What is integration and differentiation? Please suggest a few urls where i can learn them.?
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give me ten points if u need further help just mail to me.\n\nhttp://tutorial.math.lamar.edu/AllBrowsers/2413/2413.asp\n\nhttp://www.calculus.org/\n\nhttp://www.math.ucdavis.edu/~mikew/free_calc.pdf\n\nhttp://www.math.gatech.edu/~cain/notes/calculus.html\n\nhttp://www.exambot.com/cgi/topic/list.cgi/math/difc?class=hlink
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you can go to this site, you enter in the problem and it solves it for you....http://integrals.wolfram.com/index.jsp
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Calculus is a pure mathematical theory. It has 2 main applications 1 to enable calculation of the gradiant of a curve at any point on the curve, this gives you rate of change of x with respect to y. e.g. if y axis was metres and x axis was seconds this would give you the rate of change of distance with respect to time in m/s i.e. speed.\n\nIf you want to understand differentiation, I suggest you google differentiton from first principles.\n\nThis is one of the physical applications of differentiation.\n\nThe other application is to take a curve and calculate the area underneath it between two points on the curve. Say you have a curve where the y axis is speed in m/s and the x axis is time in seconds. Then the area under the curve would give you the distance travelled between points x=m and x=n. So if you integrate a velocity vs time formula you get distance travelled. This is an application of integration. To understand integration google "simpsons rule".\n\nThere are many more applications of this in physics relating to rate of change & intergration \n\ne.g Electical current is the rate of change of voltage with respect to resistance. dv/dr\n\ne.g. Power in an electical circuit is the integral of current (I) w.r.t to resistance dr
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a) dy/dx = x - 1/x^2\ndy = (x-x^(-2))dx\ny = x^2/2 + 1/x + C\ny(1) = 1^2/2 + 1/1 + C = 2\n1/2 + 1 + C = 2 implies C = 1/2\nAnswer: y = x^2/2 + 1/x + 1/2, x>0\n\nb) The integration is already finished.  Just plug in -1 for x.\ny(-1) = (-1)^2/2 + 1/-1 + C = 1\n-1/2 - 1 + C = 1 implies C = 3/2\nAnswer: x^2/2 + 1/x + 3/2, x<0\n\nc) Just take the derivative of each one.  C1 and C2 are constants, so their derivative is 0.  \n\nd) C1 = 1/2 and C2 = 3/2\n\ne) Follow parts (a) and (b) except plug in 2 for x and set it equal to -1.  Similarly for y(-2) = 2.\nCheck that C1 = -7/2 and C2 = 1/2
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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
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Try some good books.I found integration n diffrentiation the most easy ones.Solving quetions from it always used to relax me from boring linear n modern algebra.\nThere is no shortcuts to learn them but if u try hard on it u will find it the most shortcut thing in maths.\nGood Luck:)
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I don't know what it means to "integrate a derivitive [sic]", but, for a continuous function f(x), it does follow that the slope of the line tangent at point (a,b) would be equal to f'(a).
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A bit tricky to integrate, but is like this (i could mistake!):\n\na)\n9^(1/5)\nS pi ( (3 sqrt(x))^2 - x^3^2) dx\n0\n\nb)\n9^(1/5)\nS pi ( (3 sqrt(x)-6)^2 - (x^3-6)^2) dx\n0\n\nc)\n9^(3/5)\nS pi ( (y^2/9)^2 - (y^(1/3))^2) dy\n0\n\nd)\n9^(3/5)\nS pi ( (y^2/9-4)^2 - (y^(1/3)-4)^2) dy\n0\n\nRegards
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Si? Are you referring to the sine integral? If so, you need only remember two things:\nDifferentiation is the inverse of integration\nd/dx f(g(x)) = f'(g(x)) * g'(x)\n\nSo:\n\nd/dx (Si(x²)) = sin (x²)/(x²) * 2x
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using implicit differentiation:\nyou get 2y(dy/dx) - [y + x(dy/dx)] -3 = 0\nthen move 3 over and factor out (dy/dx):\n(2y-x)(dy/dx)=3 + y\ndy/dx=(3+y) / (2y-x)\nnow plug in your point (0,-1)\ndy/dx = (3+ -1)/(2(-1) - 0\ndy/dx = -1\nanswer = -1\n\n(dont forget to use your chain rule when you are implicitlly differentiating)\n\n#2\nsame thing make sure you sue multiplication rule\n[y^2 + 2xy(dx/dy)] - 2(dx/dy) + 12y^2(dx/dy) = 0\nthen move over the y^2 and factor out dx/dy\n(dx/dy)[2xy - 2 + 12y^2] = y^2\ndx/dy = y^2/ (2xy -2 +12y^2)\nPlug in 1 for y\ndx/dy = 1/(2x -2 + 12)\ndx/dy = 1/ (2x +10) or (2x + 10)^(-1)\n\nAnswer (leave in terms of x) = 1/ (2x +10) or (2x + 10)^(-1)\n\nI may have made mistakes so let me know
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Here is an excellent site:\nhttp://www.sosmath.com/calculus/calculus.html\n\nVery thorough explanations and examples provided.
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If look to the integrator more deeply, one finds that 1/x is the derivative of ln(x).\nThen this question comes up: \nWhat is the integration of uu'dx?\n\nObviously the answer is u^2/2.\n\nSo our answer is (ln(x))^2/4+c.
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You have to use integration by parts.\nLet u = ln(x) and dv = dx, then\ndu = (1/x)dx and v = x\nint(udv) = uv - int(vdu)\n= xln(x) - int(x(1/x)dx)\n= xln(x) - int(dx)\n= xln(x) - x + C, where C is a constant
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i'll do it by 2 ways \nthe first one if u know what integration by parts is:\n(int=integral)\nint[udv]=uv-int[vdu]\nlet u=lnx =>du=dx/x\n& let dv=xdx => v=(x^2)/2\nso we'll have\nint[xlnx]=((x^2)/2) lnx -1/2int[xdx]\n=((x^2)/2)lnx -((x^2)/4)+ Constante\nthe second method is\ntry derivating (x^2)lnx u'll get [(x^2)lnx]'=2xlnx +x\nthen integrate int{[(x^2)lnx]dx}=(x^2)lnx\n&int[x]=(x^2)/2\nso u'll get (x^2)lnx=2int[xlnxdx]+(x^2)/2\nso int[xlnxdx]=((x^2)/2)lnx-(x^2)/4+ C
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You use derivatives to compute how values are changing. A good example is to compute velocities.\n\nIntegration is used to compute areas and volumes.
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http://www.calculus.org/
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I haven't got a clue what you are talking about, but the fundamental theorem of Calculus is that differentiation and integration are the reverse process of each other.\n\nThis link between differentiation and integration is what Newton and Leibniz discovered independently in the late 1600's.  There is a common mis-conception that Newton and Leibniz invented calculus, but calculus dates right back to the Greeks and Eudoxus and his method of exhaustion is the earliest form of integration.\n\nI think you need to re-word your question if you want some productive help.