Calculus: Integral of F prime of x?
914813
Your correct, it is equal to f(x).\n\nthe integral and the differential cancel eachother out.!
382963
f(x)= (2x-5)(x^2+2x+2) = \n\n(2x^3+4x^2+4x)+(-5x^2-10x-10)= 2x^3 - x^2 - 6x - 10\n\nf(x) = 2x^3 - x^2 - 6x - 10\n\nIntegral of a sum, equals the sum of integrals.\n\nTake each expression separately and integrate:\n\n1) Int(2x^3dx) = (x^4)/2\n\n2)Int(x^2dx) = (x^3)/3\n\n3)Int(6xdx) = 3x^2\n\n4)Int(10dx) = 10x\n\nInt [ f(x)dx ] = 1) - 2) - 3) - 4)\n\nFinal answer:\n\nInt [ f(x)dx ] = \n\n(x^4)/2 - (x^3)/3 - 3x^2 - 10x + C (integration constant) \n\nHope this helped.
949467
You divide each term in the upper part by x^2 to get a simple polynomial ((10 x^9) / x^2) + ( 5 / x^2) = 10 x^7 + 5 x^-2\nThe integral is obtain by the rule Integrate(x^n) = (x^(n+1)) / (n + 1)
691116
x/[a^(-x^2/(2a))] dx\nRewrite bringing the denominator to the top.\nThis will change the exponent to positive.\nx(a^(x^2/(2a)) dx\nLet u = x^2/(2a), then du = (1/a)x dx or \na du = x dx\nNow the integral is\nint(a(a^u) du)\n= int(a^(u+1) du)\n= a^(u+1) + C\n= a^(x^2/(2a) + 1) + C\n(The exponent of a is x^2/(2a) + 1)
833486
int1/x^2+int1/x^5\nx^-1/-1 +x^-4/-4+C\n-1/x-1/4x^4+C
79944
The above is the derivative.\nIntegration gives x + (1/4)x^4 + c (generic constant).
337080
when you integrate 1/x you get In |x|\n\nso the answer is -1/2 * In |x| + constant term\n\n\nYou can use substitution, but there's not much point, because you'll get 1/u then (which is the same thing). 1/x is a well known integrand. You'd be better just memorising that the integral of 1/x = In(x)
574666
int((x-2)*(x-5)^(1/2)*dx)\n\nUse u sub, let:\nu=x-5\ndu=dx\nx-2=u+3\n\nResub:\n\nint((u+3)*sqrt(u)*du)\n\nDistribute:\n\nint((u^3/2 + 3sqrt(u)*du))\n\nIntegrate:\n\n0.4*u^(5/2) + 2*u^(3/2) + c\n\nResub u for x:\n\n0.4*(x-5)^(5/2) + 2*(x-5)^(3/2) + C
506663
That'll be the integral over C of k.curl(F)\ncurl(F) = 0i,-4j,-1k\nk.curl(F) = -1\narea of C = 9pi\ntherefore W = -9pi
481212
the area under the curve f(x)!
491386
Just substitute x+1 into f(x)\n\nf(x+1) = [ 6(x+1)^3 ] - [ 3(x+1)^2 ] + 4(x+1) - 9
1182078
f(x) = x^2 + 7x\n\nf(q+2) = (q+2)^2 + 7(q+2)\n\nf(q+2) = q^2 + 4q + 4 + 7q + 14\n\nf(q+2) = q^2 + 11q + 18\n\nf(q+2) = (q + 9)(q + 2)
796927
try the link below...... hope it can put an end to ur integral problems
1060628
simple find the integral of pi*(e^(-x^2))^2 from a=0 and b=2 since as x goes to infiniety is goes to zero but it is very close to zero when x=2 you can find this by graphing f(x) \nanswer should be 0.9188 good luck!
837082
Hint: Let C=X&#92;O and let f(x)=dist(x,C). The other way is trivial.
334690
The value is pi/2, but the proof of this isn't easy.  You have to use the theory of residues.\n\nThe idea is to recall Euler's formula: e^{iz} = cos(z) + i sin(z).  We will consider the function f(z) = e^{iz} / (2iz).  Fix a small radius r and a large radius R.  Define the following path in the complex plane:\n\n(1): Go from (-R,0) to (-r,0) along the x-axis,\n(2): go from (-r,0) to (r,0) clockwise along the circle of radius r,\n(3): go from (r,0) to (R,0) along the x-axis, and\n(4): go from (R,0) to (-R,0) counterclockwise along the circle of radius R.\n\nThe theory of residues says that the closed path integral of f(z) is 0:\n\nInt_{paths 1+2+3+4} f(z) dz = 0.\n\nNow let's consider each path integral separately.  The sum of the integrals along paths (1) and (3) is\n\nInt_{paths 1+3} f(z) dz = Integral from r to R of sin(x) / x.\n\nThe integrals along paths (2) and (4) are\n\nInt_{path 2} f(z) dz = (-1/2) * Integral from 0 to pi of e^{-r (sin(t) + i cos(t)} dt,\nInt_{path 4} f(z) dz = (1/2) * Integral from 0 to pi of e^{-R (sin(t) + i cos(t))} dt.\n\nAs r approaches 0, the first integral approaches -pi/2.  As R increases without bound (i.e., approaches "infinity") the second integral approaches 0.  This means\n\n{ Integral from 0 to infinity of sin(x)/x } + {-pi/2} + { 0 } = 0\n\nHence the integral from 0 to infinity of sin(x)/x is indeed pi/2.
832686
Take the integral of this function, like this:\n\n∫ (10 + x²) / x³ =\n\n∫ (10 / x³) + ∫ (x² / x³) =\n\n10 ∫ (1 / x³) + ∫ (1 / x) =\n\n10 (-1 / 2x²) + ln(x) + C = \n\n-5 / x² + ln(x) + C\n\n~ ♥ ~