Can you explain very simply how you integrate a function.?
957793
This is a huge question........here is the easiest way. Find the area between the line and the x-axis between the limits of integration. Areas above the axis are positive and below the axis are negative.
887664
let u = x^2\ndu = 2x dx\n\nThis turns the function in to:\n1 / (u^2 + 1) - which you can work from.
989233
Start by letting I = integral of [sin(x)*sin(nx)]\nUse integration by parts twice.\nLet u = sin(x) and dv = sin(nx)\nAfter using integration by parts twice, you'll end up returning to the original integral you started with, so you will have to solve for I.\n(Your instructor should have shown you this trick)\n\nAnswer: [cos(x)*sin(nx) - n*sin(x)*cos(nx)]/[n^2 - 1]\nI did this quickly, so you should check my answer by working it out.
796927
try the link below...... hope it can put an end to ur integral problems
875348
I'll call g(x) = (integral from 0 to x) f(s) ds.\nWe just need to see what g(-x) is. \ng(-x) = (integral from 0 to -x) f(s) ds.\nIf f is even, then f(s) = f(-s), so this equals:\n(integral from 0 to -x) f(-s) ds\nand that is equivalent to\n(integral from 0 to x) f(s) ds = g(x), so g is even.\nSimilarly, if f is odd, f(s) = -f(-s), so it equals\n(integral from 0 to -x) -f(-s) ds\nwhich is\n(integral from 0 to x) -f(s) ds = -g(x), so g is odd.
105440
You can try and find constants A and B such that\n\n1/(x(x-a)) = A/x + B/(x-a)\n\nIf you get this, then the integration is simple.\n\nHow do you get it?\n\nStart from A/x + B/(x-a) and calculate the common denominator:\n[A(x-a) + Bx]/[x(x-a)] = [(A+B)x - aA]/(x(x-a))\n\nNow to get the expression equal to 1/(x(x-a)) it must be:\n\nA+B = 0  ---> A= -B\naA = 1  ---> A = 1/a\n\nso \nA = 1/a (constant)\nB= - 1/a\n\nsubstitute and you get the integral as the sum of two simpler ones which get two logarithms by integration.\n\nThen you apply the property of the sum of logarithms to obtain a single one.\n\nI hope it's clear.
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.
57930
The integration by parts formula is\nint(u*dv)=u*v-int(v*du)\n\nLet u=r and dv=e^(r/2)dr\n\nNow take the derivative of u with respect to r:\ndu=dr\n\nNow integrate dv\nint(dv)=int[e^(r/2)dr]\nv=[e^(r/2)]/(1/2)  --> The chain rule was used here. \n__=2e^(r/2)\nWhen you take the derivative you multiply by "what's inside" so when you integrate, you have to divide by it.\n\nNow plug into the formula\n2re^(r/2)-int(2e^(r/2)dr)\n\n=2re^(r/2)-2(e^(r/2))/(1/2)+C\n\n=2re^(r/2)-4(e^(r/2))+C\n\n=(r-2)*2e^(r/2)+C
883154
I assume that's suppose to be an integral symbol.\n\nf(x) = [1/2 t²] from 2 to x, or 1/2 x² - 1\n\nThen f'(x) = x, and f''(x) = 1.\n\nDidn't really need to do the integration...
903814
well since a is constant well say\na(t) = a\nintegrate\nint[a(t)]=v(t)= a*t + C  where C=initial velocity v0\nintegrate again\n=s(t) = (a/2)t^2 + v0t + D where D=initial displacement s0\n\nand thats how its derived in general terms
85251
The area under any function can be got by the integration from an initial point to a final point\n\nfor example y(x)=x^2 and we are intrested to solve the area between x=1 to x=3\n\nthe integration = x^3/3 from 1 to 3\nthe area = (3^3/3) -(1^3/3) = 9 - 1/3 = 8.666667
403929
Integration by parts is done by breaking down an integral so that\n\nS u dv = uv - S v du (where S is the integration symbol)\n\nFor example, if we were to integrate xcos x dx,\n\nlet u = x\ndu = dx\n\nthen dv = cos x (all the other stuff we didn't use for u)\nv = sin x (just integrate dv)\n\nand the formula will look like\n\nS xcos x dx = xsin x - S sin x dx\n= xsin x + cos x + C\n\nIt's kinda an art form or experience in picking which part of the integral to make u and dv - there's really no rule that says u = cos x.  Experience (or trial and error) will show you that that's probably not the better choice though .
830837
If ∫f(x)dx = y(x), then f(x) = d y(x) / dx;  So to get the answer, take the derivative of the expression x^2 - 2x + 1.  The value of the integral at limit 1 is  a constant and will not appear in the derivative.
1122081
It's called the integration by parts formula.\nInt u dv=uv- int v du\n\nBoth u and v need to be absolutely continuous for this to work. A continous derivative for both is plenty.
255922
Im not sure if you need clarification on how to do this. Ill provide a few details on how to get these answers:\n\n1) ∫sinxdx = - cos(x) + C\n-This one is relatively straightforward, just think "what gives me sin(x) when I take its derivative"\n\n2)∫e3xdx = (Im assuming this means e^(3x))\n         = (e^(3x))/3 + C\n-Just think of the chain rule backwards. If you take the derivative of e^3x you'll have to multiply by 3, so when you take the derivative divide by 3.\n\n3)∫cos3xdx = (Im assuming this is (cos^3)x)\n           = ∫((cos^2)x )(cos(x))\n           = ∫(1-(sin^2)x)((cos(x))\n      Now let u=sin(x)\n             du=cos(x)dx\n           = ∫(1-u^2)du\n           = [u - (u^3)/3]\n     Sub back in for u\n           = sin(x) - (sin^3(x))/3 + C\n-Wow, that one was hard to type out. I hope you can follow it somewhat. The trick is to write cos^3(x) as cos^2(x) and cos(x). Then the identity cos^2(x) + sin^2(x)=1 is used. The u substution allows us to find the integral at the end.\n\n4) Derivative of 3x^4 + 2x^2 + x + 3\n                  = 12x^3 + 4x + 1\n-The derivative is defined by some long equation, but you can use the following fact:\nIf f(x)=x^n\nthen f'(x)=nx^n-1\n\nHopefully you understand most of what I did. Im sure someone else will probably explain a lot of this in much better detail. Like I said, I wasnt sure how much detail you were looking for. \n\nPS Ill tell you site that will help you forever in any integral calculus course: http://integrals.wolfram.com/index.jsp
954798
The integration by parts formula is:\nintegral(f'g) = fg - integral(fg').\nIf we make f' = sin2t and g = t, then that second integral will only involve sin, and it should make everything easier.\nSo, that means f = -0.5cos2t, and g = t. Then g' = 1, so we get:\nintegral(t sin 2t) = -0.5tcos2t - integral(-0.5cos2t * 1)\nThat last integral now just becomes -0.25sin2t.\nSo the answer is -0.5tcos 2t + 0.25sin2t (+c).
1130821
so for the first part, f(x)=x \nthe derivative no matter what is 1 \n\nfor the second part f(x)=1\nthe derivative no matter what is 0 \n\nso you will have combinational solution