What is the integral of 3e^x?
931775
3e^x. Pull the constant outside the integral sign\nand just let it sit there!
147855
Try http://integrals.wolfram.com/
488973
If you mean this (e^x)^x, then that is e^(x^2).  There is no closed form expression for the antiderivative of this with respect to x.\n\nHowever if you mean e^(x^x), then I don't know.\n\nUPDATE: Hey discoganya, I'm getting d/dx(x^x) to be \n(x^x)(ln x + 1), not x^x.  Now (d/dx)e^x = e^x.\n\nUPDATE (again): I think that part of the problem with that, discoganya, is that you are assuming that x is a whole number.\n\nThis is how I got my dervative.\n\n(d/dx)(x^x) = (d/dx)(exp(xln x))\n= exp(xln x)(1*ln x + x(1/x))\n= (x^x)(ln x + 1), for x > 0\n\nI have run this through Mathematica, and it confirms it.\n\nI believe the only function such that f'(x) = f(x) is f(x)=  a*exp(x), where a is an arbitrary constant.\n\nBy the way, Mathematica also could not find the integral of e^{x^x), honey, so it also probably does not exist in a closed form.
955331
3^-x=2\n-x ln3=ln2\n-x=ln2/ln3\n2x=-2ln2/ln3=-1.2618595071429148741990542286855\n3^-1.2618595071429148741990542286855=.25
526667
First of all, the characteristic equation is m^3-3m^2+3m-1=0, which factors as (m-1)^3=0, for m=1 a triply repreated root.\n\nThis gives e^x, xe^x, and x^2 e^x as three independent solutions for the *homogeneous* equation (the one with a 0 on the right hand side). Now you have to find a *particular* solution of the non-homogeneous equation. The problem is that e^x is already a solution of the homogeneous version. So now you go up in the heierarchy to try a particular solution of the form\ny_p=Ax^3 e^x.\nPlug this into the original equation to solve for A. Then add this particular solution to the general solution of the homogeneous part to get your general solution. Check to be sure there are three arbitrary constants!
689506
Since we integrate first with respect to y, x is assumed to be constant. \n\nint(from 0 to x/3) exp(x^2) dy = [y*exp(x^2)](from y=0 to x/3) = (x/3)exp(x^2) \n\nThen integrate with respect to x, then y is assumed to be constant - but of course all the y's have already dropped out because of the substitution in the earlier step. \n\nint(from 0 to 3) (x/3)exp(x^2) dx \n\nsubstitute x^2 = t, then for the boundaries: 0^2 = 0, 3^2 = 9, for the dx, dt/dx = 2x therefore dx = dt/(2x). \n\nint(from 0 to 3) (x/3)exp(x^2) dx \n= (1/3)* int(from 0 to 9) x*exp(t) dt/(2x) \n= (1/6)* int(from 0 to 9) exp(t) dt \n= (1/6)* [exp(t)](from 0 to 9) \n= (1/6)*(exp(9) - exp(0)) \n= (1/6)*(exp(9) - 1)\n\nAnd that's the exact solution... I think. :p
849119
Let u = -x^2.\n\ndu = -2xdx.\n\n-1/2 du = xdx\n\nSo Integral [x*e^(-x^2)]dx = Integral[-1/2 e^u]du\n\nWith the substitution done, I think you should be able to evaluate the integral and back-substitute.
117843
Sure.  The derivative of x^3 is 3x^2.  That should get you started.
772619
integral(x,-1,2) - 2 * integral(|x|,-1,2)\nintegral(x,-1,2) - 2 * ( integral(-x,-1,0) + integral(x,0,2))\n\n3/2 - 2( 1/2 + 2)\n3/2 - 3\n-3/2
986076
i)auxiliary equation\nL^2-2L-3=0\n(L-3)(L+1)=0\n>>L=3 or -1\ncomplementary function\n\nyc=Ce^(3x)+De^(-x)\n\nii)we try y = pe^(-3x) which has\n\ndy/dx= -3pe^(-3x),d^2y/dy^2=9pe^(-3x)\n\nsubstituting these into the DE gives\n\n9pe^(-3x)+6pe^(-3x)-3pe^(-3x)\n=5e^(-3x)\n12pe^(-3x)=5e^(-3x)\n>>>>p=5/12\n\nparticular integral\nyp=5/12e^(-3x)\n\niii)general solution\n\ny= yc+yp\n  =Ce^(3x)+De^(-x)+5/12e^(-3x)\n\ni hope that this helps
798015
y=e^x - 1 cuts the x axis at x = 1 Thus the integral needs to be done in two parts as area is a scalar quantity not a vector.\n\nA = |∫(x = -3 to 0) (e^x - 1) dx| + |∫(x = 0 to 3)  (e^x - 1) dx|\n\n= | [e^x - x] (x = -3 to 0) | + | [e^x - x] (x = 0 to 3) |\n\n= |(e^0 - 0) - (e^(-3) - -3)| + |(e^(3) - 3) - (e^0 - 0)| \n\n= 2 + e^(-3) + e^3 - 4\n\n= e^3 + e^(-3) - 2\n\n≈ 18.14 units²
1006943
Rewrite as a power series, since there is no way to do this using traditional methods:\n\nx^2 - (1/2)x^4 + (3/8)x^6 - (37/144)x^8 + (1031/5760)x^10 -.....\n\nThen integrate:\n\n(1/3)x^3 - (1/10)x^5 + (3/56)x^7 - (37/1296)x^9 + (1031/63360)x^11 -.......
773985
let 2y^4-1 be x\n8y^3dy=dx\ny^3dy=dx/8\nthe integral (1/8)rtxdx\n=(1/8)x^3/2(2/3)+C\n=(1/12)(2y^4-1)^3/2+C
755844
1 correct \n\n2 ∫x/e^x dx \n= ∫x*e(-x) dx\n= -xe^(-x) - ∫-e^(-x). 1.dx\n= -xe^(-x) - e^(-x) + c\n\nSo∫[x = 1 to ∞] x/e^x dx = lim t →∞ (-te^(-t) - e^(-t)) - (-1/e - 1/e)\n= 2/e (NB lim t →∞ (-te^(-t)) = o by d'Hopitals rule)\n\n3 correct\n\n4. Use (n² + 1)/[e^n * (n + 1)²] \n=  [(n² + 1)² - 2n]/[e^n * (n + 1)²]\n= 1/e^n - 2n/e^n(n + 1)² \nNow 1/e^n < 1/3^n and  2n/e^n(n + 1)² < 2n/3^n*n² \n= 2/n*3^n \nSo both parts converge so the difference converges\n\nNOTE: It is many, many years since I have done these BUT I am confident that I am correct ... Maybe you can tell me if that is so!
527554
Let I=Int((e^5x)*cos 3x)dx               (1)\n\nUse the formula for integration by parts twice\n\nInt(u dv/dx)dx= uv - int(v du/dx)dx\n\ntaking u = e^5x   and  dv/dx=cos 3x\n   du/dx=5(e^5x)  and      v=(1/3)*sin 3x\n\nthen I = (1/3)*e^5x(sin 3x) - (5/3)*int((e^5x)*sin 3x)dx      (2)\n\nUse integration by parts again: -\n\nint((e^5x)*sin 3x)dx = (-1/3)*(e^5x)*(cos 3x)+(5/3)*int((e^5x)*sin 3x)dx\n\nNow the int((e^5x)*sin 3x)dx we have already defined to equal I\n\ntherefore\n\nint((e^5x)*sin 3x)dx = (-1/3)*(e^5x)*(cos 3x)-(5/3)*I\n\nputting this into the equation 2 to get: -\n\nI = (1/3)*e^5x(sin 3x) + (5/9)*(e^5x)*(cos 3x)-(25/9)*I\n\nBring I over to the left hand side: -\n\n(34/9)I = (1/3)*e^5x(sin 3x) + (5/9)*(e^5x)*(cos 3x)\n\nWith a bit of careful rearranging, you can get the answer :-\n\n    I = (1/34)*(e^5x)(3sin3x+5cos3x) + C\n\nwhere C is an arbitrary constant\n\nI know it looks a bit of a bee-hatch written out here, but go through it and make sure you understand what has been done at each step. I may have made a few errors involving minus signs, so make sure you understood what i have done.\n\nI got the answer wrong first time round, but i have gone through it again and corrected the mistakes.
885310
rt2logx+C\n\n2.sectheta+C\n\n3.is it [3^(2x)+1/x^(1/3)]dx?
829669
integral=4x^-3dx-x^-2dx\n=4(x^-2)/-2-x^-1/-1\n=-2/x^2+1/x+C