpha\omega\right)=-\omega$$ Alternative method: First, redefine $\theta$ and $\psi$ using a scalar $\phi$: $$\theta=\partial_{\eta}\phi\quad\psi=-\partial_{\tau}\phi$$ This ensures that: ...

0

I also think that this integral diverges. $$\int\frac{e^{-\frac{a^2}{2 (x+y)}}}{(x+y)^{3/2}}\,dy=-\frac{\sqrt{2 \pi } }{a}\text{erf}\left(\frac{a}{\sqrt{2} \sqrt{x+y}}\right)$$ $$\int_0^\infty\frac{e^{-\frac{a^2}{2 (x+y)}}}{(x+y)^{3/2}}\,dy=\frac{\sqrt{2 \pi } }{a}\text{erf}\left(\frac{a}{\sqrt{2x} }\right)$$ provided $\Re(x)>0\lor x\notin \mathbb{R}$ ...

0

Continuing your calculations $$\int \frac{1}{49 - y^2} dy = \int \frac{1}{5} dx$$ $$\int \frac{1}{7^2 - y^2} dy = \int \frac{1}{5} dx$$ $$\int \frac{1}{7 - y} + \frac{1}{7 + y} dy = \int \frac{14}{5} dx$$ $$\log(\frac{7+y}{7-y}) = \frac{14}{5} x + C$$

2

You have, for $x>1$, $$\lim_{t\to \infty}e^{-t}t^{x-1}=0$$ and $$\lim_{t\to 0^+}e^{-t}t^{x-1}=0$$ giving $$\left[-e^{-t}t^{x-1}\right]_0^∞=0.$$

1

By using the disc method, you get to evaluate \begin{align} \int_0^5\pi (\cosh^2 2x-\sinh^2 2x)\:dx&=\int_0^5\pi (\cosh^2 2x-\sinh^2 2x)\:dx\\\\ &=\pi\int_0^5 1\:dx\\\\ &=5\pi. \end{align}

1

Let $f(x) = |x|^r$ for $|x| \le \epsilon$, $0$ otherwise. Then your integral is $\mathbb E[f(X)]$. But $f(X) \le \epsilon^r$ always, so $\mathbb E[f(X)] \le \epsilon^r$.

1

For arbitrary $r>0$, yes. For any $x$ such that $\lvert x\rvert \leq \varepsilon$, you have $\lvert x\rvert^r \leq \varepsilon^r$ since $t\mapsto t^r$ is increasing on $[0,\infty)$, and then $$\int_{\Omega} \mathbb{1}_{\{\lvert X\rvert \leq \varepsilon\}} \lvert X\rvert^r d\mathbb{P} \leq \int_{\Omega} \mathbb{1}_{\{\lvert X\rvert \leq \varepsilon\}} ... 0 Let$$Y(p) = \int_0^{\infty} dx \, y(x) \, e^{-p x}$$be the Laplace transform of y(x). By the convolution theorem, the integral equation is equivalent to$$Y(p) + \frac1{p^2} Y(p) = \frac1{p^4} $$Then$$Y(p) = \frac1{p^2 (p^2+1)} $$We may find y(x) by finding the inverse LT by either using a table or the residue theorem. Either way,$$y(x) = ...

0

You incorrectly solved for your constant. At the time of launch, $v=v_o$ and $x=a$. Thus: $\displaystyle\frac {v_0^2}{2} = \frac {ga^2}{a} + c$ $c=\displaystyle\frac {v_0^2-2ga}{2}$ Substituting $c$ back into our equation we have: $\displaystyle\frac {v^2}{2} = \frac {ga^2}{x} + \frac {v_0^2-2ga}{2}$ Letting $v=0$ and solving for $x:$ $\displaystyle0 ... 0 When you are integrating just substitute the appropriate limits. $$\int_{v_0}^0 v dv = -ga^2 \int_a^{h_{max}} x^{-2} dx$$ Following this you get the desired expression. 0 Since you want to integrate on$\;[0,1]\;$, we have$\;t>0\;$. Now, take the other factor$\;(t-1)(t-2)\;$. This is an upwards parabola which vanishes at$\;t =1,2\;$and it is thus negative precise for$\;1\le t\le 2\;$, so$\;t(t-1)(t-2)\;$in$\;[0,1]\;$is always non-negative and so is its integral. But since at$\;t=1/2\;$the function is ... 6 Notice that for all$t \in (0,1)$, $$t>0, \;\;\; (t-1)<0 \;\;\; \text{and} \;\;\; (t-2)<0$$ and it follows that $$t(t-1)(t-2)>0$$ so that every element being integrated is positive. It follows that the integral is positive. 0 A polynomial function is continuous and hence can only change signs at roots. To see whether the function is positive on$(0,1)$plug in some point in the interval. 0 It looks like you attempted to use substitution to replace $$\int \frac{\frac12 dx}{e^{x/2}}$$ with $$\int \frac{du}{u} = \ln(u) + C.$$ The fatal flaw here is that if$u = e^{x/2}$, then$du = \frac12 e^{x/2} dx \neq \frac12 dx$. The correct substitution is $$\tfrac 12 dx = \frac{1}{e^{x/2}} du = \frac{du}{u}.$$ $$\int \frac{\frac12 dx}{e^{x/2}} = \int ... 0 Notice, divide the numerator by e^{x/2} & integrate as follows$$\int \frac{e^x-2}{e^{x/2}}\ dx=\int \left(e^{x/2}-2e^{-x/2}\right)\ dx=\int e^{x/2}\ dx-2\int e^{-x/2}\ dx=2e^{x/2}\ dx-2(-2)e^{-x/2}+C=\color{red}{2e^{x/2}+4e^{-x/2}+C}$$1 The second part is false. One has$$\int \frac{2}{e^{\frac{1}{2}x}} = 2\int e^{-\frac{1}{2}x} = -4e^{-\frac{1}{2}x} + C.$\$

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