find the region of  INTEGRAL of   | (x^2 -4) | dx from x= -1 to x=3?
361536
Let's call f(x) = x^2 - 4\n\nThe graph of f(x) = x^2 - 4 looks like a smile ( U )  with the graph passing the y - axis at -4.\n\nThe roots are at x = 2 and x = -2.  This can be found by solving for x^2 - 4 = 0\nx^2 - 4 = 0\n(x + 2)(x - 2) = 0\nso x + 2 = 0, or x - 2 = 0\nso x = -2 or x = 2\n\nsince you're int'ing from x = -1 to x = 3, you have to break up the int into\n[x = -1 to x = 2, where f(x) is negative and take the absolute value] \nand\n[x = 2 to x = 3, where f(x) is positive]\n\nso int of f(x)\n= x^3/3 - 4x + c\n\nso int of f(x) from x=-1 to x=3\n= {+ve value of int of f(x) from x = -1 to 2}\n                +\n  int of f(x) from x = 2 to 3\n\nSeparately\nint of f(x), x = -1 to 2\n= | (2^3)/3  - 4(2) - [((-1)^3)/3  - 4(-1)] |\n= | (8/3) - 8 - [(-1/3) + 4] |\n= | (8/3) - 8 + (1/3) - 4 |\n= | (9/3) - 12 |\n= | 3 - 12 |\n= | -9 |\n= 9\n\nint f(x), x = 2 to 3\n= (3^3)/3  - 4(3) - [(2^3)/3  - 4(2)]\n= 27/3 - 12 - [(8/3) - 8]\n= 9 - 12 - (8/3) + 8\n= 5 - (8/3)\n= 7/3 (or 2 1/3)\n\nAdd them together and you get:\n9 + 2 1/3\n= 11 1/3 which is appromixately 11.333_
146214
D.    X (greater or equal to) 4\n      2(-2)+4 = 0\nso (-infin, -2) U [4, infin)
667880
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
909745
Focus on -(x+1).\nIt's like multiplying every term inside the bracket by -1. Hence you're going to get -x-1.\n4 - x - 1 = 8\n3 - x = 8\nAdd x to both sides: 3 = 8+x\nSubtract 8 from both sides: -5 = x\n\nHope this helps :)
101082
(x-1)/4 > 3\nx-1 > 12\nx > 13
300604
If you understand everything up through the u-substitution then you have done the tough part. If the original integral was:\n\nintegral[1,4]sqrt(1+(9/4)x)dx\n\nthen letting u = (1+(9/4)x)  and du = 9/4 dx *and* substituting the new u-values for the bounds (13/4 and 10 in place of 1 and 4).....\n\nThe new integral becomes:\n\nintegral[13/4, 10] sqrt(u) dx ....but dx = 4/9 du (see above) so now:\n\nintegral[13/4, 10] (4/9) sqrt(u) du  - I'm sure you are comfortable moving the (4/9) outside the integral to get:\n\n(4/9) integral[13/4, 10] sqrt(u) du. Rewrite sqrt(u) = u^1/2 to get:\n\n(4/9) integral[13/4, 10] u^1/2 du and integrate...\n\n(4/9) (2/3 u^3/2) from [13/4, 10]....better as\n\n(8/27)(u^3/2) from [13/4, 10] - now substitute to get:\n\n(8/27)(10^3/2 - (13/4)^3/2)\n\nIf 10^3/2 = 10sqrt(10) and 13/4^3/2 = 13sqrt(13)/8....then you have\n\n(8/27)(10sqrt(10) - 13sqrt(13)/8)\n\nDistribute the "8" from the 8/27 to get your final answer:\n\n(1/27)(80sqrt(10)-13sqrt(13))\n\nLet me know if that makes sense.
245933
(4x+3)(x-1)=0\n\nx= -3/4, x=1
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.
920343
(4/3)[x^3/(2-x)]^(1/3)\n[(2-x)(3)x^2-x^3(-1)]/\n(2-x)^2=\n\n 4x[1/(2-x)]^(1/3)\n[3x^2(2-x)+x^3]/\n3(2-x)^2
358253
-x+x^3 = 0\nx^3 - x = 0\nx(x^2-1) = 0\nx(x-1)(x+1) = 0\nx = 0, x = 1, x = -1\nThis function is odd because\n-(-x) + (-x)^3 = x - x^3 = -(-x+x^3)\nsince it is odd, the area between -1 and 0 is the same as the area between 0 and 1.  You can just find one of the areas and then multiply it by two.\nIntegrate from -1 to 0 of \n-x + x^3 dx\n= -(1/2)x^2 + (1/4)x^4 from -1 to 0\n-(1/2)(0)^2 + (1/4)(0)^4 - [-(1/2)(-1)^2 + (1/4)(-1)^4]\n= 0 + 0 - (-1/2) - (1/4)\n= 1/2 - 1/4\n= 1/4\nNow multiple this by 2\n2(1/4) = 1/2
893324
domain:  from - infinity to + infinity  (that's for x)\n\nrange: 0 to infinity (for y)\n\ny = |x+4|
1092533
The previous person almost got there.\nThe indefinite integral is (1+x)[log(1+x) - 1] + C\n\nSubstitute x=1 to get 2[log(2) - 1] + C\nSubstitute x=0 to get  -1 + C.\n\nThe definite integral is \n2[log(2) - 1]  - [-1]\n= 2log(2) -1\n=log(4) - 1\n=log(4) - log(e)\n=log(4/e)
46351
Sy   = |x| - x^2 + 2   \n     = (x^2/2) - (X^3/3) + 2X
846663
(x+3)(x-1)\n\n x= -3\nx=1
796927
try the link below...... hope it can put an end to ur integral problems
400264
split the integral up\n\nyou have:\n(1-2x^2)/x = 1/x - 2x^2/x = 1/x - 2x\n\nintegrating this gives:\n\nIn(x) - 2x^2/2 = In(x)-x^2
171039
integrate it by parts : \nint[U.dv] = [U.V] - int[V.du]\nwhere U = x, dv = sin^-1