# Tag Info

16

They share a Latin root corresponding the the concept of 'wholeness'. In the context of integers, this would be interpreted as 'whole numbers' - i.e. numbers with no fractional part. In the context of integration, this would correspond to 'summing up to create a whole' in the sense of the integral representing a continuous sum or area. I think that's why ...

9

Outline of Method Perform the substitution $u = \ln x$. You will get $$\int_{-\infty}^0 x^n e^x dx.$$ This is very nearly the Gamma function. Perform the substitution $u = -x$. You will get $$(-1)^n \int_0^\infty x^n e^{-x} dx.$$ This now is the Gamma function. In particular, you have $$(-1)^n \Gamma(n+1) = (-1)^n n!.$$ We might do a sanity check when ...

9

We have, taking $u=\sqrt{x^{2}-1}$, $$\int\frac{4}{x\sqrt{x^{2}-1}}dx=4\int\frac{1}{u^{2}+1}du.$$ I think you can get from here. Expanding a little bit, note that $$u=\sqrt{x^{2}-1}\Rightarrow du=\frac{x}{\sqrt{x^{2}-1}}dx$$ and $$x^{2}=u^{2}+1$$ then ...

7

Integrating by parts twice, we get \begin{align} &\int_0^L\sin(x)\,e^{-ax}\,\mathrm{d}x\tag{0}\\ &=-\int_0^L e^{-ax}\,\mathrm{d}\cos(x)\tag{1}\\ &=1-e^{-aL}\cos(L)-a\int_0^L\cos(x)\,e^{-ax}\,\mathrm{d}x\tag{2}\\ &=1-e^{-aL}\cos(L)-a\int_0^L e^{-ax}\,\mathrm{d}\sin(x)\tag{3}\\ ... 6 Given \int_0^1 \ln^n(x) dx $$Let us write$$ I_n = \int_0^1 \ln^n(x) dx. $$Integrate by parts$$ \begin{eqnarray} I_n &=& \int_0^1 \ln^n(x) dx\\ &=& \Bigg[ x \ln^n(x) \Bigg]_0^1 - n \int_0^1 \ln^{n-1}(x) dx\\ &=& \Bigg[ \exp(y) y^n\Bigg]_{-\infty}^0 - n \int_0^1 \ln^{n-1}(x) dx = - n I_{n-1}. \end{eqnarray} $$... 5 We have,$$\int \frac{4}{x\sqrt{x^2-1}}dx$$Let, x=sec\alpha\implies dx=\sec\alpha\tan \alpha d\alpha$$=\int \frac{4\sec\alpha\tan \alpha d\alpha}{\sec\alpha\sqrt{\sec^2\alpha-1}}dx =4\int \frac{\tan \alpha d\alpha}{\tan\alpha}dx =4\int d\alpha=4\alpha+C=4\sec^{-1}(x)+C$$5 One way is to use 'Feynamnn's trick', and differentiate under the integral. Note$$\int_0^\infty x^4e^{-\alpha x^2}=\int_0^\infty \frac{\partial^2}{\partial \alpha^2}e^{-\alpha x^2}=\frac{\partial^2}{\partial \alpha^2}\int_0^\infty e^{-\alpha x^2}$$Calculate the inner definite integral first, then differentiate it twice with respect to \alpha. 4 Let I = \displaystyle\int_{0}^{\pi}xf(\sin x)\,dx. Substitute x' = \pi - x to get I = -\displaystyle\int_{\pi}^{0}(\pi-x')f(\sin (\pi - x'))\,dx' = \displaystyle\int_{0}^{\pi}(\pi-x')f(\sin x')\,dx'. So, 2I = I+I = \displaystyle\int_{0}^{\pi}xf(\sin x)\,dx + \int_{0}^{\pi}(\pi-x)f(\sin x)\,dx Can you take it from here? 4 set t=\sqrt{x^2-1} then we get$$dx=\frac{\sqrt{x^2-1}}{x}dt$$and we get$$\int\frac{4dt}{t^2+1}$$3 Below follows an answer like the one here, but maybe with easier terminology. I will update if I find a simpler argument. Let M=\max_{x\in[a,b]}|f(x)| (which exist since f is continuous and [a,b] is compact). To show \leq, note that, since |f(x)|\leq M for all x\in[a,b],$$ \Bigl(\int_a^b |f(x)|^t\,dx\Bigr)^{1/t}\leq \Bigl(\int_a^b ...

3

I cannot think of any transformation No wonder you can't, since even its definite counterpart requires the presence of special Bessel and Struve functions: $$\int_0^1\frac{e^x}{\sqrt{1-x^2}}~dx~=~\frac\pi2\Big(I_0(1)+L_0(1)\Big),$$ $$\int_{-1}^0\frac{e^x}{\sqrt{1-x^2}}~dx~=~\frac\pi2\Big(I_0(1)-L_0(1)\Big).$$

3

Hint: Put $-\alpha x^2=u \Rightarrow -2\alpha x \mathrm dx=\mathrm du$,then $x\mathrm dx=\frac {-1}{2\alpha}\mathrm du$. you will have, $$\displaystyle\int xe^{-\alpha x^2}\,dx=\frac {-1}{2\alpha}\displaystyle\int e^u \mathrm du$$

3

Try using more segments for your quadrature formula. I've obtained $19.02$ using $h = 1$ step. Or you might have implemented Simpson's rule incorrectly. Also the integrand is not quite good for numerical integration. Recall that Simpson's rule has a truncation error $$\epsilon \sim h^4 \max_{x \in [a,b]} \left|f^{(4)}(x)\right|$$ and your function $f(x) = ... 3 Hint: I think that Stokes' theorem is overkill here. We can get a simple parametrization: $${\bf r}(t) = (\cos t, \sin t, -1+\cos t), \quad 0 \leq t< 2\pi,$$and all the functions involved are simple too. Just write${\rm d}x = -\sin t\,{\rm d}t$, etc, substitute and compute. Use and abuse of periodicity. A lot of terms should cancel out, it seems. 2 $$I(a,b,p)\equiv\int_0^{\infty}\frac{\sin\left(p\sqrt{x^2+a^2}\right)}{\sqrt{x^2+a^2}}\cos(bx)\,dx\\\\$$ Enforcing the substitution$x\to a\sinh x$and assuming that$a>0yields \begin{align} I(a,b,p)&=\int_0^{\infty}\sin\left(pa\cosh x\right)\cos(ba\sinh x)\,dx\\\\ &=\frac12\int_0^{\infty}\left(\sin\left(pa\cosh x+ba\sinh ... 2 Another way to look at it (that probably isn't simpler) : Let u=x^2, then the integral becomes:\frac{1}{2} \int_0^\infty u^{3/2}e^{-\alpha u}du$$which is the Laplace transform of u^{3/2}, times {1}/{2}. We know that the Laplace transform of u^n is n!*\alpha^{-n-1} or generally \Gamma(n+1)*\alpha^{-n-1}. Here, n=3/2 so$$\Gamma(3/2 ... 2 Here, you'll want to integrate by part to find the Gauss integral : $$I = \int_0^{\infty} x^4 e^{-\alpha x^2} dx = \int_0^{\infty} x^3 \times x e^{-\alpha x^2} dx$$ Now, you letu = x^3$and$v' = x e^{-\alpha x^2} , and this gives you $$I = \left[ x^3 \frac{e^{-\alpha x^2}}{2\alpha} \right]_0^{\infty}-\int_0^{\infty} \frac{3}{2\alpha} x^2 e^{-\alpha ... 2 You can consider that area as the sum of two integrals:$$\int_{C}r ~ dr ~d\theta = \int_{C_1}r ~ dr ~d\theta + \int_{C_2} r ~ dr ~d\theta = 2 \int_{C_1} r ~dr ~d\theta$$Where C_1 is the part of the curve which has a positive x coordinate. For this part of the curve \theta varies in the interval [-\frac{\pi}{4},\frac{\pi}{4}]. So the integral that ... 2 This is not true. It is the case that$$\int\frac{f'(x)}{f(x)}dx=\log f(x)+C$$Which can be seen by making the substitution u=f(x). 2 It's the fundamental theorem of calculus: If g is continuous, then$$\int_a^b g(x) dx = G(b)-G(a)$$With G'(x) = g(x) So, let g(x) = \tan(x), you have$$F(x) = \int_x^{x^2} g(x) dx = G(x^2)-G(x)$$This means that by the chain rule$$F'(x) = 2xG'(x^2)-G'(x)$$Hence$$F'(x) = 2xg(x^2)-g(x)=2x \tan(x^2) - \tan(x)2 There isn't one. Calc 1 doesn't come anywhere close to proving that some functions, like e^{-x^{2}}, don't have elementary integrals. 1 If x\ge 1, set x=\cosh u,\enspace u\ge0. Then you have: \begin{align*}\int\frac{4\sinh u\,\mathrm d\mkern1mu u}{\cosh u\sqrt{\cosh^2u-1 }}=4\int\frac{\mathrm d\mkern1mu u}{\cosh u}=8\int\frac{\mathrm e^u\,\mathrm d\mkern1mu u}{\mathrm e^{2u}+1}=8\arctan\mathrm e^u=8\arctan(x+\sqrt{x^2-1}). \end{align*} If x\le-1, setting x=-\cosh u,\enspace u\ge 0, ... 1 for the #1 if you want to use a geometrical solution you can say that it's the volume of a pyramid with apex at (0,0,0) intersecting the axes at (a,0,0),(0,b,0),(0,0,c) now because of the orthogonality of axes we can say the base is a right triangle having the area \frac{1}{2}ab and the length of the height is c then:volume\; of\; ... 1 Here's #2: the first equation is equivalent to $$(x-a)^2+y^2=a^2,$$ which is a cylinder centered at(a,0,0)$parallel to the$z$-axis. It's intersection with the$xy$-plane is a circle; the volume we want is the solid above this circle and below$z^2=x^2+y^2$. Let the region bounded by this circle be$C$, so the volume is $$\iint_C \sqrt{x^2+y^2}\, ... 1 The expression \int_0^\infty \frac{\sin x}{x} dx is an abuse of notation, the precise meaning of this expression should be:$$\lim_{t \to \infty} \int_0^t \frac{\sin x}{x} dx = \frac{\pi}{2} \tag{1}$$since in fact$x^{-1} \sin x$is not Lebesgue integrable over$(0, +\infty)$, because its positive and negative parts integrate to$\infty$. To show$(1)$... 1 Really not sure how to attack this one. No wonder you're not sure ! :-$)$It can only be expressed in terms the special Bessel functions.$I=\pi\Big(I_0(2a)+I_1(2a)\Big)$. 1 Hint: Take$I(k)=\displaystyle\int_0^1x^k~dx,~$and evaluate$I^{(n)}(0)$. 1 Are you familiar with integrals of the form$\int f'(x).g(f(x)) \text{d}x$, where$f$and$g$are ''well-behaved'' functions? If yes, try to see how you can apply this idea when$g(x)=e^x$and$f(x)=\alpha x^2$, where$\alpha$is a constant of course. If not, find out how to use this method because it is an essential tool. Actually, knowing about ... 1 If$\alpha=0$the integral diverges (it is$\infty$). Now if$\alpha \not =0$, the derivative of$\frac{-1}{2\alpha}e^{-\alpha x^{2}}$is$xe^{-\alpha x^{2}}$. Now$\int\limits_{0}^{\infty} xe^{-\alpha x^{2}}dx=\frac{1}{2\alpha}$. The answer should be a real number, and not a function of$x$1 Find the derivative of$x\mapsto e^{-\alpha x^2}$. You'll see that it's easy to find an antiderivative of$x\mapsto xe^{-\alpha x^2}$and thus to solve this integral, which by the way is not a function of$x\$.

Only top voted, non community-wiki answers of a minimum length are eligible