New answers tagged integration
1
Basically you $dr$ is negative , so $|dr|=-dr.$ like $|-2|=-(-2)$ , evaluate using this , the sign ambiguity will dissolve .
1
For the first part, the Wikipedia article http://en.m.wikipedia.org/wiki/Runge%E2%80%93Kutta_methods has a detailed discussion of the different members of the family along with a derivation of RK-4. The key idea is that you can do whatever you want in terms of writing down the form of your approximation; all that matters is that doing a Taylor expansion ...
2
Answer to Part 1
The fractional coefficients in RK4 sum to 1. This is in part to satisfy what is known as the order conditions of the integrator. Basically, we want RK4 to be a 4th order method. For this to be true, it has to satisfy a special relationship using the coefficients of its Butcher Tableau (see at the top here), namely
$$\mathbf{b}^T ...
0
Note that
$$ \int_{-\infty}^\infty (X-\kappa\tau)\phi(X)\,\mathrm dX=E[X-\kappa\tau]=-\kappa\tau$$
and for $\tau>0$
$$ \int_{\tau}^\infty (X-\kappa\tau)\phi(X)\,\mathrm dX\ge\int_{\tau}^\infty (1-\kappa)\tau\phi(X)\,\mathrm dX>0.$$
So for $\tau>0$ your expression is positive.
2
It suffices to see that the strange-looking integral is a particular solution. You probably came up with the following type of expression, or at least you should agree it is an obvious way to go:
$$u_p(t)=\int_0^t\int_0^v\int_0^u e(s)dsdudv.\tag{1}$$
Now this is equal to
$$\int_0^te(s)\cdot{\rm Area}(\{(u,v):s\le u\le v\le ...
6
I guess the focus of the formula is not that you can evaluate certain integrals by plugging in function values, but that you can recover a function's values by an integral along a circle.
0
(Note: $\int f$ will be used throughout in place of $\int_{-\infty}^\infty f(x)\,dx$.) Suppose $f:\Bbb{R}\to\Bbb{R}$ is integrable on $\Bbb{R}$, with $\int f=L\ne0$. By the definition of the integral $\int_{-\infty}^\infty$, we know that the one-sided limits $\lim_{y\to\infty}\int_0^y f=L_+$ and $\lim_{y\to-\infty}\int_y^0 f=L_-$ exist and $\int ...
0
A common approach is to express the integrand in the form $\exp(ax^2+bx+c)$ and then
complete the square in the argument of the exponential. Then, what remains can be
expressed in the form of an integral of a normal density over $\mathbb R$ which has value $1$, and of course anything involving $y$ and various constants is what is left. In short,
$I(y)$ ...
-1
By using the formula http://upload.wikimedia.org/wikipedia/en/math/7/2/a/72a1058ad2087aec467af24bddcf9479.png, we have $\int_0^1\sqrt{\tan^{-1}x}~dx=\sum\limits_{n=1}^\infty\sum\limits_{m=1}^{2^n-1}\dfrac{(-1)^{m+1}}{2^n}\sqrt{\tan^{-1}\dfrac{m}{2^n}}$
2
There is already a theorem which resembles your result:
Suppose $d$ is a non-negative function with the property:$$\int_{-\infty}^{+\infty }d(s)ds=1$$Then the sequence $d_k(t):=k\cdot d(kt)$ is a Dirac sequence.
A dirac sequence has the properties:
$d_k\geq0,\forall k$
$\int_{-\infty}^{+\infty }d_k(s)ds=1,\forall k$
$\forall r>0$ and $\forall ...
1
I bet your text defines a contour as a closed curve. :)
2
The fundamental theorem of calculus is applied to a special contour: intervals of the real line. And it applies to functions that have an antiderivative, or primitive. In the complex plane, we have Cauchy's Theorem which states that such nicely behaved functions ("analytic") may be integrated by evaluating its antiderivative at the endpoints of a contour, ...
5
You can have a solution as an infinite sum
$$\int_0^1 \sqrt {\tan^{-1}x}\space dx = \frac{\sqrt {\pi }}{2}-\frac{1}{\sqrt {\pi }}\sum _{n=0}^{\infty }{\frac {\psi \left( 2\,n+1
,\frac{1}{2} \right) }{ 16^n\left( 4\,n+3 \right) \left( 2\,
n+1 \right) !}} \sim 0.6298233443,$$
where $\psi(x)$ is the digamma function.
0
This is a definition. See bottom of pg. 100 of Spivak's book.
If $\omega$ is a $k$-form on $[0,1]^k$, there is a unique $f$ such that $\omega = f dx^1 \wedge \cdots \wedge dx^k$. Then define
$$
\int_{[0,1]^k} \omega := \int_{[0,1]^k} f
$$
or
$$
\int_{[0,1]^k} f dx^1 \wedge \cdots \wedge dx^k= \int_{[0,1]^k} f(x^1,\cdots,x^k)dx^1\cdots dx^k
$$
3
For continuity you can apply the Dominated convergence theorem:
Consider a $t\in\mathbb R$ and a sequence $t_n\to t$. Then $$g(t_n) = \int_{\mathbb{R}^{n}} K(t_nx)f(x)dx$$
Let $h_n(x)=K(t_nx)f(x)$. It is apparent that $h_n\to h=K(tx)f(x)$ since $K$ is continuous.
Also $|h_n(x)|=|K(t_nx)||f(x)|\leq \sup|K(x)|\cdot |f(x)|=N|f(x)|$. Now note that $N|f(x)| $ ...
4
The Riemann-Stieltjes integral $\int_a^b f(x) dg(x)$ is the limit as the step size tends to zero of $\sum_{i=0}^{n-1}f(c_i)(g(x_{i+1})-g(x_i))$ for $a=x_0<x_1<...<x_n=b$ where $c_i$ is between $x_i$ and $x_{i+1}$.
$\alpha(x_{i+1})-\alpha(x_i)$ is $0$ unless $x_i$ and $x_{i+1}$ straddle the boundary between integers, which in the limit happens at ...
2
You can find the particular solution by doing an integration by parts.
$\frac12 \int_0^t (t-s)^2 e(s)ds = [0-0] -2* \frac12 \int_0^t (t-s) E(s)ds $ where E(s) is a primitive of $e(s)$ such that $E(0)=0$
Similarly we have : $-\int_0^t (t-s) E(s)ds = - ( [0-0] - \int_0^t F(s)ds)$ where $F(s)$...
And u(t)=$\int_0^t F(s)ds$ is a solution of $u'''(t)=e(t)$
...
1
The particular solution
$$u(t)=\frac{1}{2}\int^{1}_{0}(t-s)^2e(s)ds $$
is the Green's function solution $G(t,s)$ for this problem.
2
Since $f$ is continuous, it is integrable. Hence any Darboux sum converges to integral of $f$ when partition's tightness tends to $0$. Now I suggect you to consider the simplest partition $\Delta_j:=[\frac{j}{2^n},\frac{j+1}{2^n}]$ with points $\xi_j=\frac{j}{2^n}$, where $j=0,\ldots, 2^{n}-1$
1
This is a CW answer intended to remove this question from the Unanswered queue.
As AlexE remarks, there is no general way to swap products and integrals.
This means that in general, evaluating integrals like the one stated can be quite hard, especially if the mapping $G$ is not in some way "nice".
2
The volume of your sphere is $\,\displaystyle{\frac43\pi 4^3=\frac{256\pi}{3}}\,$ . You need a line $\,y=kx+4k\,$ , which intersects $\,y=\pm\sqrt{16-x^2}\,$ at
$$16-x^2=k^2x^2+8k^2x+16k^2\implies(k^2+1)x^2+8k^2x+16(k^2-1)=0\implies$$
$$\Delta=64k^4-64(k^4-1)=8^2\implies$$
...
0
Under the condition that $H\bigl(p(0), q(0)\bigr)= 1$, we can argue as follows: Let $h(t) = H\bigl(p(t), q(t)\bigr)$ for all $t$. Differentiation using the chain rule gives
\begin{align*}
h'(t) &= \partial_p H\bigl(p(t), q(t)\bigr)p'(t) + \partial_q H\bigl(p(t), q(t)\bigr)q'(t)\\
&= -\partial_p H\bigl(p(t), q(t)\bigr)\partial_q H\bigl(p(t), ...
2
Can you tell what your proposed solution represents? I think it's far more interesting (and important) for you to be able to look at the integral, and be able tell what it means.
If you think of $x^2$ in terms of the area of a square of sides $x$, then your proposed integral calculates the volume of a solid that consists of two identical square-base ...
1
Use the substitution $z = u(x)$ on the right integral.
1
Change variables $x-y=s$, thus the new domain is $(x,x-1)$. You have
$$
\int_0^1\Xi(x-y)dy=\int_{x-1}^x\Xi(s)ds.
$$
Now, if $x-1>1$ and if $x<0$ the integral vanishes. Moreover, you have
$$
\int_{x-1}^x\Xi(s)ds=\text{length}\left((x-1,x)\cap(0,1)\right).
$$
If $0<x<1$
$$
\text{length}\left((x-1,x)\cap(0,1)\right)=x.
$$
Otherwise, if $1<x<2$
...
1
You're off to a good start. Next note that
$$
\chi_{[0,1]}(x-y) = \begin{cases} 1, & \text{for $x-1 \le y \le x$} \\ 0, & \text{otherwise}\end{cases}.
$$
Draw some graphs!
Hence, if $0< x < 1$, then $\chi_{[0,1]}(x-y)$ on the whole of $[0,1]$. Thus the integral is $0$.
If $0 < x < 1$, then $\chi_{[0,1]}(x-y) = 1$ on $[0,x]$ and $=0$ on ...
3
I am noting you a useful Theorem which I found it for you. So by using it you can defeat the second part of your question by yourself.
Theorem: The proper conic with focus-directrix distance $p$ and eccentricity $e$ has equation $$r=\frac{ep}{1-e\cos \theta}$$ in polar coordinates, if the focus is at the pole and the polar axis is perpendicular to the ...
1
The region between the ellipse and the circle is what you are asked to investigate. So it's the ellipse minus the circle; an ellipse with a circular hole in int near its center.
0
It's not enough to compute the four apexes of the ellipse, since there might be some intersection in between. Argue instead as follows:
When $x^2+y^2\leq 1$ then $|x-1|\leq 2$ and therefore
$${(x-1)^2\over 9}+{y^2\over8}\leq {4\over9}+{1\over8}<1\ .$$
This shows that the unit circle is completely contained in the interior of the given ellipse.
2
HINT:
If $$y_n=\int_0^1\frac{x^n}{x+5}dx$$
$$\int_0^1\frac{x^n}{x+5}=\int_0^1\frac{x^{n-1}(x+5-5)}{x+5}dx=\int_0^1x^{n-1}dx-5\int_0^1\frac{x^{n-1}}{x+5}dx$$ as $x+5\ne0$
Alternatively,
$$\text{So,}y_n+5y_{n-1}$$
$$=\int_0^1\frac{x^n}{x+5}dx+5\int_0^1\frac{x^{n-1}}{x+5}dx$$
$$=\int_0^1\frac{x^{n-1}(x+5)}{x+5}dx=\int_0^1x^{n-1}dx$$ as $x+5\ne0$
1
Guldin's rule, also called Pappus' centroid theorem, says the following: When you rotate a shape $S$ around an exterior axis the volume $V$ of the solid generated in this way is equal to the area of the shape times the circumference of the circle described by the centroid of $S$.
It follows that the volume of the full torus is given by $V_{\rm tot}=\pi ...
24
Actually, you have two errors there:
The minor one is that you seem to want a cube of side $2r$,
since your integral goes from $-r$ to $r$.
The major error, as others have said, is that you are finding the volume of a pyramid, not a cube. Actually, since you are integrating from $-r$ to $r$,
you are finding the volume of two pyramids, one upside-down, ...
13
The real problem here is not the endpoints of your integral, it's that the function you are integrating is not constant with respect to the variable of integration. A cube has the same cross section everywhere, while in your original integral the cross section is bigger at the ends than in the middle. See @response's solution for the right way to set this ...
35
It should be:
$$V = \int_0^a a^2 dz$$
where $a$ is the length of one of the sides of the square.
Or using your notation:
$$V = \int_0^x x^2 dz$$
where $z$ is the dimension over which you are integrating.
2
Effectively, the below constitutes a derivation of the Leibniz Integral Rule as applicable to the current setting.
Assuming sufficient conditions to allow for interchanging integration and differentiation, and existence of the integrals, you can.
Perhaps the most insightful way is to introduce some auxiliary functions. Define:
$$\begin{align}
G(x,y,b,p) ...
1
This problem popped back up in the queue, so I had a look. (I'm assuming this was never completely resolved, to judge from OP's final comment on 13 Feb, 16:10.) In the first page image posted, the construction of the integral on line 5 is largely correct:
$$\int_{-r}^0 \ t \sqrt{r^2 - t^2} \ + \ R \sqrt{r^2 - t^2} \ \ dt \ . $$
But None has neglected the ...
1
$$I=\int_0^{\dfrac{\pi}{2}}\log\sin x\,dx\cdots(1)$$
use this property:$$\int_0^a f(x)dx = \int_0^a f(a-x)dx$$
$$I=\int_0^{\dfrac{\pi}{2}}\log\cos x\,dx\cdots(2)$$
Add both eqn:
$$2I=\int_0^{\dfrac{\pi}{2}}\log\sin x+\log\cos x\,dx$$
$$2I=\int_0^{\dfrac{\pi}{2}}\log({\sin x\cdot\cos x})\,dx$$
$$2I=\int_0^{\dfrac{\pi}{2}}\log(\dfrac{2\sin x\cdot\cos ...
6
Hint: Use the fact $$\int_0^a f(x)dx = \int_0^a f(a-x)dx$$ You can show this by letting $u = a-x$.
4
Set $t = \pi/2-x$ and make use of the fact that $\displaystyle \int_a^b f(y) dy = - \int_b^a f(y) dy$ to conclude what you want.
1
Not a different answer but rather generalizing the technique Peter Tamaroff used in his answer. Notice the fact that $(u - u^{-1})^2 + 2 = u^2 + u^{-2}$, we can generalize this for any definite integral.
Say if we want to compute:
$$
\int^{\infty}_{-\infty} f(x)\,dx,
$$
knowing this is convergent. Using the trick we can see the integral is:
$$
...
2
I don't know if you wanna compute the surface integral the hard way, which is, computing the surface outward normal pointwisely and then integrating. For me this problem looks rather like a standard "applying the divergence theorem" exercise.
Denote the interior of the torus as $T$, and $\partial T = S$, then by Divergence theorem:
$$
\iint_S ...
3
We need to interpret "as accurate as possible." One common interpretation is that it is dead on for $f(x)$ identically equal to $1$, for $f(x)=x$, for $f(x)=x^2$, and so on for as long as possible.
If the formula is to give the right answer for $f(x)=1$ (and hence for $f(x)$ any constant function) we need $c_1=\frac{1}{2}$.
For the formula to give the ...
2
Let $f$ and $g$ be continuous functions on $[a, b]$ such that
$$
\int_a^b f(x)\ dx = \int_a^b g(x)\ dx.
$$
Then
$$
\int_a^b f(x)-g(x)\ dx = 0.
$$
Since $f$ and $g$ are continuous, then $f-g$ is continuous. Hence, if $f-g$ is never zero on $[a, b]$, then it must be strictly positive or negative on $[a, b]$. But then $\int_a^b f(x)-g(x)\ dx \neq 0$, which is a ...
1
Suppose that $f(x_1)=A>0$ for some $x\in [a,b]$. Then, by continuity, for $\varepsilon:=A/2$ there is a $\delta>0$ such that $x\in (x_1-\delta,\,x_1+\delta) \,\implies\, |f(x)-f(x_1)|<\varepsilon$, which in this case implies $f(x)>A/2$.
As the interval $(x_1-\delta,\,x_1+\delta)$ intersects $[a,b]$ at least on a length of $\delta$, and as ...
0
Wolfram Alpha checked that this improper integral does not converge.
2
General description. A partial fraction of the form $$\dfrac{Bx+C}{\left[ \left( x-r\right) ^{2}+s^{2}\right] ^{n}}$$ is integrable by substitution, using the change of variables $x=r+st$. With this method an integral of the form
$$\displaystyle\int \frac{Bx+C}{\left[\left( x-r\right) ^{2}+s^{2}\right] ^{n}}dx\tag{1}$$
is transformed into an integral of ...
2
To answer your specific question, here is a trick that will work. Note that
$$2x^2+2px=(2x^2+4px+2q)-2px-2q=(2x^2+4px +2q)-p(2x+2p)-2q+2p^2.$$
It follows that
$$\int \frac{dx}{x^2+2px+q}=\frac{x}{x^2+2px+q} +2\int \frac{x^2+2px+q}{(x^2+2px+q)^2}\,dx -p\int \frac{2x+2p}{(x^2+2px+q)^2}\,dx+ (2p^2-2q)\int \frac{dx}{(x^2+2px+q)^2}.$$
On the right, the first ...
1
You made a mistake there somewhere: the value of $X(0)$ makes no sense. Here's what I get for the LT:
$$X(s) = \frac{\dot{x}_0 + (s+\delta) x_0}{s^2+\delta s+\omega_0^2} + \frac{\gamma s}{(s^2+\delta s+\omega_0^2)(s^2+\omega^2)} $$
where $x_0 = x(0)$ and $\dot{x}_0 = \dot{x}(0)$. As you point out, the ILT is given by
$$\frac{1}{i 2 \pi} \lim_{R \to ...
2
1.for the integrand $\dfrac{1}{x^2+2px+q}$ complete the square to get
$\frac{1}{\dfrac{1}{4}(4q-4p)^2+(p+x)^2}$
2.then substitute $u=p+x$
3.factor out $\dfrac{1}{4}(4q-4p)^2$
The integral reduced to simple trigonometric function
9
Whenever you have a rational polynomial with a numerator of equal or larger degree than the denominator, try to factor the numerator if possible, or simply, use polynomial long division.
lab bhattacharjee noticed a nice way to simplify the rational integrand by manipulating the numerator to make life easier.
But suppose you're tired and not feeling ...
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